Explanation:
Okay, so I need to find the constant $ d $ such that the equation $ (\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x $ holds for all angles $ x $. The problem mentions that there are constants $ a $, $ b $, $ c $, and $ d $, and I need to find $ d $.

First, I remember that powers of sine can be expressed as sums of multiple angles using trigonometric identities. Since the left side is $ (\sin x)^7 $, which is an odd power, maybe I can use some kind of power-reduction formula or multiple-angle identities to express it in terms of sine functions with different arguments.

I recall that for lower powers, like $ \sin^3 x $, there are standard identities. For example, $ \sin^3 x = \frac{3 \sin x - \sin 3x}{4} $. Maybe there's a similar approach for higher powers. But $ \sin^7 x $ is a higher power, so I might need to use recursion or some other method.

Alternatively, I remember that complex exponentials can be used to express sine functions. Since $ \sin x = \frac{e^{ix} - e^{-ix}}{2i} $, maybe I can expand $ (\sin x)^7 $ using the binomial theorem and then convert back to sine terms. That might be a way to go, although it could get complicated with the exponents.

Let me try that approach. Let me write $ \sin x $ as $ \frac{e^{ix} - e^{-ix}}{2i} $, so:

$ (\sin x)^7 = \left( \frac{e^{ix} - e^{-ix}}{2i} \right)^7 $

Expanding this using the binomial theorem:

$ \frac{1}{(2i)^7} \sum_{k=0}^7 \binom{7}{k} (e^{ix})^{7 - k} (-e^{-ix})^k $

Simplify the exponents:

$ \frac{1}{(2i)^7} \sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} $

Now, $ (2i)^7 = 2^7 \cdot i^7 = 128 \cdot i^7 $. Since $ i^7 = i^{4 + 3} = i^3 = -i $, so $ (2i)^7 = 128 \cdot (-i) = -128i $. Therefore, the entire expression becomes:

$ \frac{1}{-128i} \sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} $

Simplify the coefficient:

$ \frac{1}{-128i} = \frac{i}{128} $, because multiplying numerator and denominator by $ i $:

$ \frac{1}{-128i} \cdot \frac{i}{i} = \frac{i}{-128i^2} = \frac{i}{-128(-1)} = \frac{i}{128} $

So now, the expression is:

$ \frac{i}{128} \sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} $

Now, I need to express this sum in terms of sine functions. Remember that $ e^{i\theta} = \cos \theta + i \sin \theta $, so each term in the sum is a complex exponential. However, since the original expression is real (as $ (\sin x)^7 $ is real), the imaginary parts should combine to give the sine terms.

Alternatively, since we have a sum of complex exponentials, we can pair terms to get sine functions. Let me think. Let me note that $ e^{i\theta} + e^{-i\theta} = 2 \cos \theta $, and $ e^{i\theta} - e^{-i\theta} = 2i \sin \theta $.

But in our case, the sum is over $ k $ from 0 to 7, and each term is $ \binom{7}{k} (-1)^k e^{i(7 - 2k)x} $. Let me consider the exponents $ 7 - 2k $. Let me note that for each term with exponent $ m $, there might be a corresponding term with exponent $ -m $, but since $ k $ ranges from 0 to 7, let me check the possible exponents.

Let me compute $ 7 - 2k $ for k from 0 to 7:

- k=0: 7 - 0 = 7
- k=1: 7 - 2 = 5
- k=2: 7 - 4 = 3
- k=3: 7 - 6 = 1
- k=4: 7 - 8 = -1
- k=5: 7 - 10 = -3
- k=6: 7 - 12 = -5
- k=7: 7 - 14 = -7

So the exponents are 7, 5, 3, 1, -1, -3, -5, -7.

Therefore, the sum can be written as:

$ \sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} = \sum_{m=-7}^{7} \text{something} e^{i m x} $, but with specific coefficients. However, since the exponents are symmetric around 0, except for the sign. Let me note that for each positive exponent $ m $, there is a corresponding negative exponent $ -m $, except when $ m = 0 $, but in our case, the exponents are 7,5,3,1,-1,-3,-5,-7. So, for each term with exponent $ m $, there is a term with exponent $ -m $, except when $ m = 0 $, which isn't present here.

Therefore, the sum can be grouped as pairs:

For m = 7 and m = -7,

For m = 5 and m = -5,

For m = 3 and m = -3,

For m = 1 and m = -1.

So, let me write the sum as:

$ \sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} = \sum_{m=1}^7 \left[ \text{coefficient for } m \cdot e^{i m x} + \text{coefficient for } -m \cdot e^{-i m x} \right] $

But let me compute the coefficients for each pair. Let me note that for each positive exponent $ m $, the corresponding term in the sum is when $ 7 - 2k = m $, so solving for k:

$ 7 - 2k = m \implies k = (7 - m)/2 $

Similarly, for the negative exponent $ -m $, we have $ 7 - 2k = -m \implies k = (7 + m)/2 $

Therefore, for each m from 1 to 7, the coefficients for $ e^{i m x} $ and $ e^{-i m x} $ are:

For $ e^{i m x} $: when k = (7 - m)/2, which must be integer. Let me check for m=1, 3, 5, 7:

For m=1: k=(7 - 1)/2 = 3, which is integer.

For m=3: k=(7 - 3)/2 = 2, integer.

For m=5: k=(7 - 5)/2 = 1, integer.

For m=7: k=(7 - 7)/2 = 0, integer.

Similarly, for the negative exponents:

For $ e^{-i m x} $, the corresponding k is (7 + m)/2. For m=1: (7 + 1)/2 = 4, integer. For m=3: (7 + 3)/2 = 5, integer. For m=5: (7 + 5)/2 = 6, integer. For m=7: (7 + 7)/2 = 7, integer.

Therefore, each pair of terms corresponds to m and -m. Therefore, the entire sum can be written as:

For m=1,3,5,7:

$ \binom{7}{(7 - m)/2} (-1)^{(7 - m)/2} e^{i m x} + \binom{7}{(7 + m)/2} (-1)^{(7 + m)/2} e^{-i m x} $

But since the original sum is symmetric, maybe we can combine these terms into sine functions. Let me compute each pair.

Let me compute for each m:

Starting with m=1:

For m=1:

Coefficient for $ e^{i x} $: k=(7 - 1)/2 = 3. So binomial coefficient is $ \binom{7}{3} $, and (-1)^k = (-1)^3 = -1.

Coefficient for $ e^{-i x} $: k=(7 + 1)/2 = 4. So binomial coefficient is $ \binom{7}{4} $, and (-1)^k = (-1)^4 = 1.

Therefore, the contribution for m=1 is:

$ \binom{7}{3} (-1) e^{i x} + \binom{7}{4} (1) e^{-i x} $

Similarly for m=3:

Coefficient for $ e^{i 3x} $: k=(7 - 3)/2 = 2. So binomial coefficient $ \binom{7}{2} $, (-1)^2 = 1.

Coefficient for $ e^{-i 3x} $: k=(7 + 3)/2 = 5. So binomial coefficient $ \binom{7}{5} $, (-1)^5 = -1.

Therefore, contribution for m=3:

$ \binom{7}{2} (1) e^{i 3x} + \binom{7}{5} (-1) e^{-i 3x} $

Similarly for m=5:

Coefficient for $ e^{i 5x} $: k=(7 - 5)/2 = 1. So binomial coefficient $ \binom{7}{1} $, (-1)^1 = -1.

Coefficient for $ e^{-i 5x} $: k=(7 + 5)/2 = 6. So binomial coefficient $ \binom{7}{6} $, (-1)^6 = 1.

Contribution for m=5:

$ \binom{7}{1} (-1) e^{i 5x} + \binom{7}{6} (1) e^{-i 5x} $

For m=7:

Coefficient for $ e^{i 7x} $: k=(7 - 7)/2 = 0. So binomial coefficient $ \binom{7}{0} $, (-1)^0 = 1.

Coefficient for $ e^{-i 7x} $: k=(7 + 7)/2 = 7. So binomial coefficient $ \binom{7}{7} $, (-1)^7 = -1.

Contribution for m=7:

$ \binom{7}{0} (1) e^{i 7x} + \binom{7}{7} (-1) e^{-i 7x} $

Therefore, the entire sum is:

[ $ \binom{7}{3} (-1) e^{i x} + \binom{7}{4} (1) e^{-i x} $ ] + [ $ \binom{7}{2} (1) e^{i 3x} + \binom{7}{5} (-1) e^{-i 3x} $ ] + [ $ \binom{7}{1} (-1) e^{i 5x} + \binom{7}{6} (1) e^{-i 5x} $ ] + [ $ \binom{7}{0} (1) e^{i 7x} + \binom{7}{7} (-1) e^{-i 7x} $ ]

Now, let me compute the binomial coefficients:

- $ \binom{7}{0} = 1 $
- $ \binom{7}{1} = 7 $
- $ \binom{7}{2} = 21 $
- $ \binom{7}{3} = 35 $
- $ \binom{7}{4} = 35 $
- $ \binom{7}{5} = 21 $
- $ \binom{7}{6} = 7 $
- $ \binom{7}{7} = 1 $

So substituting these values:

For m=1:

$ 35 \times (-1) e^{i x} + 35 \times 1 e^{-i x} = -35 e^{i x} + 35 e^{-i x} $

For m=3:

$ 21 \times 1 e^{i 3x} + 21 \times (-1) e^{-i 3x} = 21 e^{i 3x} - 21 e^{-i 3x} $

For m=5:

$ 7 \times (-1) e^{i 5x} + 7 \times 1 e^{-i 5x} = -7 e^{i 5x} + 7 e^{-i 5x} $

For m=7:

$ 1 \times 1 e^{i 7x} + 1 \times (-1) e^{-i 7x} = e^{i 7x} - e^{-i 7x} $

Therefore, the entire sum becomes:

[ -35 e^{i x} + 35 e^{-i x} ] + [ 21 e^{i 3x} - 21 e^{-i 3x} ] + [ -7 e^{i 5x} + 7 e^{-i 5x} ] + [ e^{i 7x} - e^{-i 7x} ]

Now, let me factor out the terms:

Grouping the positive and negative exponents:

For $ e^{i x} $: -35 e^{i x}

For $ e^{-i x} $: +35 e^{-i x}

Similarly for others:

For $ e^{i 3x} $: +21 e^{i 3x}

For $ e^{-i 3x} $: -21 e^{-i 3x}

For $ e^{i 5x} $: -7 e^{i 5x}

For $ e^{-i 5x} $: +7 e^{-i 5x}

For $ e^{i 7x} $: +1 e^{i 7x}

For $ e^{-i 7x} $: -1 e^{-i 7x}

Now, recall that $ e^{i\theta} - e^{-i\theta} = 2i \sin \theta $, so if I have terms like $ A e^{i\theta} + B e^{-i\theta} $, I can write them as combinations of sine or cosine. However, in our case, the coefficients for $ e^{i\theta} $ and $ e^{-i\theta} $ are not necessarily equal. However, since the original expression is real, the imaginary parts must cancel out. Let me think.

But since we have already expressed the entire sum as a combination of complex exponentials, and we have multiplied by $ \frac{i}{128} $, let me recall that the entire expression for $ (\sin x)^7 $ is:

$ \frac{i}{128} \times [ \text{sum} ] $

So let me compute that. Let me write the sum as:

Sum = (-35 e^{i x} + 35 e^{-i x}) + (21 e^{i 3x} - 21 e^{-i 3x}) + (-7 e^{i 5x} + 7 e^{-i 5x}) + (e^{i 7x} - e^{-i 7x})

Let me factor out the coefficients:

Sum = -35 e^{i x} + 35 e^{-i x} + 21 e^{i 3x} - 21 e^{-i 3x} -7 e^{i 5x} + 7 e^{-i 5x} + e^{i 7x} - e^{-i 7x}

Now, let me group terms with $ e^{i\theta} $ and $ e^{-i\theta} $:

For $ e^{i x} $: -35 e^{i x}

For $ e^{-i x} $: +35 e^{-i x}

Similarly:

For $ e^{i 3x} $: +21 e^{i 3x}

For $ e^{-i 3x} $: -21 e^{-i 3x}

For $ e^{i 5x} $: -7 e^{i 5x}

For $ e^{-i 5x} $: +7 e^{-i 5x}

For $ e^{i 7x} $: +1 e^{i 7x}

For $ e^{-i 7x} $: -1 e^{-i 7x}

Now, let me write each pair as $ A e^{i\theta} + B e^{-i\theta} $, and then convert to sine terms. However, since we have coefficients for both $ e^{i\theta} $ and $ e^{-i\theta} $, perhaps we can write them as:

For each pair, let me consider:

For example, take the first pair: -35 e^{i x} + 35 e^{-i x} = 35 (e^{-i x} - e^{i x}) = 35 (-2i \sin x) = -70i \sin x

Similarly, the second pair: 21 e^{i 3x} - 21 e^{-i 3x} = 21 (e^{i 3x} - e^{-i 3x}) = 21 (2i \sin 3x) = 42i \sin 3x

Third pair: -7 e^{i 5x} + 7 e^{-i 5x} = 7 (e^{-i 5x} - e^{i 5x}) = 7 (-2i \sin 5x) = -14i \sin 5x

Fourth pair: e^{i 7x} - e^{-i 7x} = 2i \sin 7x

Therefore, the entire sum becomes:

-70i \sin x + 42i \sin 3x -14i \sin 5x + 2i \sin 7x

Therefore, the entire expression for $ (\sin x)^7 $ is:

$ \frac{i}{128} \times [ -70i \sin x + 42i \sin 3x -14i \sin 5x + 2i \sin 7x ] $

Let me compute this multiplication:

First, factor out the i:

$ \frac{i}{128} \times i [ -70 \sin x + 42 \sin 3x -14 \sin 5x + 2 \sin 7x ] $

Since $ i \times i = i^2 = -1 $, so:

$ \frac{-1}{128} [ -70 \sin x + 42 \sin 3x -14 \sin 5x + 2 \sin 7x ] $

Multiply the negative sign:

$ \frac{1}{128} [ 70 \sin x - 42 \sin 3x + 14 \sin 5x - 2 \sin 7x ] $

Therefore, $ (\sin x)^7 = \frac{70}{128} \sin x - \frac{42}{128} \sin 3x + \frac{14}{128} \sin 5x - \frac{2}{128} \sin 7x $

Simplify the fractions:

Let me note that 70, 42, 14, 2 are all divisible by 2:

So:

$ \frac{70}{128} = \frac{35}{64} $

$ \frac{42}{128} = \frac{21}{64} $

$ \frac{14}{128} = \frac{7}{64} $

$ \frac{2}{128} = \frac{1}{64} $

Therefore, the expression becomes:

$ (\sin x)^7 = \frac{35}{64} \sin x - \frac{21}{64} \sin 3x + \frac{7}{64} \sin 5x - \frac{1}{64} \sin 7x $

Comparing this with the original equation:

$ (\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x $

So, matching coefficients:

- Coefficient of $ \sin 7x $: $ a = -\frac{1}{64} $
- Coefficient of $ \sin 5x $: $ b = \frac{7}{64} $
- Coefficient of $ \sin 3x $: $ c = -\frac{21}{64} $
- Coefficient of $ \sin x $: $ d = \frac{35}{64} $

Therefore, the value of $ d $ is $ \frac{35}{64} $.

But let me check if this is correct. Let me verify with a specific value of x. For example, take x = 0. Then, sin(0) = 0, so left side is 0. Right side is a sin 0 + b sin 0 + c sin 0 + d sin 0 = 0. So that's okay. How about x = π/2? Let me compute both sides.

Left side: sin^7(π/2) = 1^7 = 1.

Right side: a sin(7π/2) + b sin(5π/2) + c sin(3π/2) + d sin(π/2)

Compute each term:

sin(7π/2) = sin(3π + π/2) = -sin(π/2) = -1

sin(5π/2) = sin(2π + π/2) = sin(π/2) = 1

sin(3π/2) = -1

sin(π/2) = 1

Therefore, right side:

a*(-1) + b*(1) + c*(-1) + d*(1) = -a + b - c + d

Substituting the values:

-a = -(-1/64) = 1/64

b = 7/64

-c = -(-21/64) = 21/64

d = 35/64

Adding them up: 1/64 + 7/64 + 21/64 + 35/64 = (1 + 7 + 21 + 35)/64 = 64/64 = 1. Which matches the left side. So that checks out.

Another check: x = π/4. Let me compute left side and right side.

Left side: sin^7(π/4) = (sqrt(2)/2)^7 = (2^{1/2}/2)^7 = 2^{7/2}/2^7 = 2^{-7/2} = 1/(2^{7/2}) = 1/(8 * sqrt(2)) ≈ 1/(11.3137) ≈ 0.0884.

Right side: a sin(7π/4) + b sin(5π/4) + c sin(3π/4) + d sin(π/4)

Compute each term:

sin(7π/4) = -sqrt(2)/2

sin(5π/4) = -sqrt(2)/2

sin(3π/4) = sqrt(2)/2

sin(π/4) = sqrt(2)/2

Therefore, right side:

a*(-sqrt(2)/2) + b*(-sqrt(2)/2) + c*(sqrt(2)/2) + d*(sqrt(2)/2)

Factor out sqrt(2)/2:

[ -a - b + c + d ] * sqrt(2)/2

Substituting the values:

-a = 1/64, -b = -7/64, +c = 21/64, +d = 35/64

Adding them: 1/64 -7/64 +21/64 +35/64 = (1 -7 +21 +35)/64 = (1 -7 = -6; -6 +21=15; 15 +35=50)/64 = 50/64 = 25/32

Multiply by sqrt(2)/2: (25/32) * sqrt(2)/2 = 25 sqrt(2)/64 ≈ 25 * 1.4142 / 64 ≈ 35.355 / 64 ≈ 0.5524. Wait, but the left side was approximately 0.0884. That's a discrepancy. Wait, that can't be. Did I make a mistake here?

Wait, hold on. Wait, my calculation must be wrong. Let me check again.

Wait, let me recalculate the right side for x = π/4.

First, compute left side: sin^7(π/4) = (sqrt(2)/2)^7. Let me compute that:

(sqrt(2)/2)^2 = (2)/4 = 1/2. Then, (sqrt(2)/2)^4 = (1/2)^2 = 1/4. Then, (sqrt(2)/2)^6 = (1/2)^3 = 1/8. Then, (sqrt(2)/2)^7 = (sqrt(2)/2)^6 * (sqrt(2)/2) = (1/8) * (sqrt(2)/2) = sqrt(2)/16 ≈ 1.4142 / 16 ≈ 0.0884. Correct.

Now, right side:

a sin(7π/4) + b sin(5π/4) + c sin(3π/4) + d sin(π/4)

Compute each term:

a = -1/64, sin(7π/4) = -sqrt(2)/2, so term1: (-1/64)*(-sqrt(2)/2) = (sqrt(2))/128

b = 7/64, sin(5π/4) = -sqrt(2)/2, term2: (7/64)*(-sqrt(2)/2) = -7 sqrt(2)/128

c = -21/64, sin(3π/4) = sqrt(2)/2, term3: (-21/64)*(sqrt(2)/2) = -21 sqrt(2)/128

d = 35/64, sin(π/4) = sqrt(2)/2, term4: (35/64)*(sqrt(2)/2) = 35 sqrt(2)/128

Adding all terms:

term1 + term2 + term3 + term4 = [sqrt(2)/128 -7 sqrt(2)/128 -21 sqrt(2)/128 +35 sqrt(2)/128]

Combine coefficients:

1 -7 -21 +35 = (1 -7) + (-21 +35) = (-6) + 14 = 8

Therefore, total is 8 sqrt(2)/128 = sqrt(2)/16 ≈ 0.0884, which matches the left side. So my previous mistake was in the sign when I calculated the coefficients. So that checks out. Therefore, the expression is correct.

Therefore, the value of $ d $ is $ \frac{35}{64} $.

But let me check if there's another way to get this result, maybe using recurrence relations or other identities, just to confirm.

Alternatively, I remember that for $ \sin^n x $, there is a formula involving Chebyshev polynomials or using recursive relations. For example, using the identity:

$ \sin^{2n+1}x = \frac{1}{2^{2n}} \sum_{k=0}^{n} (-1)^{n - k} \binom{2n + 1}{k} \sin((2n + 1 - 2k)x) $

Wait, not sure. Alternatively, there is a formula for expressing $ \sin^7 x $ as a combination of multiple angles.

Alternatively, using the identity for $ \sin^7 x $, we can use the formula for $ \sin^7 x $ in terms of multiple angles.

Alternatively, using the formula for $ \sin^7 x $, which can be derived by using the identity for $ \sin^3 x $, and then multiplying by $ \sin^4 x $, but that might be more complicated.

Alternatively, using the identity:

$ \sin^7 x = \sin x \cdot \sin^6 x $

And then express $ \sin^6 x $ in terms of cosines, then use product-to-sum formulas. Let me try that.

First, $ \sin^6 x = (\sin^2 x)^3 = \left( \frac{1 - \cos 2x}{2} \right)^3 $

Expanding that:

$ \frac{1}{8} (1 - 3 \cos 2x + 3 \cos^2 2x - \cos^3 2x) $

Then, we can express $ \cos^2 2x $ and $ \cos^3 2x $ in terms of multiple angles.

First, $ \cos^2 2x = \frac{1 + \cos 4x}{2} $

Then, $ \cos^3 2x $ can be expressed using the identity:

$ \cos^3 \theta = \frac{3 \cos \theta + \cos 3\theta}{4} $

Therefore, substituting:

$ \cos^3 2x = \frac{3 \cos 2x + \cos 6x}{4} $

Therefore, putting it all together:

$ \sin^6 x = \frac{1}{8} \left[ 1 - 3 \cos 2x + 3 \cdot \frac{1 + \cos 4x}{2} - \frac{3 \cos 2x + \cos 6x}{4} \right] $

Let me compute this step by step:

First, expand the terms inside the brackets:

1 - 3 cos 2x + 3*(1 + cos 4x)/2 - (3 cos 2x + cos 6x)/4

Compute each term:

1 is 1.

-3 cos 2x is -3 cos 2x.

3*(1 + cos 4x)/2 = 3/2 + (3/2) cos 4x.

- (3 cos 2x + cos 6x)/4 = -3/4 cos 2x - 1/4 cos 6x.

Now, combine all terms:

1 + 3/2 - 3 cos 2x - 3/4 cos 2x + (3/2) cos 4x - 1/4 cos 6x

Compute constants: 1 + 3/2 = 5/2

Compute cos 2x terms: -3 cos 2x - 3/4 cos 2x = - (12/4 + 3/4) cos 2x = -15/4 cos 2x

cos 4x term: + 3/2 cos 4x

cos 6x term: -1/4 cos 6x

Therefore, the entire expression inside the brackets is:

5/2 - (15/4) cos 2x + (3/2) cos 4x - (1/4) cos 6x

Therefore, $ \sin^6 x = \frac{1}{8} [ 5/2 - (15/4) cos 2x + (3/2) cos 4x - (1/4) cos 6x ] $

Simplify:

Multiply each term by 1/8:

$ \frac{5}{16} - \frac{15}{32} \cos 2x + \frac{3}{16} \cos 4x - \frac{1}{32} \cos 6x $

Therefore, $ \sin^7 x = \sin x \cdot \sin^6 x = \sin x \cdot \left( \frac{5}{16} - \frac{15}{32} \cos 2x + \frac{3}{16} \cos 4x - \frac{1}{32} \cos 6x \right) $

Now, multiply each term by $ \sin x $:

First term: $ \frac{5}{16} \sin x $

Second term: $ -\frac{15}{32} \sin x \cos 2x $

Third term: $ \frac{3}{16} \sin x \cos 4x $

Fourth term: $ -\frac{1}{32} \sin x \cos 6x $

Now, use the identity $ \sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)] $

Apply this to each term:

First term remains as is: $ \frac{5}{16} \sin x $

Second term: $ -\frac{15}{32} \cdot \frac{1}{2} [\sin(x + 2x) + \sin(x - 2x)] = -\frac{15}{64} [\sin 3x + \sin(-x)] = -\frac{15}{64} [\sin 3x - \sin x] $

Third term: $ \frac{3}{16} \cdot \frac{1}{2} [\sin(x + 4x) + \sin(x - 4x)] = \frac{3}{32} [\sin 5x + \sin(-3x)] = \frac{3}{32} [\sin 5x - \sin 3x] $

Fourth term: $ -\frac{1}{32} \cdot \frac{1}{2} [\sin(x + 6x) + \sin(x - 6x)] = -\frac{1}{64} [\sin 7x + \sin(-5x)] = -\frac{1}{64} [\sin 7x - \sin 5x] $

Now, expand all terms:

First term: $ \frac{5}{16} \sin x $

Second term: $ -\frac{15}{64} \sin 3x + \frac{15}{64} \sin x $

Third term: $ \frac{3}{32} \sin 5x - \frac{3}{32} \sin 3x $

Fourth term: $ -\frac{1}{64} \sin 7x + \frac{1}{64} \sin 5x $

Now, combine like terms:

For $ \sin x $:

$ \frac{5}{16} \sin x + \frac{15}{64} \sin x = \left( \frac{20}{64} + \frac{15}{64} \right) \sin x = \frac{35}{64} \sin x $

For $ \sin 3x $:

$ -\frac{15}{64} \sin 3x - \frac{3}{32} \sin 3x = -\frac{15}{64} \sin 3x - \frac{6}{64} \sin 3x = -\frac{21}{64} \sin 3x $

For $ \sin 5x $:

$ \frac{3}{32} \sin 5x + \frac{1}{64} \sin 5x = \frac{6}{64} \sin 5x + \frac{1}{64} \sin 5x = \frac{7}{64} \sin 5x $

For $ \sin 7x $:

$ -\frac{1}{64} \sin 7x $

Therefore, combining all terms:

$ \sin^7 x = \frac{35}{64} \sin x - \frac{21}{64} \sin 3x + \frac{7}{64} \sin 5x - \frac{1}{64} \sin 7x $

Which matches exactly the result I obtained earlier using complex exponentials. Therefore, this confirms that the coefficients are correct, and so $ d = \frac{35}{64} $.

Therefore, the answer is $ \boxed{\dfrac{35}{64}} $.

**Final Answer**
\boxed{\dfrac{35}{64}}

To solve the equation

$$
(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x
$$

for all $ x $, we aim to express $ (\sin x)^7 $ as a linear combination of sine functions with different arguments. This is a classic problem in trigonometric identities, and one effective method is to use **complex exponentials** to expand the power of sine.

---

### Step 1: Express $ \sin x $ in terms of complex exponentials

We use the identity:

$$
\sin x = \frac{e^{ix} - e^{-ix}}{2i}
$$

Then:

$$
(\sin x)^7 = \left( \frac{e^{ix} - e^{-ix}}{2i} \right)^7
$$

Expanding this using the binomial theorem:

$$
(\sin x)^7 = \frac{1}{(2i)^7} \sum_{k=0}^{7} \binom{7}{k} (-1)^k e^{i(7 - 2k)x}
$$

Note that $ (2i)^7 = 128i^7 = 128(-i) = -128i $, so:

$$
(\sin x)^7 = \frac{1}{-128i} \sum_{k=0}^{7} \binom{7}{k} (-1)^k e^{i(7 - 2k)x}
$$

Multiplying numerator and denominator by $ i $, we get:

$$
(\sin x)^7 = \frac{i}{128} \sum_{k=0}^{7} \binom{7}{k} (-1)^k e^{i(7 - 2k)x}
$$

---

### Step 2: Group terms and convert to sine functions

The exponents $ 7 - 2k $ for $ k = 0 $ to $ 7 $ are:

$$
7, 5, 3, 1, -1, -3, -5, -7
$$

So the sum becomes:

$$
\sum_{k=0}^{7} \binom{7}{k} (-1)^k e^{i(7 - 2k)x} = \sum_{m=-7}^{7} \text{terms with } e^{imx}
$$

Grouping terms with $ e^{imx} $ and $ e^{-imx} $, and using the identity:

$$
e^{i\theta} - e^{-i\theta} = 2i \sin \theta
$$

we can express the entire sum in terms of sine functions. After simplifying, we find:

$$
(\sin x)^7 = \frac{35}{64} \sin x - \frac{21}{64} \sin 3x + \frac{7}{64} \sin 5x - \frac{1}{64} \sin 7x
$$

---

### Step 3: Match coefficients

Comparing this with the original equation:

$$
(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x
$$

We identify the coefficients:

- $ a = -\frac{1}{64} $
- $ b = \frac{7}{64} $
- $ c = -\frac{21}{64} $
- $ d = \frac{35}{64} $

---

### Final Answer

$$
\boxed{\dfrac{35}{64}}
$$

[Assistant]
Explanation:
Okay, so I need to find the constant $ d $ such that the equation $ (\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x $ holds for all angles $ x $. The problem mentions that there are constants $ a $, $ b $, $ c $, and $ d $, and I need to find $ d $.

First, I remember that powers of sine can be expressed as sums of multiple angles using trigonometric identities. Since the left side is $ (\sin x)^7 $, which is an odd power, maybe I can use some kind of power-reduction formula or multiple-angle identities to express it in terms of sine functions with different arguments.

I recall that for lower powers, like $ \sin^3 x $, there are standard identities. For example, $ \sin^3 x = \frac{3 \sin x - \sin 3x}{4} $. Maybe there's a similar approach for higher powers. But $ \sin^7 x $ is a higher power, so I might need to use recursion or some other method.

Alternatively, I remember that complex exponentials can be used to express sine functions. Since $ \sin x = \frac{e^{ix} - e^{-ix}}{2i} $, maybe I can expand $ (\sin x)^7 $ using the binomial theorem and then convert back to sine terms. That might be a way to go, although it could get complicated with the exponents.

Let me try that approach. Let me write $ \sin x $ as $ \frac{e^{ix} - e^{-ix}}{2i} $, so:

$ (\sin x)^7 = \left( \frac{e^{ix} - e^{-ix}}{2i} \right)^7 $

Expanding this using the binomial theorem:

$ \frac{1}{(2i)^7} \sum_{k=0}^7 \binom{7}{k} (e^{ix})^{7 - k} (-e^{-ix})^k $

Simplify the exponents:

$ \frac{1}{(2i)^7} \sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} $

Now, $ (2i)^7 = 2^7 \cdot i^7 = 128 \cdot i^7 $. Since $ i^7 = i^{4 + 3} = i^3 = -i $, so $ (2i)^7 = 128 \cdot (-i) = -128i $. Therefore, the entire expression becomes:

$ \frac{1}{-128i} \sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} $

Simplify the coefficient:

$ \frac{1}{-128i} = \frac{i}{128} $, because multiplying numerator and denominator by $ i $:

$ \frac{1}{-128i} \cdot \frac{i}{i} = \frac{i}{-128i^2} = \frac{i}{-128(-1)} = \frac{i}{128} $

So now, the expression is:

$ \frac{i}{128} \sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} $

Now, I need to express this sum in terms of sine functions. Remember that $ e^{i\theta} = \cos \theta + i \sin \theta $, so each term in the sum is a complex exponential. However, since the original expression is real (as $ (\sin x)^7 $ is real), the imaginary parts should combine to give the sine terms.

Alternatively, since we have a sum of complex exponentials, we can pair terms to get sine functions. Let me think. Let me note that $ e^{i\theta} + e^{-i\theta} = 2 \cos \theta $, and $ e^{i\theta} - e^{-i\theta} = 2i \sin \theta $.

But in our case, the sum is over $ k $ from 0 to 7, and each term is $ \binom{7}{k} (-1)^k e^{i(7 - 2k)x} $. Let me consider the exponents $ 7 - 2k $. Let me note that for each term with exponent $ m $, there might be a corresponding term with exponent $ -m $, but since $ k $ ranges from 0 to 7, let me check the possible exponents.

Let me compute $ 7 - 2k $ for k from 0 to 7:

- k=0: 7 - 0 = 7
- k=1: 7 - 2 = 5
- k=2: 7 - 4 = 3
- k=3: 7 - 6 = 1
- k=4: 7 - 8 = -1
- k=5: 7 - 10 = -3
- k=6: 7 - 12 = -5
- k=7: 7 - 14 = -7

So the exponents are 7, 5, 3, 1, -1, -3, -5, -7.

Therefore, the sum can be written as:

$ \sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} = \sum_{m=-7}^{7} \text{something} e^{i m x} $, but with specific coefficients. However, since the exponents are symmetric around 0, except for the sign. Let me note that for each positive exponent $ m $, there is a corresponding negative exponent $ -m $, except when $ m = 0 $, but in our case, the exponents are 7,5,3,1,-1,-3,-5,-7. So, for each term with exponent $ m $, there is a term with exponent $ -m $, except when $ m = 0 $, which isn't present here.

Therefore, the sum can be grouped as pairs:

For m = 7 and m = -7,

For m = 5 and m = -5,

For m = 3 and m = -3,

For m = 1 and m = -1.

So, let me write the sum as:

$ \sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} = \sum_{m=1}^7 \left[ \text{coefficient for } m \cdot e^{i m x} + \text{coefficient for } -m \cdot e^{-i m x} \right] $

But let me compute the coefficients for each pair. Let me note that for each positive exponent $ m $, the corresponding term in the sum is when $ 7 - 2k = m $, so solving for k:

$ 7 - 2k = m \implies k = (7 - m)/2 $

Similarly, for the negative exponent $ -m $, we have $ 7 - 2k = -m \implies k = (7 + m)/2 $

Therefore, for each m from 1 to 7, the coefficients for $ e^{i m x} $ and $ e^{-i m x} $ are:

For $ e^{i m x} $: when k = (7 - m)/2, which must be integer. Let me check for m=1, 3, 5, 7:

For m=1: k=(7 - 1)/2 = 3, which is integer.

For m=3: k=(7 - 3)/2 = 2, integer.

For m=5: k=(7 - 5)/2 = 1, integer.

For m=7: k=(7 - 7)/2 = 0, integer.

Similarly, for the negative exponents:

For $ e^{-i m x} $, the corresponding k is (7 + m)/2. For m=1: (7 + 1)/2 = 4, integer. For m=3: (7 + 3)/2 = 5, integer. For m=5: (7 + 5)/2 = 6, integer. For m=7: (7 + 7)/2 = 7, integer.

Therefore, each pair of terms corresponds to m and -m. Therefore, the entire sum can be written as:

For m=1,3,5,7:

$ \binom{7}{(7 - m)/2} (-1)^{(7 - m)/2} e^{i m x} + \binom{7}{(7 + m)/2} (-1)^{(7 + m)/2} e^{-i m x} $

But since the original sum is symmetric, maybe we can combine these terms into sine functions. Let me compute each pair.

Let me compute for each m:

Starting with m=1:

For m=1:

Coefficient for $ e^{i x} $: k=(7 - 1)/2 = 3. So binomial coefficient is $ \binom{7}{3} $, and (-1)^k = (-1)^3 = -1.

Coefficient for $ e^{-i x} $: k=(7 + 1)/2 = 4. So binomial coefficient is $ \binom{7}{4} $, and (-1)^k = (-1)^4 = 1.

Therefore, the contribution for m=1 is:

$ \binom{7}{3} (-1) e^{i x} + \binom{7}{4} (1) e^{-i x} $

Similarly for m=3:

Coefficient for $ e^{i 3x} $: k=(7 - 3)/2 = 2. So binomial coefficient $ \binom{7}{2} $, (-1)^2 = 1.

Coefficient for $ e^{-i 3x} $: k=(7 + 3)/2 = 5. So binomial coefficient $ \binom{7}{5} $, (-1)^5 = -1.

Therefore, contribution for m=3:

$ \binom{7}{2} (1) e^{i 3x} + \binom{7}{5} (-1) e^{-i 3x} $

Similarly for m=5:

Coefficient for $ e^{i 5x} $: k=(7 - 5)/2 = 1. So binomial coefficient $ \binom{7}{1} $, (-1)^1 = -1.

Coefficient for $ e^{-i 5x} $: k=(7 + 5)/2 = 6. So binomial coefficient $ \binom{7}{6} $, (-1)^6 = 1.

Contribution for m=5:

$ \binom{7}{1} (-1) e^{i 5x} + \binom{7}{6} (1) e^{-i 5x} $

For m=7:

Coefficient for $ e^{i 7x} $: k=(7 - 7)/2 = 0. So binomial coefficient $ \binom{7}{0} $, (-1)^0 = 1.

Coefficient for $ e^{-i 7x} $: k=(7 + 7)/2 = 7. So binomial coefficient $ \binom{7}{7} $, (-1)^7 = -1.

Contribution for m=7:

$ \binom{7}{0} (1) e^{i 7x} + \binom{7}{7} (-1) e^{-i 7x} $

Therefore, the entire sum is:

[ $ \binom{7}{3} (-1) e^{i x} + \binom{7}{4} (1) e^{-i x} $ ] + [ $ \binom{7}{2} (1) e^{i 3x} + \binom{7}{5} (-1) e^{-i 3x} $ ] + [ $ \binom{7}{1} (-1) e^{i 5x} + \binom{7}{6} (1) e^{-i 5x} $ ] + [ $ \binom{7}{0} (1) e^{i 7x} + \binom{7}{7} (-1) e^{-i 7x} $ ]

Now, let me compute the binomial coefficients:

- $ \binom{7}{0} = 1 $
- $ \binom{7}{1} = 7 $
- $ \binom{7}{2} = 21 $
- $ \binom{7}{3} = 35 $
- $ \binom{7}{4} = 35 $
- $ \binom{7}{5} = 21 $
- $ \binom{7}{6} = 7 $
- $ \binom{7}{7} = 1 $

So substituting these values:

For m=1:

$ 35 \times (-1) e^{i x} + 35 \times 1 e^{-i x} = -35 e^{i x} + 35 e^{-i x} $

For m=3:

$ 21 \times 1 e^{i 3x} + 21 \times (-1) e^{-i 3x} = 21 e^{i 3x} - 21 e^{-i 3x} $

For m=5:

$ 7 \times (-1) e^{i 5x} + 7 \times 1 e^{-i 5x} = -7 e^{i 5x} + 7 e^{-i 5x} $

For m=7:

$ 1 \times 1 e^{i 7x} + 1 \times (-1) e^{-i 7x} = e^{i 7x} - e^{-i 7x} $

Therefore, the entire sum becomes:

[ -35 e^{i x} + 35 e^{-i x} ] + [ 21 e^{i 3x} - 21 e^{-i 3x} ] + [ -7 e^{i 5x} + 7 e^{-i 5x} ] + [ e^{i 7x} - e^{-i 7x} ]

Now, let me factor out the terms:

Grouping the positive and negative exponents:

For $ e^{i x} $: -35 e^{i x}

For $ e^{-i x} $: +35 e^{-i x}

Similarly for others:

For $ e^{i 3x} $: +21 e^{i 3x}

For $ e^{-i 3x} $: -21 e^{-i 3x}

For $ e^{i 5x} $: -7 e^{i 5x}

For $ e^{-i 5x} $: +7 e^{-i 5x}

For $ e^{i 7x} $: +1 e^{i 7x}

For $ e^{-i 7x} $: -1 e^{-i 7x}

Now, recall that $ e^{i\theta} - e^{-i\theta} = 2i \sin \theta $, so if I have terms like $ A e^{i\theta} + B e^{-i\theta} $, I can write them as combinations of sine or cosine. However, in our case, the coefficients for $ e^{i\theta} $ and $ e^{-i\theta} $ are not necessarily equal. However, since the original expression is real, the imaginary parts must cancel out. Let me think.

But since we have already expressed the entire sum as a combination of complex exponentials, and we have multiplied by $ \frac{i}{128} $, let me recall that the entire expression for $ (\sin x)^7 $ is:

$ \frac{i}{128} \times [ \text{sum} ] $

So let me compute that. Let me write the sum as:

Sum = (-35 e^{i x} + 35 e^{-i x}) + (21 e^{i 3x} - 21 e^{-i 3x}) + (-7 e^{i 5x} + 7 e^{-i 5x}) + (e^{i 7x} - e^{-i 7x})

Let me factor out the coefficients:

Sum = -35 e^{i x} + 35 e^{-i x} + 21 e^{i 3x} - 21 e^{-i 3x} -7 e^{i 5x} + 7 e^{-i 5x} + e^{i 7x} - e^{-i 7x}

Now, let me group terms with $ e^{i\theta} $ and $ e^{-i\theta} $:

For $ e^{i x} $: -35 e^{i x}

For $ e^{-i x} $: +35 e^{-i x}

Similarly:

For $ e^{i 3x} $: +21 e^{i 3x}

For $ e^{-i 3x} $: -21 e^{-i 3x}

For $ e^{i 5x} $: -7 e^{i 5x}

For $ e^{-i 5x} $: +7 e^{-i 5x}

For $ e^{i 7x} $: +1 e^{i 7x}

For $ e^{-i 7x} $: -1 e^{-i 7x}

Now, let me write each pair as $ A e^{i\theta} + B e^{-i\theta} $, and then convert to sine terms. However, since we have coefficients for both $ e^{i\theta} $ and $ e^{-i\theta} $, perhaps we can write them as:

For each pair, let me consider:

For example, take the first pair: -35 e^{i x} + 35 e^{-i x} = 35 (e^{-i x} - e^{i x}) = 35 (-2i \sin x) = -70i \sin x

Similarly, the second pair: 21 e^{i 3x} - 21 e^{-i 3x} = 21 (e^{i 3x} - e^{-i 3x}) = 21 (2i \sin 3x) = 42i \sin 3x

Third pair: -7 e^{i 5x} + 7 e^{-i 5x} = 7 (e^{-i 5x} - e^{i 5x}) = 7 (-2i \sin 5x) = -14i \sin 5x

Fourth pair: e^{i 7x} - e^{-i 7x} = 2i \sin 7x

Therefore, the entire sum becomes:

-70i \sin x + 42i \sin 3x -14i \sin 5x + 2i \sin 7x

Therefore, the entire expression for $ (\sin x)^7 $ is:

$ \frac{i}{128} \times [ -70i \sin x + 42i \sin 3x -14i \sin 5x + 2i \sin 7x ] $

Let me compute this multiplication:

First, factor out the i:

$ \frac{i}{128} \times i [ -70 \sin x + 42 \sin 3x -14 \sin 5x + 2 \sin 7x ] $

Since $ i \times i = i^2 = -1 $, so:

$ \frac{-1}{128} [ -70 \sin x + 42 \sin 3x -14 \sin 5x + 2 \sin 7x ] $

Multiply the negative sign:

$ \frac{1}{128} [ 70 \sin x - 42 \sin 3x + 14 \sin 5x - 2 \sin 7x ] $

Therefore, $ (\sin x)^7 = \frac{70}{128} \sin x - \frac{42}{128} \sin 3x + \frac{14}{128} \sin 5x - \frac{2}{128} \sin 7x $

Simplify the fractions:

Let me note that 70, 42, 14, 2 are all divisible by 2:

So:

$ \frac{70}{128} = \frac{35}{64} $

$ \frac{42}{128} = \frac{21}{64} $

$ \frac{14}{128} = \frac{7}{64} $

$ \frac{2}{128} = \frac{1}{64} $

Therefore, the expression becomes:

$ (\sin x)^7 = \frac{35}{64} \sin x - \frac{21}{64} \sin 3x + \frac{7}{64} \sin 5x - \frac{1}{64} \sin 7x $

Comparing this with the original equation:

$ (\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x $

So, matching coefficients:

- Coefficient of $ \sin 7x $: $ a = -\frac{1}{64} $
- Coefficient of $ \sin 5x $: $ b = \frac{7}{64} $
- Coefficient of $ \sin 3x $: $ c = -\frac{21}{64} $
- Coefficient of $ \sin x $: $ d = \frac{35}{64} $

Therefore, the value of $ d $ is $ \frac{35}{64} $.

But let me check if this is correct. Let me verify with a specific value of x. For example, take x = 0. Then, sin(0) = 0, so left side is 0. Right side is a sin 0 + b sin 0 + c sin 0 + d sin 0 = 0. So that's okay. How about x = π/2? Let me compute both sides.

Left side: sin^7(π/2) = 1^7 = 1.

Right side: a sin(7π/2) + b sin(5π/2) + c sin(3π/2) + d sin(π/2)

Compute each term:

sin(7π/2) = sin(3π + π/2) = -sin(π/2) = -1

sin(5π/2) = sin(2π + π/2) = sin(π/2) = 1

sin(3π/2) = -1

sin(π/2) = 1

Therefore, right side:

a*(-1) + b*(1) + c*(-1) + d*(1) = -a + b - c + d

Substituting the values:

-a = -(-1/64) = 1/64

b = 7/64

-c = -(-21/64) = 21/64

d = 35/64

Adding them up: 1/64 + 7/64 + 21/64 + 35/64 = (1 + 7 + 21 + 35)/64 = 64/64 = 1. Which matches the left side. So that checks out.

Another check: x = π/4. Let me compute left side and right side.

Left side: sin^7(π/4) = (sqrt(2)/2)^7 = (2^{1/2}/2)^7 = 2^{7/2}/2^7 = 2^{-7/2} = 1/(2^{7/2}) = 1/(8 * sqrt(2)) ≈ 1/(11.3137) ≈ 0.0884.

Right side: a sin(7π/4) + b sin(5π/4) + c sin(3π/4) + d sin(π/4)

Compute each term:

sin(7π/4) = -sqrt(2)/2

sin(5π/4) = -sqrt(2)/2

sin(3π/4) = sqrt(2)/2

sin(π/4) = sqrt(2)/2

Therefore, right side:

a*(-sqrt(2)/2) + b*(-sqrt(2)/2) + c*(sqrt(2)/2) + d*(sqrt(2)/2)

Factor out sqrt(2)/2:

[ -a - b + c + d ] * sqrt(2)/2

Substituting the values:

-a = 1/64, -b = -7/64, +c = 21/64, +d = 35/64

Adding them: 1/64 -7/64 +21/64 +35/64 = (1 -7 +21 +35)/64 = (1 -7 = -6; -6 +21=15; 15 +35=50)/64 = 50/64 = 25/32

Multiply by sqrt(2)/2: (25/32) * sqrt(2)/2 = 25 sqrt(2)/64 ≈ 25 * 1.4142 / 64 ≈ 35.355 / 64 ≈ 0.5524. Wait, but the left side was approximately 0.0884. That's a discrepancy. Wait, that can't be. Did I make a mistake here?

Wait, hold on. Wait, my calculation must be wrong. Let me check again.

Wait, let me recalculate the right side for x = π/4.

First, compute left side: sin^7(π/4) = (sqrt(2)/2)^7. Let me compute that:

(sqrt(2)/2)^2 = (2)/4 = 1/2. Then, (sqrt(2)/2)^4 = (1/2)^2 = 1/4. Then, (sqrt(2)/2)^6 = (1/2)^3 = 1/8. Then, (sqrt(2)/2)^7 = (sqrt(2)/2)^6 * (sqrt(2)/2) = (1/8) * (sqrt(2)/2) = sqrt(2)/16 ≈ 1.4142 / 16 ≈ 0.0884. Correct.

Now, right side:

a sin(7π/4) + b sin(5π/4) + c sin(3π/4) + d sin(π/4)

Compute each term:

a = -1/64, sin(7π/4) = -sqrt(2)/2, so term1: (-1/64)*(-sqrt(2)/2) = (sqrt(2))/128

b = 7/64, sin(5π/4) = -sqrt(2)/2, term2: (7/64)*(-sqrt(2)/2) = -7 sqrt(2)/128

c = -21/64, sin(3π/4) = sqrt(2)/2, term3: (-21/64)*(sqrt(2)/2) = -21 sqrt(2)/128

d = 35/64, sin(π/4) = sqrt(2)/2, term4: (35/64)*(sqrt(2)/2) = 35 sqrt(2)/128

Adding all terms:

term1 + term2 + term3 + term4 = [sqrt(2)/128 -7 sqrt(2)/128 -21 sqrt(2)/128 +35 sqrt(2)/128]

Combine coefficients:

1 -7 -21 +35 = (1 -7) + (-21 +35) = (-6) + 14 = 8

Therefore, total is 8 sqrt(2)/128 = sqrt(2)/16 ≈ 0.0884, which matches the left side. So my previous mistake was in the sign when I calculated the coefficients. So that checks out. Therefore, the expression is correct.

Therefore, the value of $ d $ is $ \frac{35}{64} $.

But let me check if there's another way to get this result, maybe using recurrence relations or other identities, just to confirm.

Alternatively, I remember that for $ \sin^n x $, there is a formula involving Chebyshev polynomials or using recursive relations. For example, using the identity:

$ \sin^{2n+1}x = \frac{1}{2^{2n}} \sum_{k=0}^{n} (-1)^{n - k} \binom{2n + 1}{k} \sin((2n + 1 - 2k)x) $

Wait, not sure. Alternatively, there is a formula for expressing $ \sin^7 x $ as a combination of multiple angles.

Alternatively, using the identity for $ \sin^7 x $, we can use the formula for $ \sin^7 x $ in terms of multiple angles.

Alternatively, using the formula for $ \sin^7 x $, which can be derived by using the identity for $ \sin^3 x $, and then multiplying by $ \sin^4 x $, but that might be more complicated.

Alternatively, using the identity:

$ \sin^7 x = \sin x \cdot \sin^6 x $

And then express $ \sin^6 x $ in terms of cosines, then use product-to-sum formulas. Let me try that.

First, $ \sin^6 x = (\sin^2 x)^3 = \left( \frac{1 - \cos 2x}{2} \right)^3 $

Expanding that:

$ \frac{1}{8} (1 - 3 \cos 2x + 3 \cos^2 2x - \cos^3 2x) $

Then, we can express $ \cos^2 2x $ and $ \cos^3 2x $ in terms of multiple angles.

First, $ \cos^2 2x = \frac{1 + \cos 4x}{2} $

Then, $ \cos^3 2x $ can be expressed using the identity:

$ \cos^3 \theta = \frac{3 \cos \theta + \cos 3\theta}{4} $

Therefore, substituting:

$ \cos^3 2x = \frac{3 \cos 2x + \cos 6x}{4} $

Therefore, putting it all together:

$ \sin^6 x = \frac{1}{8} \left[ 1 - 3 \cos 2x + 3 \cdot \frac{1 + \cos 4x}{2} - \frac{3 \cos 2x + \cos 6x}{4} \right] $

Let me compute this step by step:

First, expand the terms inside the brackets:

1 - 3 cos 2x + 3*(1 + cos 4x)/2 - (3 cos 2x + cos 6x)/4

Compute each term:

1 is 1.

-3 cos 2x is -3 cos 2x.

3*(1 + cos 4x)/2 = 3/2 + (3/2) cos 4x.

- (3 cos 2x + cos 6x)/4 = -3/4 cos 2x - 1/4 cos 6x.

Now, combine all terms:

1 + 3/2 - 3 cos 2x - 3/4 cos 2x + (3/2) cos 4x - 1/4 cos 6x

Compute constants: 1 + 3/2 = 5/2

Compute cos 2x terms: -3 cos 2x - 3/4 cos 2x = - (12/4 + 3/4) cos 2x = -15/4 cos 2x

cos 4x term: + 3/2 cos 4x

cos 6x term: -1/4 cos 6x

Therefore, the entire expression inside the brackets is:

5/2 - (15/4) cos 2x + (3/2) cos 4x - (1/4) cos 6x

Therefore, $ \sin^6 x = \frac{1}{8} [ 5/2 - (15/4) cos 2x + (3/2) cos 4x - (1/4) cos 6x ] $

Simplify:

Multiply each term by 1/8:

$ \frac{5}{16} - \frac{15}{32} \cos 2x + \frac{3}{16} \cos 4x - \frac{1}{32} \cos 6x $

Therefore, $ \sin^7 x = \sin x \cdot \sin^6 x = \sin x \cdot \left( \frac{5}{16} - \frac{15}{32} \cos 2x + \frac{3}{16} \cos 4x - \frac{1}{32} \cos 6x \right) $

Now, multiply each term by $ \sin x $:

First term: $ \frac{5}{16} \sin x $

Second term: $ -\frac{15}{32} \sin x \cos 2x $

Third term: $ \frac{3}{16} \sin x \cos 4x $

Fourth term: $ -\frac{1}{32} \sin x \cos 6x $

Now, use the identity $ \sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)] $

Apply this to each term:

First term remains as is: $ \frac{5}{16} \sin x $

Second term: $ -\frac{15}{32} \cdot \frac{1}{2} [\sin(x + 2x) + \sin(x - 2x)] = -\frac{15}{64} [\sin 3x + \sin(-x)] = -\frac{15}{64} [\sin 3x - \sin x] $

Third term: $ \frac{3}{16} \cdot \frac{1}{2} [\sin(x + 4x) + \sin(x - 4x)] = \frac{3}{32} [\sin 5x + \sin(-3x)] = \frac{3}{32} [\sin 5x - \sin 3x] $

Fourth term: $ -\frac{1}{32} \cdot \frac{1}{2} [\sin(x + 6x) + \sin(x - 6x)] = -\frac{1}{64} [\sin 7x + \sin(-5x)] = -\frac{1}{64} [\sin 7x - \sin 5x] $

Now, expand all terms:

First term: $ \frac{5}{16} \sin x $

Second term: $ -\frac{15}{64} \sin 3x + \frac{15}{64} \sin x $

Third term: $ \frac{3}{32} \sin 5x - \frac{3}{32} \sin 3x $

Fourth term: $ -\frac{1}{64} \sin 7x + \frac{1}{64} \sin 5x $

Now, combine like terms:

For $ \sin x $:

$ \frac{5}{16} \sin x + \frac{15}{64} \sin x = \left( \frac{20}{64} + \frac{15}{64} \right) \sin x = \frac{35}{64} \sin x $

For $ \sin 3x $:

$ -\frac{15}{64} \sin 3x - \frac{3}{32} \sin 3x = -\frac{15}{64} \sin 3x - \frac{6}{64} \sin 3x = -\frac{21}{64} \sin 3x $

For $ \sin 5x $:

$ \frac{3}{32} \sin 5x + \frac{1}{64} \sin 5x = \frac{6}{64} \sin 5x + \frac{1}{64} \sin 5x = \frac{7}{64} \sin 5x $

For $ \sin 7x $:

$ -\frac{1}{64} \sin 7x $

Therefore, combining all terms:

$ \sin^7 x = \frac{35}{64} \sin x - \frac{21}{64} \sin 3x + \frac{7}{64} \sin 5x - \frac{1}{64} \sin 7x $

Which matches exactly the result I obtained earlier using complex exponentials. Therefore, this confirms that the coefficients are correct, and so $ d = \frac{35}{64} $.

Therefore, the answer is $ \boxed{\dfrac{35}{64}} $.

**Final Answer**
\boxed{\dfrac{35}{64}}

To solve the equation

$$
(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x
$$

for all $ x $, we aim to express $ (\sin x)^7 $ as a linear combination of sine functions with different arguments. This is a classic problem in trigonometric identities, and one effective method is to use **complex exponentials** to expand the power of sine.

---

### Step 1: Express $ \sin x $ in terms of complex exponentials

We use the identity:

$$
\sin x = \frac{e^{ix} - e^{-ix}}{2i}
$$

Then:

$$
(\sin x)^7 = \left( \frac{e^{ix} - e^{-ix}}{2i} \right)^7
$$

Expanding this using the binomial theorem:

$$
(\sin x)^7 = \frac{1}{(2i)^7} \sum_{k=0}^{7} \binom{7}{k} (-1)^k e^{i(7 - 2k)x}
$$

Note that $ (2i)^7 = 128i^7 = 128(-i) = -128i $, so:

$$
(\sin x)^7 = \frac{1}{-128i} \sum_{k=0}^{7} \binom{7}{k} (-1)^k e^{i(7 - 2k)x}
$$

Multiplying numerator and denominator by $ i $, we get:

$$
(\sin x)^7 = \frac{i}{128} \sum_{k=0}^{7} \binom{7}{k} (-1)^k e^{i(7 - 2k)x}
$$

---

### Step 2: Group terms and convert to sine functions

The exponents $ 7 - 2k $ for $ k = 0 $ to $ 7 $ are:

$$
7, 5, 3, 1, -1, -3, -5, -7
$$

So the sum becomes:

$$
\sum_{k=0}^{7} \binom{7}{k} (-1)^k e^{i(7 - 2k)x} = \sum_{m=-7}^{7} \text{terms with } e^{imx}
$$

Grouping terms with $ e^{imx} $ and $ e^{-imx} $, and using the identity:

$$
e^{i\theta} - e^{-i\theta} = 2i \sin \theta
$$

we can express the entire sum in terms of sine functions. After simplifying, we find:

$$
(\sin x)^7 = \frac{35}{64} \sin x - \frac{21}{64} \sin 3x + \frac{7}{64} \sin 5x - \frac{1}{64} \sin 7x
$$

---

### Step 3: Match coefficients

Comparing this with the original equation:

$$
(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x
$$

We identify the coefficients:

- $ a = -\frac{1}{64} $
- $ b = \frac{7}{64} $
- $ c = -\frac{21}{64} $
- $ d = \frac{35}{64} $

---

### Final Answer

$$
\boxed{\dfrac{35}{64}}
$$

Solution:
We have that
\[\sin x = \frac{e^{ix} - e^{-ix}}{2i},\]so by the Binomial Theorem,
\begin{align*}
\sin^7 x &= \left( \frac{e^{ix} - e^{-ix}}{2i} \right)^7 \\
&= \frac{1}{128i^7} (e^{7ix} - 7 e^{5ix} + 21 e^{3ix} - 35 e^{ix} + 35 e^{-ix} - 21 e^{-3ix} + 7e^{-5ix} - e^{-7ix}) \\
&= \frac{i}{128} [(e^{7ix} - e^{-7ix}) - 7(e^{5ix} - e^{-5ix}) + 21(e^{3ix} - e^{-3ix}) - 35(e^{ix} - e^{-ix})] \\
&= \frac{i}{128} (2i \sin 7x - 14i \sin 5x + 42i \sin 3x - 70i \sin x) \\
&= -\frac{1}{64} \sin 7x + \frac{7}{64} \sin 5x - \frac{21}{64} \sin 3x + \frac{35}{64} \sin x.
\end{align*}Thus, the constant $d$ we seek is $\boxed{\frac{35}{64}}$.


Explanation:
Okay, so I need to figure out by what percentage the area of a square decreases when two opposite sides are decreased by 40% and the other two sides are increased by 50% to form a rectangle. Let me start by recalling some basic concepts.

First, a square has all sides equal. Let me denote the length of each side of the square as 's'. Therefore, the original area of the square would be s squared, right? So, Area_original = s².

Now, the problem says that two opposite sides are decreased by 40%, and the other two sides are increased by 50%. Let me visualize this. If I imagine a square, let's say the top and bottom sides are decreased by 40%, and the left and right sides are increased by 50%. Wait, but actually, in a square, all sides are the same, so decreasing two opposite sides and increasing the other two would result in a rectangle. Let me confirm that.

So, if the original square has sides of length 's', then after the changes:

- The length of the sides that are decreased by 40% would be s - 40% of s. Let me calculate that. 40% of s is 0.4s, so the new length is s - 0.4s = 0.6s. So, those sides are now 0.6s.

- The other two sides are increased by 50%. 50% of s is 0.5s, so the new length is s + 0.5s = 1.5s. Therefore, those sides are now 1.5s.

Therefore, the resulting figure is a rectangle with length and width of 0.6s and 1.5s? Wait, but which sides are which? Let me think. If we consider the square as having length and width both equal to s, then if we decrease two opposite sides (let's say the length) by 40%, and increase the other two sides (the width) by 50%, then the new rectangle would have length 0.6s and width 1.5s. Alternatively, it could be the other way around, but since length and width are just labels, it doesn't matter. The area would be the same either way.

So, the new area would be length × width = 0.6s × 1.5s. Let me compute that. 0.6 multiplied by 1.5. Let me do that calculation. 0.6 * 1.5. Well, 0.6 * 1 is 0.6, and 0.6 * 0.5 is 0.3, so adding them together gives 0.6 + 0.3 = 0.9. Therefore, the new area is 0.9s².

Wait, so the original area was s², and the new area is 0.9s². Therefore, the area has decreased by 0.1s². To find the percentage decrease, I need to calculate (decrease / original area) × 100%.

So, that would be (0.1s² / s²) × 100% = 0.1 × 100% = 10%. Therefore, the area decreases by 10%? Hmm, that seems straightforward, but let me check again to make sure I didn't make a mistake.

Let me go through the steps again. Original square: side length s, area s². Two opposite sides decreased by 40%: so each of those sides becomes 60% of original, which is 0.6s. The other two sides increased by 50%: each becomes 150% of original, which is 1.5s. Therefore, the rectangle has sides 0.6s and 1.5s. Multiplying them gives 0.6 * 1.5 = 0.9, so 0.9s². Original area was s², so the difference is s² - 0.9s² = 0.1s². Therefore, the percentage decrease is (0.1s² / s²) * 100% = 10%. So, 10% decrease. That seems correct.

But wait, let me think again. Maybe I need to consider that when they say "two opposite sides are decreased by 40%", does that mean that both sides are decreased by 40%? For example, if the square has sides AB, BC, CD, DA. If we decrease AB and CD by 40%, then AB and CD become 0.6s, and BC and DA are increased by 50% to 1.5s. So, the rectangle would have length AB = 0.6s and BC = 1.5s. Therefore, area is 0.6s * 1.5s = 0.9s². So, same as before. So, the area is 10% less. So, the answer is 10%?

But wait, maybe I need to check with actual numbers. Let me take a specific example. Let me assume the original square has side length 10 units. Then original area is 10*10=100.

Decrease two opposite sides by 40%: 40% of 10 is 4, so new length is 10 - 4 = 6. Increase the other two sides by 50%: 50% of 10 is 5, so new length is 10 + 5 = 15. Therefore, the rectangle has sides 6 and 15. Area is 6*15=90. Original area was 100, so decrease is 10, which is 10% decrease. So, same result. Therefore, 10% decrease. So, seems correct.

But wait, maybe there's a trick here? Let me think again. The problem says "the lengths of two opposite sides of a square are decreased by 40% while the lengths of the other two sides are increased by 50% to form a rectangle." So, is there a possibility that the sides are not just scaled independently? For example, maybe when you decrease two opposite sides, does that affect the other sides? But no, in a square, all sides are independent. If you change two opposite sides, the other two sides are separate. So, if you have a square, and you change two opposite sides, you have to change the other two sides as well. So, the rectangle is formed by changing two pairs of sides. So, the calculation seems correct.

Alternatively, maybe the problem is referring to changing the length and width? Wait, but in a square, length and width are the same. So, if you decrease the length by 40% and increase the width by 50%, but since they are the same, you have to adjust accordingly. But that's essentially what I did. So, if you consider length and width as the two different sides, then changing them by different percentages would result in a rectangle. So, the area would be (original length * 0.6) * (original width * 1.5). But since original length and width are both s, then it's 0.6s * 1.5s = 0.9s². So, same result.

Alternatively, maybe the problem is in three dimensions? But no, it's talking about a square, which is two-dimensional, so it's a square, then changing sides to form a rectangle. So, no, that's not the case.

Alternatively, maybe the problem is referring to changing the sides in a different way? For example, if you decrease two opposite sides by 40%, does that mean that the total length of those sides is decreased by 40%? But no, the problem says "the lengths of two opposite sides are decreased by 40%", which would mean each of those sides is decreased by 40%, not the total. So, each side is multiplied by 0.6. Similarly, the other two sides are increased by 50%, so each is multiplied by 1.5. Therefore, the area is multiplied by 0.6 * 1.5 = 0.9, so 10% decrease. So, that seems correct.

Alternatively, maybe the problem is in the interpretation of "decreased by 40%". For example, sometimes people confuse percentage decrease versus percentage of the original. But in this case, decreasing by 40% means subtracting 40% of the original value, so 60% of the original. Similarly, increasing by 50% means adding 50% of the original value, so 150% of the original. So, that's correct.

Alternatively, maybe the problem is expecting me to consider that when you decrease two sides and increase the other two, the area is calculated differently? For example, if you have a square, and you stretch it into a rectangle, maybe there's some other factor? But no, the area of a rectangle is just length times width, regardless of how it's formed. So, if you have two different lengths, then area is their product.

Alternatively, maybe I need to check if the problem is referring to the perimeter? But the question specifically says "the square's area decrease", so no, it's about area.

Alternatively, maybe I made a mistake in calculation? Let me check again:

Original area: s².

New area: (0.6s)(1.5s) = 0.6 * 1.5 * s². Let me compute 0.6 * 1.5. 0.6 * 1.5 is equal to (3/5) * (3/2) = 9/10 = 0.9. So, 0.9s². Therefore, the area is 90% of the original, so a 10% decrease. So, that's correct.

Alternatively, maybe the problem is in the fact that when you decrease two sides and increase the other two, the shape is not a rectangle? But no, if you have a square and you change two opposite sides to different lengths, then it's a rectangle. For example, if you have a square with sides AB, BC, CD, DA. If AB and CD are changed to 0.6s, and BC and DA are changed to 1.5s, then the figure is a rectangle with length AB = 0.6s and width BC = 1.5s. So, yes, it's a rectangle.

Alternatively, maybe the problem is referring to changing the sides in a way that affects the other sides? For example, if you decrease the length, does that affect the width? But no, in a square, the sides are independent. If you change two sides, the other two sides are changed independently. So, the problem says "the lengths of two opposite sides... are decreased... while the lengths of the other two sides... are increased...", so they are changed independently. Therefore, the calculation is correct.

Alternatively, maybe the problem is in the percentage calculation? Let me check again. If the original area is 100, and the new area is 90, then the decrease is 10, so 10/100 = 10%. So, 10% decrease. So, that's correct.

Alternatively, maybe the problem is expecting a different answer? Let me think. Suppose someone thought that decreasing two sides by 40% would mean decreasing the total length of those sides by 40%, but that would be different. For example, if you have two sides of length s each, total length is 2s. If you decrease that total by 40%, you get 2s * 0.6 = 1.2s. Then, if you increase the other two sides by 50%, total length would be 2s * 1.5 = 3s. But that approach is wrong because the problem says "the lengths of two opposite sides are decreased by 40%", which refers to each side, not the total. So, that approach is incorrect. So, that would be a wrong interpretation. But since the problem says "the lengths of two opposite sides", meaning each of those sides is decreased by 40%, not the total length of those sides. Therefore, the correct calculation is as before.

Alternatively, maybe someone might think that decreasing two sides by 40% and increasing the other two by 50% would result in a different area? For example, if you think that the area is calculated by some other formula? But no, area of a rectangle is length times width. So, if you have two different sides, then it's just the product.

Alternatively, maybe the problem is in the fact that when you change the sides, the shape is not a rectangle? But no, as long as opposite sides are equal, it's a rectangle. So, if you have two sides of 0.6s and two sides of 1.5s, then it's a rectangle.

Alternatively, maybe the problem is in the fact that the square is transformed into a rectangle, so maybe there's some other consideration? For example, if you are changing the sides, maybe the area is affected by some other factor? But no, the area is purely based on the product of the length and width.

Alternatively, maybe I need to check with another example. Let me take s = 2. Original area is 4. Decrease two sides by 40%: 2 - 0.4*2 = 2 - 0.8 = 1.2. Increase the other two sides by 50%: 2 + 0.5*2 = 2 + 1 = 3. Then, area is 1.2 * 3 = 3.6. Original area was 4, so decrease is 0.4, which is 0.4 / 4 = 0.1, so 10% decrease. Same result. So, that's correct.

Alternatively, maybe the problem is in the percentage decrease formula? Let me recall: percentage decrease is ((original - new)/original)*100%. So, ((s² - 0.9s²)/s²)*100% = 0.1*100% = 10%. Correct.

Alternatively, maybe the problem is expecting me to consider that decreasing two sides and increasing the other two would have some compounding effect? But no, because the area is a product of the two different sides. So, if you have two different scaling factors, you multiply them. So, 0.6 * 1.5 = 0.9. So, 10% decrease. So, that's straightforward.

Alternatively, maybe the problem is in the fact that the question says "the lengths of two opposite sides of a square are decreased by 40% while the lengths of the other two sides are increased by 50% to form a rectangle." So, maybe the sides are not just scaled, but something else? For example, maybe the sides are changed in a way that affects the other sides? But no, the problem says "the lengths of two opposite sides... are decreased... while the lengths of the other two sides... are increased...", so they are changed independently. Therefore, the calculation is correct.

Alternatively, maybe the problem is in the fact that when you decrease two sides, you have to adjust the other sides accordingly? For example, if you decrease the length, does that affect the width? But in a square, the sides are independent. If you have a square, and you change two opposite sides, you can change them independently. So, for example, if you have a square, and you make the top and bottom sides shorter, and the left and right sides longer, then it's a rectangle. So, the sides are independent. Therefore, the area is just the product of the new length and new width.

Therefore, after checking multiple times, I think my answer is correct. The area decreases by 10%.

But just to be thorough, let me think if there's another way to approach this problem. For example, using algebra.

Let me denote the original side length as s. Original area: A = s².

After the changes:

- Two opposite sides are decreased by 40%, so their new length is s - 0.4s = 0.6s.

- The other two sides are increased by 50%, so their new length is s + 0.5s = 1.5s.

Therefore, the new area is 0.6s * 1.5s = 0.9s².

Therefore, the decrease in area is A - A' = s² - 0.9s² = 0.1s².

Percentage decrease is (0.1s² / s²) * 100% = 10%.

Yes, same result. Therefore, I think this is solid.

Alternatively, maybe the problem is in the interpretation of "decreased by 40%". For example, if someone thought that decreasing by 40% means multiplying by 40%, but that would be incorrect. But in standard terms, decreasing by 40% means subtracting 40% of the original value, so 60% of the original. Similarly, increasing by 50% means adding 50% of the original value, so 150% of the original. So, that's correct.

Alternatively, maybe the problem is in the fact that the question says "the lengths of two opposite sides of a square are decreased by 40% while the lengths of the other two sides are increased by 50% to form a rectangle." So, maybe the sides are not just scaled, but the entire figure is transformed in some other way? For example, maybe stretching or something else? But no, the problem says "the lengths of two opposite sides... are decreased... while the lengths of the other two sides... are increased...", so it's a straightforward scaling of the sides.

Alternatively, maybe the problem is in the fact that when you decrease two sides and increase the other two, the area is not just the product? But no, area of a rectangle is always length times width, regardless of how you get there. So, if you have a rectangle with sides a and b, area is ab. So, if a is 0.6s and b is 1.5s, then area is 0.9s².

Therefore, I think my answer is correct. The area decreases by 10%.

But just to make sure, let me check with another example. Let me take s = 1. Original area is 1. New area is 0.6 * 1.5 = 0.9. So, decrease is 0.1, which is 10%. So, same result. Therefore, regardless of the value of s, the percentage decrease is always 10%. Therefore, the answer is 10%.

Therefore, I think the answer is 10%.

**Final Answer**
The square's area decreases by \boxed{10} percent.

To determine the percentage decrease in the area of a square after modifying its sides, we begin by analyzing the original and modified dimensions.

---

### **Step 1: Original Square**

Let the side length of the square be $ s $.
- The area of the square is:
  $$
  A_{\text{original}} = s^2
  $$

---

### **Step 2: Modify the Sides**

- Two opposite sides are **decreased by 40%**.
  So, the new length of these sides is:
  $$
  s - 0.4s = 0.6s
  $$

- The other two sides are **increased by 50%**.
  So, the new length of these sides is:
  $$
  s + 0.5s = 1.5s
  $$

These modified side lengths form a **rectangle** with dimensions $ 0.6s $ and $ 1.5s $.

---

### **Step 3: Calculate the New Area**

The area of the resulting rectangle is:
$$
A_{\text{new}} = (0.6s)(1.5s) = 0.9s^2
$$

---

### **Step 4: Determine the Percentage Decrease**

The decrease in area is:
$$
\Delta A = A_{\text{original}} - A_{\text{new}} = s^2 - 0.9s^2 = 0.1s^2
$$

The percentage decrease is:
$$
\text{Percentage decrease} = \left( \frac{\Delta A}{A_{\text{original}}} \right) \times 100\% = \left( \frac{0.1s^2}{s^2} \right) \times 100\% = 10\%
$$

---

### **Final Answer**

$$
\boxed{10}
$$

[Assistant]
Explanation:
Okay, so I need to figure out by what percentage the area of a square decreases when two opposite sides are decreased by 40% and the other two sides are increased by 50% to form a rectangle. Let me start by recalling some basic concepts.

First, a square has all sides equal. Let me denote the length of each side of the square as 's'. Therefore, the original area of the square would be s squared, right? So, Area_original = s².

Now, the problem says that two opposite sides are decreased by 40%, and the other two sides are increased by 50%. Let me visualize this. If I imagine a square, let's say the top and bottom sides are decreased by 40%, and the left and right sides are increased by 50%. Wait, but actually, in a square, all sides are the same, so decreasing two opposite sides and increasing the other two would result in a rectangle. Let me confirm that.

So, if the original square has sides of length 's', then after the changes:

- The length of the sides that are decreased by 40% would be s - 40% of s. Let me calculate that. 40% of s is 0.4s, so the new length is s - 0.4s = 0.6s. So, those sides are now 0.6s.

- The other two sides are increased by 50%. 50% of s is 0.5s, so the new length is s + 0.5s = 1.5s. Therefore, those sides are now 1.5s.

Therefore, the resulting figure is a rectangle with length and width of 0.6s and 1.5s? Wait, but which sides are which? Let me think. If we consider the square as having length and width both equal to s, then if we decrease two opposite sides (let's say the length) by 40%, and increase the other two sides (the width) by 50%, then the new rectangle would have length 0.6s and width 1.5s. Alternatively, it could be the other way around, but since length and width are just labels, it doesn't matter. The area would be the same either way.

So, the new area would be length × width = 0.6s × 1.5s. Let me compute that. 0.6 multiplied by 1.5. Let me do that calculation. 0.6 * 1.5. Well, 0.6 * 1 is 0.6, and 0.6 * 0.5 is 0.3, so adding them together gives 0.6 + 0.3 = 0.9. Therefore, the new area is 0.9s².

Wait, so the original area was s², and the new area is 0.9s². Therefore, the area has decreased by 0.1s². To find the percentage decrease, I need to calculate (decrease / original area) × 100%.

So, that would be (0.1s² / s²) × 100% = 0.1 × 100% = 10%. Therefore, the area decreases by 10%? Hmm, that seems straightforward, but let me check again to make sure I didn't make a mistake.

Let me go through the steps again. Original square: side length s, area s². Two opposite sides decreased by 40%: so each of those sides becomes 60% of original, which is 0.6s. The other two sides increased by 50%: each becomes 150% of original, which is 1.5s. Therefore, the rectangle has sides 0.6s and 1.5s. Multiplying them gives 0.6 * 1.5 = 0.9, so 0.9s². Original area was s², so the difference is s² - 0.9s² = 0.1s². Therefore, the percentage decrease is (0.1s² / s²) * 100% = 10%. So, 10% decrease. That seems correct.

But wait, let me think again. Maybe I need to consider that when they say "two opposite sides are decreased by 40%", does that mean that both sides are decreased by 40%? For example, if the square has sides AB, BC, CD, DA. If we decrease AB and CD by 40%, then AB and CD become 0.6s, and BC and DA are increased by 50% to 1.5s. So, the rectangle would have length AB = 0.6s and BC = 1.5s. Therefore, area is 0.6s * 1.5s = 0.9s². So, same as before. So, the area is 10% less. So, the answer is 10%?

But wait, maybe I need to check with actual numbers. Let me take a specific example. Let me assume the original square has side length 10 units. Then original area is 10*10=100.

Decrease two opposite sides by 40%: 40% of 10 is 4, so new length is 10 - 4 = 6. Increase the other two sides by 50%: 50% of 10 is 5, so new length is 10 + 5 = 15. Therefore, the rectangle has sides 6 and 15. Area is 6*15=90. Original area was 100, so decrease is 10, which is 10% decrease. So, same result. Therefore, 10% decrease. So, seems correct.

But wait, maybe there's a trick here? Let me think again. The problem says "the lengths of two opposite sides of a square are decreased by 40% while the lengths of the other two sides are increased by 50% to form a rectangle." So, is there a possibility that the sides are not just scaled independently? For example, maybe when you decrease two opposite sides, does that affect the other sides? But no, in a square, all sides are independent. If you change two opposite sides, the other two sides are separate. So, if you have a square, and you change two opposite sides, you have to change the other two sides as well. So, the rectangle is formed by changing two pairs of sides. So, the calculation seems correct.

Alternatively, maybe the problem is referring to changing the length and width? Wait, but in a square, length and width are the same. So, if you decrease the length by 40% and increase the width by 50%, but since they are the same, you have to adjust accordingly. But that's essentially what I did. So, if you consider length and width as the two different sides, then changing them by different percentages would result in a rectangle. So, the area would be (original length * 0.6) * (original width * 1.5). But since original length and width are both s, then it's 0.6s * 1.5s = 0.9s². So, same result.

Alternatively, maybe the problem is in three dimensions? But no, it's talking about a square, which is two-dimensional, so it's a square, then changing sides to form a rectangle. So, no, that's not the case.

Alternatively, maybe the problem is referring to changing the sides in a different way? For example, if you decrease two opposite sides by 40%, does that mean that the total length of those sides is decreased by 40%? But no, the problem says "the lengths of two opposite sides are decreased by 40%", which would mean each of those sides is decreased by 40%, not the total. So, each side is multiplied by 0.6. Similarly, the other two sides are increased by 50%, so each is multiplied by 1.5. Therefore, the area is multiplied by 0.6 * 1.5 = 0.9, so 10% decrease. So, that seems correct.

Alternatively, maybe the problem is in the interpretation of "decreased by 40%". For example, sometimes people confuse percentage decrease versus percentage of the original. But in this case, decreasing by 40% means subtracting 40% of the original value, so 60% of the original. Similarly, increasing by 50% means adding 50% of the original value, so 150% of the original. So, that's correct.

Alternatively, maybe the problem is expecting me to consider that when you decrease two sides and increase the other two, the area is calculated differently? For example, if you have a square, and you stretch it into a rectangle, maybe there's some other factor? But no, the area of a rectangle is just length times width, regardless of how it's formed. So, if you have two different lengths, then area is their product.

Alternatively, maybe I need to check if the problem is referring to the perimeter? But the question specifically says "the square's area decrease", so no, it's about area.

Alternatively, maybe I made a mistake in calculation? Let me check again:

Original area: s².

New area: (0.6s)(1.5s) = 0.6 * 1.5 * s². Let me compute 0.6 * 1.5. 0.6 * 1.5 is equal to (3/5) * (3/2) = 9/10 = 0.9. So, 0.9s². Therefore, the area is 90% of the original, so a 10% decrease. So, that's correct.

Alternatively, maybe the problem is in the fact that when you decrease two sides and increase the other two, the shape is not a rectangle? But no, if you have a square and you change two opposite sides to different lengths, then it's a rectangle. For example, if you have a square with sides AB, BC, CD, DA. If AB and CD are changed to 0.6s, and BC and DA are changed to 1.5s, then the figure is a rectangle with length AB = 0.6s and width BC = 1.5s. So, yes, it's a rectangle.

Alternatively, maybe the problem is referring to changing the sides in a way that affects the other sides? For example, if you decrease the length, does that affect the width? But no, in a square, the sides are independent. If you change two sides, the other two sides are changed independently. So, the problem says "the lengths of two opposite sides... are decreased... while the lengths of the other two sides... are increased...", so they are changed independently. Therefore, the calculation is correct.

Alternatively, maybe the problem is in the percentage calculation? Let me check again. If the original area is 100, and the new area is 90, then the decrease is 10, so 10/100 = 10%. So, 10% decrease. So, that's correct.

Alternatively, maybe the problem is expecting a different answer? Let me think. Suppose someone thought that decreasing two sides by 40% would mean decreasing the total length of those sides by 40%, but that would be different. For example, if you have two sides of length s each, total length is 2s. If you decrease that total by 40%, you get 2s * 0.6 = 1.2s. Then, if you increase the other two sides by 50%, total length would be 2s * 1.5 = 3s. But that approach is wrong because the problem says "the lengths of two opposite sides are decreased by 40%", which refers to each side, not the total. So, that approach is incorrect. So, that would be a wrong interpretation. But since the problem says "the lengths of two opposite sides", meaning each of those sides is decreased by 40%, not the total length of those sides. Therefore, the correct calculation is as before.

Alternatively, maybe someone might think that decreasing two sides by 40% and increasing the other two by 50% would result in a different area? For example, if you think that the area is calculated by some other formula? But no, area of a rectangle is length times width. So, if you have two different sides, then it's just the product.

Alternatively, maybe the problem is in the fact that when you change the sides, the shape is not a rectangle? But no, as long as opposite sides are equal, it's a rectangle. So, if you have two sides of 0.6s and two sides of 1.5s, then it's a rectangle.

Alternatively, maybe the problem is in the fact that the square is transformed into a rectangle, so maybe there's some other consideration? For example, if you are changing the sides, maybe the area is affected by some other factor? But no, the area is purely based on the product of the length and width.

Alternatively, maybe I need to check with another example. Let me take s = 2. Original area is 4. Decrease two sides by 40%: 2 - 0.4*2 = 2 - 0.8 = 1.2. Increase the other two sides by 50%: 2 + 0.5*2 = 2 + 1 = 3. Then, area is 1.2 * 3 = 3.6. Original area was 4, so decrease is 0.4, which is 0.4 / 4 = 0.1, so 10% decrease. Same result. So, that's correct.

Alternatively, maybe the problem is in the percentage decrease formula? Let me recall: percentage decrease is ((original - new)/original)*100%. So, ((s² - 0.9s²)/s²)*100% = 0.1*100% = 10%. Correct.

Alternatively, maybe the problem is expecting me to consider that decreasing two sides and increasing the other two would have some compounding effect? But no, because the area is a product of the two different sides. So, if you have two different scaling factors, you multiply them. So, 0.6 * 1.5 = 0.9. So, 10% decrease. So, that's straightforward.

Alternatively, maybe the problem is in the fact that the question says "the lengths of two opposite sides of a square are decreased by 40% while the lengths of the other two sides are increased by 50% to form a rectangle." So, maybe the sides are not just scaled, but something else? For example, maybe the sides are changed in a way that affects the other sides? But no, the problem says "the lengths of two opposite sides... are decreased... while the lengths of the other two sides... are increased...", so they are changed independently. Therefore, the calculation is correct.

Alternatively, maybe the problem is in the fact that when you decrease two sides, you have to adjust the other sides accordingly? For example, if you decrease the length, does that affect the width? But in a square, the sides are independent. If you have a square, and you change two opposite sides, you can change them independently. So, for example, if you have a square, and you make the top and bottom sides shorter, and the left and right sides longer, then it's a rectangle. So, the sides are independent. Therefore, the area is just the product of the new length and new width.

Therefore, after checking multiple times, I think my answer is correct. The area decreases by 10%.

But just to be thorough, let me think if there's another way to approach this problem. For example, using algebra.

Let me denote the original side length as s. Original area: A = s².

After the changes:

- Two opposite sides are decreased by 40%, so their new length is s - 0.4s = 0.6s.

- The other two sides are increased by 50%, so their new length is s + 0.5s = 1.5s.

Therefore, the new area is 0.6s * 1.5s = 0.9s².

Therefore, the decrease in area is A - A' = s² - 0.9s² = 0.1s².

Percentage decrease is (0.1s² / s²) * 100% = 10%.

Yes, same result. Therefore, I think this is solid.

Alternatively, maybe the problem is in the interpretation of "decreased by 40%". For example, if someone thought that decreasing by 40% means multiplying by 40%, but that would be incorrect. But in standard terms, decreasing by 40% means subtracting 40% of the original value, so 60% of the original. Similarly, increasing by 50% means adding 50% of the original value, so 150% of the original. So, that's correct.

Alternatively, maybe the problem is in the fact that the question says "the lengths of two opposite sides of a square are decreased by 40% while the lengths of the other two sides are increased by 50% to form a rectangle." So, maybe the sides are not just scaled, but the entire figure is transformed in some other way? For example, maybe stretching or something else? But no, the problem says "the lengths of two opposite sides... are decreased... while the lengths of the other two sides... are increased...", so it's a straightforward scaling of the sides.

Alternatively, maybe the problem is in the fact that when you decrease two sides and increase the other two, the area is not just the product? But no, area of a rectangle is always length times width, regardless of how you get there. So, if you have a rectangle with sides a and b, area is ab. So, if a is 0.6s and b is 1.5s, then area is 0.9s².

Therefore, I think my answer is correct. The area decreases by 10%.

But just to make sure, let me check with another example. Let me take s = 1. Original area is 1. New area is 0.6 * 1.5 = 0.9. So, decrease is 0.1, which is 10%. So, same result. Therefore, regardless of the value of s, the percentage decrease is always 10%. Therefore, the answer is 10%.

Therefore, I think the answer is 10%.

**Final Answer**
The square's area decreases by \boxed{10} percent.

To determine the percentage decrease in the area of a square after modifying its sides, we begin by analyzing the original and modified dimensions.

---

### **Step 1: Original Square**

Let the side length of the square be $ s $.
- The area of the square is:
  $$
  A_{\text{original}} = s^2
  $$

---

### **Step 2: Modify the Sides**

- Two opposite sides are **decreased by 40%**.
  So, the new length of these sides is:
  $$
  s - 0.4s = 0.6s
  $$

- The other two sides are **increased by 50%**.
  So, the new length of these sides is:
  $$
  s + 0.5s = 1.5s
  $$

These modified side lengths form a **rectangle** with dimensions $ 0.6s $ and $ 1.5s $.

---

### **Step 3: Calculate the New Area**

The area of the resulting rectangle is:
$$
A_{\text{new}} = (0.6s)(1.5s) = 0.9s^2
$$

---

### **Step 4: Determine the Percentage Decrease**

The decrease in area is:
$$
\Delta A = A_{\text{original}} - A_{\text{new}} = s^2 - 0.9s^2 = 0.1s^2
$$

The percentage decrease is:
$$
\text{Percentage decrease} = \left( \frac{\Delta A}{A_{\text{original}}} \right) \times 100\% = \left( \frac{0.1s^2}{s^2} \right) \times 100\% = 10\%
$$

---

### **Final Answer**

$$
\boxed{10}
$$

Solution: Let $A$ be the area of the square. The lengths of one pair of opposite sides was decreased by $40\%$, so the area became $.6A$. The other pair of sides were increased by $50\%$, so the area became $1.5\cdot .6 A = .9A$. Thus, the area decreased by $\boxed{10}$ percent.


Explanation:
Okay, so I need to solve this equation: tan 53° tan 81° tan x° = tan 53° + tan 81° + tan x°, where x is between 0 and 180 degrees. Hmm, let me think about how to approach this.

First, I remember that there are some trigonometric identities involving tangent functions. Maybe I can use the tangent addition formula or something related? Let me recall... The tangent addition formula is tan(A + B) = (tan A + tan B) / (1 - tan A tan B). But I'm not sure if that's directly applicable here.

Alternatively, maybe I can rearrange the given equation to make it look like something familiar. Let me write down the equation again:

tan 53° tan 81° tan x° = tan 53° + tan 81° + tan x°.

Let me denote tan 53° as T and tan 81° as S for simplicity. Then the equation becomes:

T * S * tan x = T + S + tan x.

Hmm, maybe I can rearrange terms to group tan x terms together. Let me try moving all terms to one side:

T * S * tan x - tan x - T - S = 0.

Factor tan x from the first two terms:

tan x (T * S - 1) - (T + S) = 0.

So, tan x (T * S - 1) = T + S.

Therefore, tan x = (T + S) / (T * S - 1).

Wait, that looks similar to the formula for tan(A + B). Let me recall:

tan(A + B) = (tan A + tan B) / (1 - tan A tan B).

But here, the denominator is T * S - 1, which is similar but with a negative sign. Let me write that:

If I have tan(A + B) = (tan A + tan B) / (1 - tan A tan B), then if I take negative of denominator:

(tan A + tan B) / (tan A tan B - 1) = - tan(A + B).

So, tan x = - tan(A + B), where A and B are 53° and 81°? Let me check:

Let me compute tan(53° + 81°). 53 + 81 is 134°, so tan(134°). But tan(134°) is tan(180° - 46°) = -tan(46°). So tan(134°) = -tan(46°).

But according to the previous equation, tan x = (T + S)/(T S - 1) = - tan(A + B) = - tan(134°) = tan(46°). Wait, because tan(134°) is -tan(46°), so negative of that is tan(46°). Therefore, tan x = tan(46°).

Therefore, x could be 46°, or 46° + 180°, but since x is between 0 and 180°, the only possible solution is 46°.

Wait, but let me check if that's correct. Let me verify by plugging x = 46° into the original equation.

First, compute tan 53° tan 81° tan 46° and compare it with tan 53° + tan 81° + tan 46°.

But maybe I can check if tan x = tan(46°) is indeed the solution. Let me compute tan(53° + 81°) = tan(134°) = -tan(46°). Therefore, according to the identity:

tan(A + B) = (tan A + tan B)/(1 - tan A tan B). Therefore, rearranged:

tan A + tan B = tan(A + B)(1 - tan A tan B).

But in our case, we had tan x = (tan A + tan B)/(tan A tan B - 1) = - tan(A + B). Therefore, tan x = - tan(A + B) = tan(- (A + B)). But since tan is periodic with period 180°, tan(- (A + B)) = tan(180° - (A + B)).

Wait, let me think again. If tan x = - tan(A + B), then tan x = tan(- (A + B)), so x could be - (A + B) + k*180°, but since x is between 0 and 180°, we need to find an appropriate k.

So A + B is 53 + 81 = 134°, so -134° + 180° = 46°, which is within 0 and 180°, so x = 46°.

Therefore, the solution is 46°.

But let me verify with actual numbers to make sure. Let me calculate tan 53°, tan 81°, tan 46°, and check if the equation holds.

First, I need approximate values. Let me recall that tan 45° = 1, tan 60° ≈ 1.732, tan 30° ≈ 0.577.

But exact values for 53°, 81°, 46°? Maybe I can use calculator approximations.

But since I don't have a calculator here, maybe I can use some known relationships. Wait, 53° and 37° are complementary in some contexts, like in 3-4-5 triangles. Wait, tan 37° ≈ 3/4 = 0.75, tan 53° ≈ 4/3 ≈ 1.333. Similarly, tan 81° is a large value, since tan 60° ≈ 1.732, tan 81° is close to tan 90°, which is infinity. So tan 81° is about 6.3138 (I remember that tan 80° ≈ 5.671, tan 81° ≈ 6.3138).

Similarly, tan 46° is tan(45° + 1°) ≈ 1 + tan 1° / (1 - tan 1°) ??? Wait, maybe better to recall that tan 45° = 1, tan 46° is slightly larger. Let me approximate tan 46° ≈ 1.0355.

But maybe I can use exact relations? Wait, if x = 46°, then let me check:

Left side: tan53 * tan81 * tan46

Right side: tan53 + tan81 + tan46

But if tan53 * tan81 * tan46 equals tan53 + tan81 + tan46?

Alternatively, since we derived that tan x = tan(46°), so if that's the case, then the equation should hold. But maybe there's a mistake in my derivation?

Wait, let me check again. Starting from:

tan53 tan81 tanx = tan53 + tan81 + tanx.

Let me rearrange:

tan53 tan81 tanx - tanx = tan53 + tan81.

tanx (tan53 tan81 - 1) = tan53 + tan81.

Therefore, tanx = (tan53 + tan81)/(tan53 tan81 - 1).

But from the tangent addition formula:

tan(A + B) = (tanA + tanB)/(1 - tanA tanB). Therefore, (tanA + tanB)/(tanA tanB - 1) = - tan(A + B).

Therefore, tanx = - tan(A + B) = tan(- (A + B)).

So tanx = tan(- (53 + 81)) = tan(-134°). But tan(-134°) = - tan(134°). However, tan(134°) = tan(180° - 46°) = - tan(46°). Therefore, tan(-134°) = - (- tan46°) = tan46°. Therefore, tanx = tan46°, so x = 46° + k*180°. Since x is between 0 and 180°, x = 46°.

Therefore, the answer is 46°.

But let me check with approximate values. Let me try to compute both sides numerically.

First, compute tan53°, tan81°, tan46°.

Using calculator approximations:

tan(53°) ≈ 1.3270

tan(81°) ≈ 6.3138

tan(46°) ≈ 1.0355

Left side: 1.3270 * 6.3138 * 1.0355 ≈ Let me compute step by step:

First, 1.3270 * 6.3138 ≈ Let me compute 1.327 * 6.3138.

Well, 1.327 * 6 = 7.962, 1.327 * 0.3138 ≈ approx 1.327 * 0.3 = 0.3981, 1.327 * 0.0138 ≈ ~0.0183. So total approx 0.3981 + 0.0183 ≈ 0.4164. Therefore total approx 7.962 + 0.4164 ≈ 8.3784.

Then multiply by 1.0355: 8.3784 * 1.0355 ≈ 8.3784 * 1 + 8.3784 * 0.0355 ≈ 8.3784 + approx 0.297 ≈ 8.6754.

Now compute the right side: tan53 + tan81 + tan46 ≈ 1.3270 + 6.3138 + 1.0355 ≈ 1.3270 + 6.3138 = 7.6408 + 1.0355 ≈ 8.6763.

Wow, that's very close! The left side was approx 8.6754 and the right side approx 8.6763. The slight difference is due to my approximated values of the tangents. Therefore, this confirms that x = 46° is indeed the solution.

Therefore, the answer is 46 degrees.

But just to make sure there are no other solutions. Since tan x = tan 46°, and x is between 0 and 180°, the solutions are x = 46° and x = 46° + 180°, but 46 + 180 = 226°, which is outside the given range. Therefore, only x = 46° is valid.

Alternatively, could there be another angle in 0 to 180° where tan x equals tan 46°? Well, tan is positive in first and third quadrants. But since x is between 0 and 180°, third quadrant would be between 180° and 270°, but our upper limit is 180°, so only the first quadrant. Therefore, only 46°.

Therefore, the answer is 46°.

**Final Answer**
\boxed{46}

To solve the equation:

$$
\tan 53^\circ \tan 81^\circ \tan x^\circ = \tan 53^\circ + \tan 81^\circ + \tan x^\circ
$$

for some value of $ x $ in the interval $ 0 < x < 180 $, we begin by simplifying and analyzing the structure of the equation.

---

### Step 1: Introduce Variables for Simplicity

Let:
- $ T = \tan 53^\circ $
- $ S = \tan 81^\circ $
- $ t = \tan x^\circ $

Then the equation becomes:

$$
T S t = T + S + t
$$

---

### Step 2: Rearranging the Equation

Bring all terms to one side:

$$
T S t - t = T + S
$$

Factor out $ t $ on the left-hand side:

$$
t (T S - 1) = T + S
$$

Now solve for $ t $:

$$
t = \frac{T + S}{T S - 1}
$$

---

### Step 3: Recognize a Tangent Identity

Recall the tangent addition identity:

$$
\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}
$$

This is similar to the expression we have, except the denominator is $ T S - 1 $ instead of $ 1 - T S $. So we can write:

$$
\frac{T + S}{T S - 1} = -\tan(A + B)
$$

where $ A = 53^\circ $, $ B = 81^\circ $, and $ A + B = 134^\circ $. Therefore:

$$
t = -\tan(134^\circ)
$$

But $ \tan(134^\circ) = \tan(180^\circ - 46^\circ) = -\tan(46^\circ) $, so:

$$
t = -(-\tan(46^\circ)) = \tan(46^\circ)
$$

---

### Step 4: Determine the Value of $ x $

We now have:

$$
\tan x = \tan 46^\circ
$$

Since $ \tan x = \tan 46^\circ $, and $ x \in (0^\circ, 180^\circ) $, the only solution is:

$$
x = 46^\circ
$$

---

### Step 5: Verification (Optional but Helpful)

To confirm, we can plug $ x = 46^\circ $ back into the original equation and verify that both sides are equal. Using approximate values:

- $ \tan 53^\circ \approx 1.3270 $
- $ \tan 81^\circ \approx 6.3138 $
- $ \tan 46^\circ \approx 1.0355 $

Compute left-hand side:

$$
1.3270 \times 6.3138 \times 1.0355 \approx 8.6754
$$

Compute right-hand side:

$$
1.3270 + 6.3138 + 1.0355 \approx 8.6763
$$

The values are very close, confirming the correctness of the solution.

---

### Final Answer

$$
\boxed{46}
$$

[Assistant]
Explanation:
Okay, so I need to solve this equation: tan 53° tan 81° tan x° = tan 53° + tan 81° + tan x°, where x is between 0 and 180 degrees. Hmm, let me think about how to approach this.

First, I remember that there are some trigonometric identities involving tangent functions. Maybe I can use the tangent addition formula or something related? Let me recall... The tangent addition formula is tan(A + B) = (tan A + tan B) / (1 - tan A tan B). But I'm not sure if that's directly applicable here.

Alternatively, maybe I can rearrange the given equation to make it look like something familiar. Let me write down the equation again:

tan 53° tan 81° tan x° = tan 53° + tan 81° + tan x°.

Let me denote tan 53° as T and tan 81° as S for simplicity. Then the equation becomes:

T * S * tan x = T + S + tan x.

Hmm, maybe I can rearrange terms to group tan x terms together. Let me try moving all terms to one side:

T * S * tan x - tan x - T - S = 0.

Factor tan x from the first two terms:

tan x (T * S - 1) - (T + S) = 0.

So, tan x (T * S - 1) = T + S.

Therefore, tan x = (T + S) / (T * S - 1).

Wait, that looks similar to the formula for tan(A + B). Let me recall:

tan(A + B) = (tan A + tan B) / (1 - tan A tan B).

But here, the denominator is T * S - 1, which is similar but with a negative sign. Let me write that:

If I have tan(A + B) = (tan A + tan B) / (1 - tan A tan B), then if I take negative of denominator:

(tan A + tan B) / (tan A tan B - 1) = - tan(A + B).

So, tan x = - tan(A + B), where A and B are 53° and 81°? Let me check:

Let me compute tan(53° + 81°). 53 + 81 is 134°, so tan(134°). But tan(134°) is tan(180° - 46°) = -tan(46°). So tan(134°) = -tan(46°).

But according to the previous equation, tan x = (T + S)/(T S - 1) = - tan(A + B) = - tan(134°) = tan(46°). Wait, because tan(134°) is -tan(46°), so negative of that is tan(46°). Therefore, tan x = tan(46°).

Therefore, x could be 46°, or 46° + 180°, but since x is between 0 and 180°, the only possible solution is 46°.

Wait, but let me check if that's correct. Let me verify by plugging x = 46° into the original equation.

First, compute tan 53° tan 81° tan 46° and compare it with tan 53° + tan 81° + tan 46°.

But maybe I can check if tan x = tan(46°) is indeed the solution. Let me compute tan(53° + 81°) = tan(134°) = -tan(46°). Therefore, according to the identity:

tan(A + B) = (tan A + tan B)/(1 - tan A tan B). Therefore, rearranged:

tan A + tan B = tan(A + B)(1 - tan A tan B).

But in our case, we had tan x = (tan A + tan B)/(tan A tan B - 1) = - tan(A + B). Therefore, tan x = - tan(A + B) = tan(- (A + B)). But since tan is periodic with period 180°, tan(- (A + B)) = tan(180° - (A + B)).

Wait, let me think again. If tan x = - tan(A + B), then tan x = tan(- (A + B)), so x could be - (A + B) + k*180°, but since x is between 0 and 180°, we need to find an appropriate k.

So A + B is 53 + 81 = 134°, so -134° + 180° = 46°, which is within 0 and 180°, so x = 46°.

Therefore, the solution is 46°.

But let me verify with actual numbers to make sure. Let me calculate tan 53°, tan 81°, tan 46°, and check if the equation holds.

First, I need approximate values. Let me recall that tan 45° = 1, tan 60° ≈ 1.732, tan 30° ≈ 0.577.

But exact values for 53°, 81°, 46°? Maybe I can use calculator approximations.

But since I don't have a calculator here, maybe I can use some known relationships. Wait, 53° and 37° are complementary in some contexts, like in 3-4-5 triangles. Wait, tan 37° ≈ 3/4 = 0.75, tan 53° ≈ 4/3 ≈ 1.333. Similarly, tan 81° is a large value, since tan 60° ≈ 1.732, tan 81° is close to tan 90°, which is infinity. So tan 81° is about 6.3138 (I remember that tan 80° ≈ 5.671, tan 81° ≈ 6.3138).

Similarly, tan 46° is tan(45° + 1°) ≈ 1 + tan 1° / (1 - tan 1°) ??? Wait, maybe better to recall that tan 45° = 1, tan 46° is slightly larger. Let me approximate tan 46° ≈ 1.0355.

But maybe I can use exact relations? Wait, if x = 46°, then let me check:

Left side: tan53 * tan81 * tan46

Right side: tan53 + tan81 + tan46

But if tan53 * tan81 * tan46 equals tan53 + tan81 + tan46?

Alternatively, since we derived that tan x = tan(46°), so if that's the case, then the equation should hold. But maybe there's a mistake in my derivation?

Wait, let me check again. Starting from:

tan53 tan81 tanx = tan53 + tan81 + tanx.

Let me rearrange:

tan53 tan81 tanx - tanx = tan53 + tan81.

tanx (tan53 tan81 - 1) = tan53 + tan81.

Therefore, tanx = (tan53 + tan81)/(tan53 tan81 - 1).

But from the tangent addition formula:

tan(A + B) = (tanA + tanB)/(1 - tanA tanB). Therefore, (tanA + tanB)/(tanA tanB - 1) = - tan(A + B).

Therefore, tanx = - tan(A + B) = tan(- (A + B)).

So tanx = tan(- (53 + 81)) = tan(-134°). But tan(-134°) = - tan(134°). However, tan(134°) = tan(180° - 46°) = - tan(46°). Therefore, tan(-134°) = - (- tan46°) = tan46°. Therefore, tanx = tan46°, so x = 46° + k*180°. Since x is between 0 and 180°, x = 46°.

Therefore, the answer is 46°.

But let me check with approximate values. Let me try to compute both sides numerically.

First, compute tan53°, tan81°, tan46°.

Using calculator approximations:

tan(53°) ≈ 1.3270

tan(81°) ≈ 6.3138

tan(46°) ≈ 1.0355

Left side: 1.3270 * 6.3138 * 1.0355 ≈ Let me compute step by step:

First, 1.3270 * 6.3138 ≈ Let me compute 1.327 * 6.3138.

Well, 1.327 * 6 = 7.962, 1.327 * 0.3138 ≈ approx 1.327 * 0.3 = 0.3981, 1.327 * 0.0138 ≈ ~0.0183. So total approx 0.3981 + 0.0183 ≈ 0.4164. Therefore total approx 7.962 + 0.4164 ≈ 8.3784.

Then multiply by 1.0355: 8.3784 * 1.0355 ≈ 8.3784 * 1 + 8.3784 * 0.0355 ≈ 8.3784 + approx 0.297 ≈ 8.6754.

Now compute the right side: tan53 + tan81 + tan46 ≈ 1.3270 + 6.3138 + 1.0355 ≈ 1.3270 + 6.3138 = 7.6408 + 1.0355 ≈ 8.6763.

Wow, that's very close! The left side was approx 8.6754 and the right side approx 8.6763. The slight difference is due to my approximated values of the tangents. Therefore, this confirms that x = 46° is indeed the solution.

Therefore, the answer is 46 degrees.

But just to make sure there are no other solutions. Since tan x = tan 46°, and x is between 0 and 180°, the solutions are x = 46° and x = 46° + 180°, but 46 + 180 = 226°, which is outside the given range. Therefore, only x = 46° is valid.

Alternatively, could there be another angle in 0 to 180° where tan x equals tan 46°? Well, tan is positive in first and third quadrants. But since x is between 0 and 180°, third quadrant would be between 180° and 270°, but our upper limit is 180°, so only the first quadrant. Therefore, only 46°.

Therefore, the answer is 46°.

**Final Answer**
\boxed{46}

To solve the equation:

$$
\tan 53^\circ \tan 81^\circ \tan x^\circ = \tan 53^\circ + \tan 81^\circ + \tan x^\circ
$$

for some value of $ x $ in the interval $ 0 < x < 180 $, we begin by simplifying and analyzing the structure of the equation.

---

### Step 1: Introduce Variables for Simplicity

Let:
- $ T = \tan 53^\circ $
- $ S = \tan 81^\circ $
- $ t = \tan x^\circ $

Then the equation becomes:

$$
T S t = T + S + t
$$

---

### Step 2: Rearranging the Equation

Bring all terms to one side:

$$
T S t - t = T + S
$$

Factor out $ t $ on the left-hand side:

$$
t (T S - 1) = T + S
$$

Now solve for $ t $:

$$
t = \frac{T + S}{T S - 1}
$$

---

### Step 3: Recognize a Tangent Identity

Recall the tangent addition identity:

$$
\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}
$$

This is similar to the expression we have, except the denominator is $ T S - 1 $ instead of $ 1 - T S $. So we can write:

$$
\frac{T + S}{T S - 1} = -\tan(A + B)
$$

where $ A = 53^\circ $, $ B = 81^\circ $, and $ A + B = 134^\circ $. Therefore:

$$
t = -\tan(134^\circ)
$$

But $ \tan(134^\circ) = \tan(180^\circ - 46^\circ) = -\tan(46^\circ) $, so:

$$
t = -(-\tan(46^\circ)) = \tan(46^\circ)
$$

---

### Step 4: Determine the Value of $ x $

We now have:

$$
\tan x = \tan 46^\circ
$$

Since $ \tan x = \tan 46^\circ $, and $ x \in (0^\circ, 180^\circ) $, the only solution is:

$$
x = 46^\circ
$$

---

### Step 5: Verification (Optional but Helpful)

To confirm, we can plug $ x = 46^\circ $ back into the original equation and verify that both sides are equal. Using approximate values:

- $ \tan 53^\circ \approx 1.3270 $
- $ \tan 81^\circ \approx 6.3138 $
- $ \tan 46^\circ \approx 1.0355 $

Compute left-hand side:

$$
1.3270 \times 6.3138 \times 1.0355 \approx 8.6754
$$

Compute right-hand side:

$$
1.3270 + 6.3138 + 1.0355 \approx 8.6763
$$

The values are very close, confirming the correctness of the solution.

---

### Final Answer

$$
\boxed{46}
$$

Solution:
Isolating $\tan x^\circ,$ we find
\begin{align*}
\tan x &= \frac{\tan 53^\circ + \tan 81^\circ}{\tan 53^\circ \tan 81^\circ - 1} \\
&= -\frac{\tan 53^\circ + \tan 81^\circ}{1 - \tan 53^\circ \tan 81^\circ}.
\end{align*}From the angle addition formula, this is equal to
\[-\tan (53^\circ + 81^\circ) = -\tan 134^\circ = \tan 46^\circ.\]Therefore, $x = \boxed{46}.$