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. Maybe I can use some kind of power-reduction formulas or multiple-angle identities to expand $ (\sin x)^7 $ into a combination of sine functions with different arguments.

I know that for lower powers, like $ \sin^2 x $, we can use the identity $ \sin^2 x = \frac{1 - \cos 2x}{2} $. Similarly, higher powers can be expressed using more complex identities. However, for the seventh power, this might get complicated. Maybe there's a recursive way or a formula for expressing $ \sin^n x $ as a sum of multiple angles?

Alternatively, I recall that complex exponentials can be useful for such expansions. Euler's formula says that $ \sin x = \frac{e^{ix} - e^{-ix}}{2i} $. So, perhaps I can express $ (\sin x)^7 $ in terms of exponentials and then convert back to sine terms?

Let me try that approach. Let me write $ \sin x $ as $ \frac{e^{ix} - e^{-ix}}{2i} $. Then $ (\sin x)^7 $ would be $ \left( \frac{e^{ix} - e^{-ix}}{2i} \right)^7 $. Expanding this using the binomial theorem might give me terms involving $ e^{i k x} $ for various integers $ k $, which I can then convert back to sine functions.

Let me write that out:

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

Wait, let me check that. The binomial expansion of $ (a + b)^n $ is $ \sum_{k=0}^n \binom{n}{k} a^{n - k} b^k $. In this case, $ a = e^{ix} $, $ b = -e^{-ix} $, and $ n = 7 $. So:

$ (e^{ix} - e^{-ix})^7 = \sum_{k=0}^7 \binom{7}{k} (e^{ix})^{7 - k} (-e^{-ix})^k $.

Simplify the exponents:

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

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

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

Simplify the constants: $ \frac{1}{-128i} = \frac{i}{128} $, since multiplying numerator and denominator by $ i $ gives $ \frac{i}{-128i^2} = \frac{i}{-128(-1)} = \frac{i}{128} $.

So, $ (\sin x)^7 = \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 must cancel out, leaving only sine terms.

Alternatively, perhaps I can pair terms to get sine functions. Let me consider that each term $ e^{i(7 - 2k)x} $ can be written as $ \cos((7 - 2k)x) + i \sin((7 - 2k)x) $. Therefore, when we take the sum over $ k $, and multiply by $ \frac{i}{128} $, the result should be purely imaginary, since the original expression is real. Therefore, the imaginary parts will combine to give sine terms.

Alternatively, maybe I can write the entire expression as a sum of sine terms. Let me think.

Alternatively, since $ e^{i\theta} + e^{-i\theta} = 2\cos \theta $, and $ e^{i\theta} - e^{-i\theta} = 2i \sin \theta $. However, in our case, the sum is over different exponents. Let me try to compute the sum.

Let me write out the sum:

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

Let me make a substitution. Let me let $ m = 7 - 2k $. Then, when $ k = 0 $, $ m = 7 $; when $ k = 1 $, $ m = 5 $; $ k = 2 $, $ m = 3 $; $ k = 3 $, $ m = 1 $; $ k = 4 $, $ m = -1 $; $ k = 5 $, $ m = -3 $; $ k = 6 $, $ m = -5 $; $ k = 7 $, $ m = -7 $.

So, the exponents go from 7 down to -7 in steps of -2. Therefore, the sum can be rewritten as:

$ \sum_{m = -7, -5, -3, -1, 1, 3, 5, 7} \binom{7}{(7 - m)/2} (-1)^{(7 - m)/2} e^{i m x} $.

Wait, maybe that's more complicated. Alternatively, note that the terms for $ k $ and $ 7 - k $ might be related. Let me check for symmetry.

Let me consider that when $ k $ ranges from 0 to 7, $ 7 - 2k $ ranges from 7 to -7. Let me note that for each term with exponent $ m $, there is a corresponding term with exponent $ -m $, except when $ m = 0 $. However, since 7 is odd, $ 7 - 2k = 0 $ would require $ 2k = 7 $, which is not possible for integer $ k $. Therefore, all exponents are non-zero and come in pairs $ m $ and $ -m $.

Therefore, perhaps I can pair terms. Let me consider that:

For each $ k $, there is a term with exponent $ m = 7 - 2k $, and another term with exponent $ -m $ when $ k' $ satisfies $ 7 - 2k' = -m $. Let me check for $ k' $:

$ 7 - 2k' = -m \Rightarrow 2k' = 7 + m \Rightarrow k' = (7 + m)/2 $.

But since $ m = 7 - 2k $, substituting:

$ k' = (7 + 7 - 2k)/2 = (14 - 2k)/2 = 7 - k $.

Therefore, each term with index $ k $ has a corresponding term with index $ 7 - k $. Therefore, the sum can be written as:

$ \sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} = \sum_{k=0}^3 \left[ \binom{7}{k} (-1)^k e^{i(7 - 2k)x} + \binom{7}{7 - k} (-1)^{7 - k} e^{i(7 - 2(7 - k))x} \right] $.

Wait, since when $ k $ goes from 0 to 7, pairing $ k $ and $ 7 - k $, but since 7 is odd, the middle term would be when $ k = 3.5 $, which isn't an integer. Therefore, the sum is symmetric around $ k = 3.5 $, so pairing terms from $ k = 0 $ and $ k = 7 $, $ k = 1 $ and $ k = 6 $, $ k = 2 $ and $ k = 5 $, $ k = 3 $ and $ k = 4 $.

Let me compute each pair:

First pair: $ k = 0 $ and $ k = 7 $:

Term for $ k = 0 $: $ \binom{7}{0} (-1)^0 e^{i(7 - 0)x} = 1 \cdot 1 \cdot e^{i7x} $.

Term for $ k = 7 $: $ \binom{7}{7} (-1)^7 e^{i(7 - 14)x} = 1 \cdot (-1)^7 e^{-i7x} = -e^{-i7x} $.

Adding them together: $ e^{i7x} - e^{-i7x} = 2i \sin 7x $.

Second pair: $ k = 1 $ and $ k = 6 $:

Term for $ k = 1 $: $ \binom{7}{1} (-1)^1 e^{i(7 - 2)x} = 7 \cdot (-1) \cdot e^{i5x} = -7 e^{i5x} $.

Term for $ k = 6 $: $ \binom{7}{6} (-1)^6 e^{i(7 - 12)x} = 7 \cdot 1 \cdot e^{-i5x} = 7 e^{-i5x} $.

Adding them together: $ -7 e^{i5x} + 7 e^{-i5x} = 7(-e^{i5x} + e^{-i5x}) = -14i \sin 5x $.

Third pair: $ k = 2 $ and $ k = 5 $:

Term for $ k = 2 $: $ \binom{7}{2} (-1)^2 e^{i(7 - 4)x} = 21 \cdot 1 \cdot e^{i3x} = 21 e^{i3x} $.

Term for $ k = 5 $: $ \binom{7}{5} (-1)^5 e^{i(7 - 10)x} = 21 \cdot (-1)^5 e^{-i3x} = -21 e^{-i3x} $.

Adding them together: $ 21 e^{i3x} - 21 e^{-i3x} = 42i \sin 3x $.

Fourth pair: $ k = 3 $ and $ k = 4 $:

Term for $ k = 3 $: $ \binom{7}{3} (-1)^3 e^{i(7 - 6)x} = 35 \cdot (-1)^3 e^{i1x} = -35 e^{ix} $.

Term for $ k = 4 $: $ \binom{7}{4} (-1)^4 e^{i(7 - 8)x} = 35 \cdot 1 \cdot e^{-ix} = 35 e^{-ix} $.

Adding them together: $ -35 e^{ix} + 35 e^{-ix} = 70i \sin x $.

Therefore, the entire sum is:

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

Therefore, going back to the original expression:

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

Let me compute this:

Multiply $ \frac{i}{128} $ by each term:

First term: $ \frac{i}{128} \times 2i \sin 7x = \frac{2i^2}{128} \sin 7x = \frac{2(-1)}{128} \sin 7x = -\frac{1}{64} \sin 7x $.

Second term: $ \frac{i}{128} \times (-14i) \sin 5x = \frac{-14i^2}{128} \sin 5x = \frac{-14(-1)}{128} \sin 5x = \frac{14}{128} \sin 5x = \frac{7}{64} \sin 5x $.

Third term: $ \frac{i}{128} \times 42i \sin 3x = \frac{42i^2}{128} \sin 3x = \frac{42(-1)}{128} \sin 3x = -\frac{21}{64} \sin 3x $.

Fourth term: $ \frac{i}{128} \times 70i \sin x = \frac{70i^2}{128} \sin x = \frac{70(-1)}{128} \sin x = -\frac{35}{64} \sin x $.

Therefore, combining all these terms:

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

But according to the problem statement, this should be equal to $ a \sin 7x + b \sin 5x + c \sin 3x + d \sin x $. Comparing 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} $.

Wait, but let me check if I did the calculations correctly. Let me verify each step again.

First, when I expanded the binomial, I think that was correct. Then, pairing the terms, for each pair, I calculated the sum. Let me check the coefficients again:

First pair: $ e^{i7x} - e^{-i7x} = 2i \sin 7x $. Then multiplied by $ \frac{i}{128} $ gives $ \frac{i}{128} \times 2i \sin 7x = \frac{2i^2}{128} \sin 7x = -2/128 \sin 7x = -1/64 \sin 7x $. Correct.

Second pair: $ -7 e^{i5x} + 7 e^{-i5x} = -14i \sin 5x $. Then multiplied by $ \frac{i}{128} $: $ \frac{i}{128} \times (-14i) \sin 5x = \frac{-14i^2}{128} \sin 5x = \frac{14}{128} \sin 5x = 7/64 \sin 5x $. Correct.

Third pair: $ 21 e^{i3x} - 21 e^{-i3x} = 42i \sin 3x $. Multiply by $ \frac{i}{128} $: $ \frac{i}{128} \times 42i \sin 3x = \frac{42i^2}{128} \sin 3x = -42/128 \sin 3x = -21/64 \sin 3x $. Correct.

Fourth pair: $ -35 e^{ix} + 35 e^{-ix} = 70i \sin x $. Multiply by $ \frac{i}{128} $: $ \frac{i}{128} \times 70i \sin x = \frac{70i^2}{128} \sin x = -70/128 \sin x = -35/64 \sin x $. Correct.

Therefore, the coefficients are as above. Therefore, $ d = -35/64 $.

But let me check if there's another way to verify this. Maybe using known identities or another method?

Alternatively, I can use the formula for $ \sin^n x $ in terms of multiple angles. For example, I know that for $ \sin^7 x $, there is a formula. Alternatively, I can use recurrence relations or Chebyshev polynomials?

Alternatively, I can use the identity for $ \sin^7 x $ by writing it as $ \sin x \cdot \sin^6 x $, and then using the identity for $ \sin^6 x $, which can be expressed in terms of lower powers. But that might take longer. Let me try.

First, recall that $ \sin^2 x = \frac{1 - \cos 2x}{2} $.

Then $ \sin^4 x = (\sin^2 x)^2 = \left( \frac{1 - \cos 2x}{2} \right)^2 = \frac{1 - 2\cos 2x + \cos^2 2x}{4} $.

Then $ \cos^2 2x = \frac{1 + \cos 4x}{2} $, so:

$ \sin^4 x = \frac{1 - 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} = \frac{2 - 4\cos 2x + 1 + \cos 4x}{8} = \frac{3 - 4\cos 2x + \cos 4x}{8} $.

Then $ \sin^6 x = (\sin^2 x)^3 = \left( \frac{1 - \cos 2x}{2} \right)^3 = \frac{1 - 3\cos 2x + 3\cos^2 2x - \cos^3 2x}{8} $.

Now, $ \cos^2 2x = \frac{1 + \cos 4x}{2} $, and $ \cos^3 2x $ can be expressed using the identity $ \cos^3 \theta = \frac{3\cos \theta + \cos 3\theta}{4} $.

Therefore:

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

Therefore, substituting back into $ \sin^6 x $:

$ \sin^6 x = \frac{1 - 3\cos 2x + 3 \cdot \frac{1 + \cos 4x}{2} - \frac{3\cos 2x + \cos 6x}{4}}{8} $.

Let me compute numerator step by step:

First term: 1

Second term: -3 cos 2x

Third term: 3*(1 + cos 4x)/2 = 3/2 + (3/2) cos 4x

Fourth term: - [3 cos 2x + cos 6x]/4 = -3/4 cos 2x - 1/4 cos 6x

So combining all terms:

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

Combine constants: 1 + 3/2 = 5/2.

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

Then, (3/2) cos 4x and -1/4 cos 6x.

Therefore, numerator is:

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

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

Simplify:

Multiply numerator and denominator:

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

Therefore, $ \sin^6 x = \frac{5}{16} - \frac{15}{32} \cos 2x + \frac{3}{16} \cos 4x - \frac{1}{32} \cos 6x $.

Now, $ \sin^7 x = \sin x \cdot \sin^6 x $. Therefore, multiply this expression by $ \sin x $:

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

Now, distribute $ \sin x $:

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

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

Let me apply this to each term:

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

Second term: $ -\frac{15}{32} \sin x \cos 2x = -\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} \sin x \cos 4x = \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} \sin x \cos 6x = -\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, let me write all terms together:

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 7x $: only the fourth term has it: $ -\frac{1}{64} \sin 7x $.

For $ \sin 5x $: third term $ \frac{3}{32} \sin 5x $ and fourth term $ \frac{1}{64} \sin 5x $. Let me convert to 64 denominator:

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

For $ \sin 3x $: second term $ -\frac{15}{64} \sin 3x $ and third term $ -\frac{3}{32} \sin 3x $. Convert $ -\frac{3}{32} = -\frac{6}{64} $, so total $ -\frac{15}{64} - \frac{6}{64} = -\frac{21}{64} \sin 3x $.

For $ \sin x $: first term $ \frac{5}{16} \sin x $ and second term $ \frac{15}{64} \sin x $. Convert $ \frac{5}{16} = \frac{20}{64} $, so total $ \frac{20}{64} + \frac{15}{64} = \frac{35}{64} \sin x $.

Therefore, combining all:

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

Wait, but earlier with the complex exponential method, I had $ -\frac{35}{64} \sin x $, but here it's $ +\frac{35}{64} \sin x $. There's a discrepancy here. Which one is correct?

Wait, let me check the fourth term again. The fourth term was:

Fourth term: $ -\frac{1}{64} [\sin 7x - \sin 5x] $, which is $ -\frac{1}{64} \sin 7x + \frac{1}{64} \sin 5x $. That's correct.

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

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

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

Then combining $ \sin x $ terms: $ \frac{5}{16} + \frac{15}{64} = \frac{20 + 15}{64} = \frac{35}{64} $. So positive 35/64. But in the complex exponential method, I had negative 35/64. Which is conflicting. Therefore, there must be a mistake in one of the methods.

Wait, let me check the complex exponential approach again. Let me go back.

Earlier, I had:

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

But in the second method, I have:

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

So the difference is in the sign of the $ \sin x $ term. Which one is correct?

Let me test with a specific value of $ x $. Let me take $ x = 0 $. Then $ \sin 0 = 0 $, so $ (\sin 0)^7 = 0 $. Let me compute both expressions.

First, complex exponential method:

$ -\frac{1}{64} \sin 0 + \frac{7}{64} \sin 0 - \frac{21}{64} \sin 0 - \frac{35}{64} \sin 0 = 0 $. Correct.

Second method:

$ -\frac{1}{64} \sin 0 + \frac{7}{64} \sin 0 - \frac{21}{64} \sin 0 + \frac{35}{64} \sin 0 = 0 $. Also correct. So both expressions give 0. Not helpful.

Let me try $ x = \pi/2 $. Then $ \sin(\pi/2) = 1 $, so $ (\sin x)^7 = 1 $.

Compute using complex exponential method:

$ -\frac{1}{64} \sin(7\pi/2) + \frac{7}{64} \sin(5\pi/2) - \frac{21}{64} \sin(3\pi/2) - \frac{35}{64} \sin(\pi/2) $.

Compute each term:

- $ \sin(7\pi/2) = \sin(3\pi + \pi/2) = -\sin(\pi/2) = -1 $
- $ \sin(5\pi/2) = \sin(2\pi + \pi/2) = \sin(\pi/2) = 1 $
- $ \sin(3\pi/2) = -1 $
- $ \sin(\pi/2) = 1 $

Therefore:

$ -\frac{1}{64}(-1) + \frac{7}{64}(1) - \frac{21}{64}(-1) - \frac{35}{64}(1) $

= $ \frac{1}{64} + \frac{7}{64} + \frac{21}{64} - \frac{35}{64} $

= $ (1 + 7 + 21 - 35)/64 = (-6)/64 = -3/32 $. But the actual value is 1. So this is wrong. Therefore, the complex exponential method must have an error.

Wait, that's a problem. So according to this, the complex exponential method gives -3/32 instead of 1. Therefore, there must be a mistake in that approach. Let me check.

Wait, let me check the complex exponential approach again. Let me go back.

Original expression:

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

Wait, let me recompute that multiplication:

Multiply $ \frac{i}{128} $ by each term:

First term: $ \frac{i}{128} \times 2i \sin 7x = \frac{2i^2}{128} \sin 7x = \frac{-2}{128} \sin 7x = -\frac{1}{64} \sin 7x $. Correct.

Second term: $ \frac{i}{128} \times (-14i) \sin 5x = \frac{-14i^2}{128} \sin 5x = \frac{14}{128} \sin 5x = \frac{7}{64} \sin 5x $. Correct.

Third term: $ \frac{i}{128} \times 42i \sin 3x = \frac{42i^2}{128} \sin 3x = \frac{-42}{128} \sin 3x = -\frac{21}{64} \sin 3x $. Correct.

Fourth term: $ \frac{i}{128} \times 70i \sin x = \frac{70i^2}{128} \sin x = \frac{-70}{128} \sin x = -\frac{35}{64} \sin x $. Correct.

But when I plug in $ x = \pi/2 $, I get -3/32 instead of 1. Therefore, there must be an error in the complex exponential approach. Let me check the earlier steps.

Wait, when I expanded $ (\sin x)^7 $, I had:

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

But let me check the initial expansion. The original expression was:

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

Which is equal to $ \frac{1}{(2i)^7} (e^{ix} - e^{-ix})^7 $.

But $ (2i)^7 = 128 i^7 = 128 (-i) $, so $ \frac{1}{(2i)^7} = \frac{1}{-128i} = \frac{i}{128} $. Correct.

Then expanding $ (e^{ix} - e^{-ix})^7 $, which is $ \sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} $. Correct.

Then pairing terms and getting the sum as $ 2i \sin 7x - 14i \sin 5x + 42i \sin 3x + 70i \sin x $. Then multiplying by $ \frac{i}{128} $ gives the expression. But when I plug in $ x = \pi/2 $, it doesn't match. Therefore, there must be an error in the pairing step.

Wait, let me check the pairing again. Let me recompute the sum:

Original sum after expansion:

For k from 0 to 7:

k=0: binom(7,0)*(-1)^0*e^{i7x} = 1*1*e^{i7x}

k=1: binom(7,1)*(-1)^1*e^{i5x} = 7*(-1)*e^{i5x}

k=2: binom(7,2)*(-1)^2*e^{i3x} = 21*1*e^{i3x}

k=3: binom(7,3)*(-1)^3*e^{ix} = 35*(-1)*e^{ix}

k=4: binom(7,4)*(-1)^4*e^{-ix} = 35*1*e^{-ix}

k=5: binom(7,5)*(-1)^5*e^{-i3x} = 21*(-1)*e^{-i3x}

k=6: binom(7,6)*(-1)^6*e^{-i5x} = 7*1*e^{-i5x}

k=7: binom(7,7)*(-1)^7*e^{-i7x} = 1*(-1)*e^{-i7x}

Therefore, the entire sum is:

e^{i7x} -7e^{i5x} +21e^{i3x} -35e^{ix} +35e^{-ix} -21e^{-i3x} +7e^{-i5x} -e^{-i7x}

Now, grouping terms:

(e^{i7x} - e^{-i7x}) + (-7e^{i5x} +7e^{-i5x}) + (21e^{i3x} -21e^{-i3x}) + (-35e^{ix} +35e^{-ix})

Which is:

2i sin7x + (-14i sin5x) + 42i sin3x + 70i sinx.

Wait, that's what I had before. Then multiplying by i/128:

i/128*(2i sin7x -14i sin5x +42i sin3x +70i sinx) = i/128*(2i sin7x) + ...

But let me compute this again:

First term: 2i sin7x * i/128 = 2i^2 sin7x /128 = -2 sin7x /128 = -sin7x /64

Second term: -14i sin5x * i/128 = -14i^2 sin5x /128 = 14 sin5x /128 = 7 sin5x /64

Third term: 42i sin3x * i/128 = 42i^2 sin3x /128 = -42 sin3x /128 = -21 sin3x /64

Fourth term: 70i sinx * i/128 = 70i^2 sinx /128 = -70 sinx /128 = -35 sinx /64

So the expression is:

- (1/64) sin7x + (7/64) sin5x - (21/64) sin3x - (35/64) sinx.

But when I plug in x = π/2, this gives:

- (1/64) sin(7π/2) + (7/64) sin(5π/2) - (21/64) sin(3π/2) - (35/64) 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

So substituting:

- (1/64)(-1) + (7/64)(1) - (21/64)(-1) - (35/64)(1)

= 1/64 + 7/64 + 21/64 - 35/64

= (1 + 7 + 21 - 35)/64

= (-6)/64 = -3/32

But the actual value of (sin(π/2))^7 is 1. So clearly, there is a mistake here. Therefore, my complex exponential approach must have an error.

But where?

Wait, let me check the original expansion. Let me compute (sinx)^7 at x = π/2. It's 1. Let me compute the expression from the second method:

From the second method, we had:

sin^7x = -1/64 sin7x + 7/64 sin5x -21/64 sin3x + 35/64 sinx.

At x = π/2:

sin7x = sin(7π/2) = -1

sin5x = sin(5π/2) = 1

sin3x = sin(3π/2) = -1

sinx = sin(π/2) = 1

Therefore:

-1/64*(-1) + 7/64*(1) -21/64*(-1) + 35/64*(1)

= 1/64 + 7/64 + 21/64 + 35/64

= (1 + 7 + 21 + 35)/64 = 64/64 = 1. Correct!

So the second method gives the correct result, while the first method gives -3/32. Therefore, there must be a mistake in the first method.

Wait, so why is there a discrepancy? Let me check the first method again.

Wait, in the first method, I had:

(\sin x)^7 = \frac{i}{128} [2i \sin 7x - 14i \sin 5x + 42i \sin 3x + 70i \sin x]

But when I compute this expression, it's:

\frac{i}{128} * [2i sin7x -14i sin5x +42i sin3x +70i sinx]

= \frac{i}{128} * [i(2 sin7x -14 sin5x +42 sin3x +70 sinx)]

= \frac{i^2}{128} [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= \frac{-1}{128} [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= -\frac{2}{128} sin7x + \frac{14}{128} sin5x - \frac{42}{128} sin3x - \frac{70}{128} sinx

Simplify:

= -\frac{1}{64} sin7x + \frac{7}{64} sin5x - \frac{21}{64} sin3x - \frac{35}{64} sinx

But according to the second method, it should be:

- \frac{1}{64} sin7x + \frac{7}{64} sin5x - \frac{21}{64} sin3x + \frac{35}{64} sinx

So the difference is the sign of the last term. Therefore, there must be a mistake in the first method's calculation of the sum.

Wait, let me check the pairing again. When I paired the terms, I had:

For k=0 and k=7: e^{i7x} - e^{-i7x} = 2i sin7x. Correct.

For k=1 and k=6: -7e^{i5x} +7e^{-i5x} = -14i sin5x. Correct.

For k=2 and k=5: 21e^{i3x} -21e^{-i3x} = 42i sin3x. Correct.

For k=3 and k=4: -35e^{ix} +35e^{-ix} = 70i sinx. Correct.

Therefore, the sum is 2i sin7x -14i sin5x +42i sin3x +70i sinx.

But then multiplying by i/128 gives:

i/128*(2i sin7x -14i sin5x +42i sin3x +70i sinx)

= i/128 * [i(2 sin7x -14 sin5x +42 sin3x +70 sinx)]

= i^2 / 128 * [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= -1/128 * [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= -2/128 sin7x +14/128 sin5x -42/128 sin3x -70/128 sinx

= -1/64 sin7x +7/64 sin5x -21/64 sin3x -35/64 sinx

But according to the second method, it should be +35/64 sinx. So why is there a discrepancy?

Wait, but in the second method, the expression was:

sin^7x = -1/64 sin7x +7/64 sin5x -21/64 sin3x +35/64 sinx

So the difference is the sign of the last term. Therefore, there must be an error in the first method's calculation.

Wait, let me check the original expansion again. The original expansion was:

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

But (2i)^7 = 128 i^7 = 128*(-i). Therefore, 1/(2i)^7 = 1/( -128i ) = i/128. Correct.

But then the sum is:

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

But when I expanded this sum, I got:

e^{i7x} -7e^{i5x} +21e^{i3x} -35e^{ix} +35e^{-ix} -21e^{-i3x} +7e^{-i5x} -e^{-i7x}

But when I group terms:

(e^{i7x} - e^{-i7x}) + (-7e^{i5x} +7e^{-i5x}) + (21e^{i3x} -21e^{-i3x}) + (-35e^{ix} +35e^{-ix})

Which is:

2i sin7x -14i sin5x +42i sin3x +70i sinx

But if I instead consider that the original expansion is:

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

But when I expand this, the terms are:

For k=0: binom(7,0)*(-1)^0*e^{i7x} = 1*1*e^{i7x}

k=1: binom(7,1)*(-1)^1*e^{i(7 - 2)x} = 7*(-1)*e^{i5x}

k=2: binom(7,2)*(-1)^2*e^{i(7 - 4)x} = 21*1*e^{i3x}

k=3: binom(7,3)*(-1)^3*e^{i(7 - 6)x} = 35*(-1)*e^{i1x}

k=4: binom(7,4)*(-1)^4*e^{i(7 - 8)x} = 35*1*e^{-i1x}

k=5: binom(7,5)*(-1)^5*e^{i(7 - 10)x} = 21*(-1)*e^{-i3x}

k=6: binom(7,6)*(-1)^6*e^{i(7 - 12)x} = 7*1*e^{-i5x}

k=7: binom(7,7)*(-1)^7*e^{i(7 - 14)x} = 1*(-1)*e^{-i7x}

So the sum is:

e^{i7x} -7e^{i5x} +21e^{i3x} -35e^{ix} +35e^{-ix} -21e^{-i3x} +7e^{-i5x} -e^{-i7x}

Now, grouping terms:

(e^{i7x} - e^{-i7x}) + (-7e^{i5x} +7e^{-i5x}) + (21e^{i3x} -21e^{-i3x}) + (-35e^{ix} +35e^{-ix})

Which is:

2i sin7x -14i sin5x +42i sin3x +70i sinx

But when I multiply by i/128, I get:

i/128*(2i sin7x -14i sin5x +42i sin3x +70i sinx) = -1/64 sin7x +7/64 sin5x -21/64 sin3x -35/64 sinx

But according to the second method, it should be +35/64 sinx. So why is there a discrepancy?

Wait, but in the second method, we had:

sin^7x = -1/64 sin7x +7/64 sin5x -21/64 sin3x +35/64 sinx

So the difference is the sign of the last term. Let me check the second method again.

In the second method, when I calculated the fourth term, I had:

Fourth term: -1/32 sinx cos6x = -1/32 * [sin(x + 6x) + sin(x - 6x)] / 2 = -1/64 [sin7x + sin(-5x)] = -1/64 [sin7x - sin5x].

Wait, but in the expansion, this term is -1/64 sin7x + 1/64 sin5x.

Then, when combining all terms, the sinx terms were:

First term: 5/16 sinx

Second term: +15/64 sinx

Total: 5/16 + 15/64 = 20/64 + 15/64 = 35/64 sinx.

But in the complex exponential method, it's -35/64 sinx.

So why is there a sign difference?

Wait, let me check the fourth term in the second method again.

Fourth term: -1/32 sinx cos6x.

Using identity: sinA cosB = [sin(A+B) + sin(A-B)] / 2.

So sinx cos6x = [sin(x + 6x) + sin(x - 6x)] / 2 = [sin7x + sin(-5x)] / 2 = [sin7x - sin5x]/2.

Therefore, -1/32 sinx cos6x = -1/32 * [sin7x - sin5x]/2 = -1/64 sin7x + 1/64 sin5x.

Yes, that's correct.

Then, when combining all terms, the sinx terms are from first term (5/16 sinx) and second term (15/64 sinx).

But where does the 15/64 sinx come from?

Second term was: -15/32 sinx cos2x = -15/32 * [sin3x - sinx]/2 = -15/64 sin3x + 15/64 sinx.

Yes, so that gives +15/64 sinx.

Therefore, total sinx terms: 5/16 sinx + 15/64 sinx = 35/64 sinx.

Therefore, the second method gives +35/64 sinx.

But the complex exponential method gives -35/64 sinx.

So there must be a sign error in the complex exponential method.

Wait, let me check the original expansion again.

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

But (2i)^7 = 128 i^7 = 128 * (-i). Therefore, 1/(2i)^7 = 1/( -128i ) = i/128.

But let me check this calculation again.

i^7 = i^(4 + 3) = i^3 = -i.

Yes, so (2i)^7 = 2^7 * i^7 = 128 * (-i).

Therefore, 1/(2i)^7 = 1/( -128i ) = multiply numerator and denominator by i: i / (-128i^2) = i / (-128 * -1) = i / 128.

Yes, correct.

Then, the sum is 2i sin7x -14i sin5x +42i sin3x +70i sinx.

Multiply by i/128:

i/128 * [2i sin7x -14i sin5x +42i sin3x +70i sinx]

= i/128 * i [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= i^2 / 128 [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= -1/128 [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= -2/128 sin7x +14/128 sin5x -42/128 sin3x -70/128 sinx

= -1/64 sin7x +7/64 sin5x -21/64 sin3x -35/64 sinx

But according to the second method, it should be +35/64 sinx.

Wait, but if I take the negative of this expression, I get the second method's result.

Wait, what if there was a sign error in the initial expansion? Let me check the original expansion of (\sin x)^7.

(\sin x)^7 = [ (e^{ix} - e^{-ix}) / (2i) ]^7

= (e^{ix} - e^{-ix})^7 / (2i)^7

= [sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} ] / (2i)^7

Yes, that's correct.

But if I instead had [ (e^{-ix} - e^{ix}) / (2i) ]^7, which is [ - (e^{ix} - e^{-ix}) / (2i) ]^7 = (-1)^7 [ (e^{ix} - e^{-ix}) / (2i) ]^7 = - (\sin x)^7.

But no, the original expression is correct.

Wait, but maybe there is a mistake in the sign when expanding (e^{ix} - e^{-ix})^7? Let me check.

The binomial expansion is (a + b)^n = sum_{k=0}^n \binom{n}{k} a^{n - k} b^k.

Here, a = e^{ix}, b = -e^{-ix}.

So (e^{ix} - e^{-ix})^7 = sum_{k=0}^7 \binom{7}{k} (e^{ix})^{7 - k} (-e^{-ix})^k

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

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

Yes, correct.

So the expansion is correct.

But then why is there a discrepancy?

Wait, let me check with x = π/2.

From the complex exponential method, we have:

-1/64 sin7x +7/64 sin5x -21/64 sin3x -35/64 sinx

At x = π/2, this is -1/64*(-1) +7/64*(1) -21/64*(-1) -35/64*(1) = 1/64 +7/64 +21/64 -35/64 = (1 +7 +21 -35)/64 = -6/64 = -3/32.

But the correct value is 1.

From the second method, it's -1/64 sin7x +7/64 sin5x -21/64 sin3x +35/64 sinx, which at x = π/2 is 1/64 +7/64 +21/64 +35/64 = 64/64 = 1.

So the second method is correct, and the first method is wrong.

But why? Where is the error in the first method?

Wait, let me check the initial step.

(\sin x)^7 = [ (e^{ix} - e^{-ix}) / (2i) ]^7

= (e^{ix} - e^{-ix})^7 / (2i)^7

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

But if I instead write this as [ - (e^{-ix} - e^{ix}) ]^7 = (-1)^7 (e^{-ix} - e^{ix})^7 = - (e^{-ix} - e^{ix})^7

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

Wait, this seems complicated.

Alternatively, perhaps there is a mistake in the sign when converting the exponential terms to sine.

Wait, let me recall that e^{iθ} - e^{-iθ} = 2i sinθ.

But in the complex exponential method, we have:

The sum is 2i sin7x -14i sin5x +42i sin3x +70i sinx.

But if I multiply this by i/128, I get:

i/128 * [2i sin7x -14i sin5x +42i sin3x +70i sinx]

= i/128 * i [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= i^2 / 128 [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= -1/128 [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= -2/128 sin7x +14/128 sin5x -42/128 sin3x -70/128 sinx

= -1/64 sin7x +7/64 sin5x -21/64 sin3x -35/64 sinx

But according to the second method, it should be +35/64 sinx.

Wait, but if I instead had multiplied by -i/128 instead of i/128, would that fix it? Let me check.

If I had a sign error in the initial step, like if the original expression was [ (e^{ix} - e^{-ix}) / (2i) ]^7 = [ - (e^{-ix} - e^{ix}) / (2i) ]^7 = (-1)^7 [ (e^{-ix} - e^{ix}) / (2i) ]^7 = - [ (e^{-ix} - e^{ix}) / (2i) ]^7.

But (e^{-ix} - e^{ix}) / (2i) = [ - (e^{ix} - e^{-ix}) ] / (2i) = - [ (e^{ix} - e^{-ix}) / (2i) ] = -sinx.

So [ (e^{-ix} - e^{ix}) / (2i) ]^7 = (-sinx)^7 = -sin^7x.

Therefore, the original expression is - [ (e^{-ix} - e^{ix}) / (2i) ]^7 = -(-sin^7x) = sin^7x.

Wait, this seems to suggest that the original expansion is correct.

But why is there a discrepancy?

Wait, let me try to compute the complex exponential expression at x = π/2.

If I use the second method's result, which is correct, and compare it to the complex exponential method's result, which is wrong, there must be a miscalculation in the complex exponential method.

Wait, let me try to compute the complex exponential expression again.

Let me compute the complex exponential expression:

-1/64 sin7x +7/64 sin5x -21/64 sin3x -35/64 sinx

At x = π/2, this is:

-1/64 * sin(7π/2) +7/64 * sin(5π/2) -21/64 * sin(3π/2) -35/64 * sin(π/2)

= -1/64 * (-1) +7/64 * 1 -21/64 * (-1) -35/64 * 1

= 1/64 +7/64 +21/64 -35/64

= (1 +7 +21 -35)/64 = -6/64 = -3/32

But the correct value is 1.

However, if I take the negative of this expression, I get 3/32, which is still not 1.

Wait, but if I had the expression as -1/64 sin7x +7/64 sin5x -21/64 sin3x +35/64 sinx, then at x = π/2, it would be 1/64 +7/64 +21/64 +35/64 = 64/64 = 1.

So the error is in the sign of the last term in the complex exponential method.

But why? Where is the mistake?

Wait, let me check the original expansion again.

The sum was:

e^{i7x} -7e^{i5x} +21e^{i3x} -35e^{ix} +35e^{-ix} -21e^{-i3x} +7e^{-i5x} -e^{-i7x}

When I grouped terms, I said this is:

(e^{i7x} - e^{-i7x}) + (-7e^{i5x} +7e^{-i5x}) + (21e^{i3x} -21e^{-i3x}) + (-35e^{ix} +35e^{-ix})

But let me check the last term: -35e^{ix} +35e^{-ix} = 35(e^{-ix} - e^{ix}) = -35(e^{ix} - e^{-ix}) = -35 * 2i sinx = -70i sinx.

Wait, but earlier I said this is 70i sinx.

Wait, this is the mistake!

Wait, -35e^{ix} +35e^{-ix} = 35(e^{-ix} - e^{ix}) = -35(e^{ix} - e^{-ix}) = -35 * 2i sinx = -70i sinx.

But earlier, I said this is 70i sinx.

Ah! Here is the error!

Previously, I said that -35e^{ix} +35e^{-ix} = 70i sinx, but it should be -70i sinx.

Let me recheck:

-35e^{ix} +35e^{-ix} = 35(e^{-ix} - e^{ix}) = 35 * (- (e^{ix} - e^{-ix})) = -35 * 2i sinx = -70i sinx.

Yes! So previously, I had said this equals 70i sinx, but it should be -70i sinx.

Therefore, the sum is:

(e^{i7x} - e^{-i7x}) + (-7e^{i5x} +7e^{-i5x}) + (21e^{i3x} -21e^{-i3x}) + (-35e^{ix} +35e^{-ix})

= 2i sin7x -14i sin5x +42i sin3x -70i sinx

Therefore, the sum is 2i sin7x -14i sin5x +42i sin3x -70i sinx.

Previously, I had +70i sinx, but it should be -70i sinx.

Therefore, the correct sum is 2i sin7x -14i sin5x +42i sin3x -70i sinx.

Then, multiplying by i/128:

i/128 * [2i sin7x -14i sin5x +42i sin3x -70i sinx]

= i/128 * i [2 sin7x -14 sin5x +42 sin3x -70 sinx]

= i^2 / 128 [2 sin7x -14 sin5x +42 sin3x -70 sinx]

= -1/128 [2 sin7x -14 sin5x +42 sin3x -70 sinx]

= -2/128 sin7x +14/128 sin5x -42/128 sin3x +70/128 sinx

= -1/64 sin7x +7/64 sin5x -21/64 sin3x +35/64 sinx

Which matches the second method!

So the error was in the sign of the last term in the pairing step. I had incorrectly said that -35e^{ix} +35e^{-ix} = 70i sinx, but it should be -70i sinx.

Therefore, the correct expression is:

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

Therefore, the coefficient d is 35/64.

But wait, in the second method, we had the same result. So the correct value of d is 35/64.

But initially, I thought it was -35/64, but that was due to a sign error in the complex exponential method.

Therefore, the correct value of d is 35/64.

But let me confirm with x = π/2.

Using the corrected complex exponential expression:

-1/64 sin7x +7/64 sin5x -21/64 sin3x +35/64 sinx

At x = π/2:

-1/64*(-1) +7/64*(1) -21/64*(-1) +35/64*(1) = 1/64 +7/64 +21/64 +35/64 = (1 +7 +21 +35)/64 = 64/64 = 1. Correct.

Therefore, the correct value of d is 35/64.

But in the initial complex exponential calculation, I had mistakenly taken the last term as +70i sinx instead of -70i sinx, leading to the wrong sign.

Therefore, the correct answer is d = 35/64.

But let me check with the second method's result. In the second method, we had:

sin^7x = -1/64 sin7x +7/64 sin5x -21/64 sin3x +35/64 sinx

So yes, d = 35/64.

Therefore, the correct value of d is 35/64.

But the problem says "Find d." So the answer is 35/64.

But let me check if this matches with any known formulas.

Alternatively, I can recall that the general formula for sin^n x can be expressed as a combination of multiple angles. For example, for odd powers, it's a combination of sines of odd multiples.

Alternatively, I can use the formula from the second method, which is reliable, and it gives d = 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 find the constant $ d $.

---

### Step 1: Use Complex Exponentials

We begin by expressing $ \sin x $ using Euler's formula:

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

Then,

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

Now, compute $ (2i)^7 $:

$$
(2i)^7 = 2^7 \cdot i^7 = 128 \cdot i^7 = 128 \cdot (-i) = -128i
$$

So,

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

Multiply numerator and denominator by $ i $ to simplify:

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

---

### Step 2: Group Terms and Convert to Sine

We now evaluate the sum:

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

This sum can be grouped into pairs of terms that are complex conjugates, and each pair can be expressed as a sine function. After careful pairing and simplification, we find:

$$
\sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} = 2i \sin 7x - 14i \sin 5x + 42i \sin 3x - 70i \sin x
$$

Now multiply by $ \frac{i}{128} $:

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

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

$$
= \frac{-1}{128} \left( 2 \sin 7x - 14 \sin 5x + 42 \sin 3x - 70 \sin x \right)
$$

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

---

### Step 3: Match with Given Form

We are given:

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

Comparing 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. Maybe I can use some kind of power-reduction formulas or multiple-angle identities to expand $ (\sin x)^7 $ into a combination of sine functions with different arguments.

I know that for lower powers, like $ \sin^2 x $, we can use the identity $ \sin^2 x = \frac{1 - \cos 2x}{2} $. Similarly, higher powers can be expressed using more complex identities. However, for the seventh power, this might get complicated. Maybe there's a recursive way or a formula for expressing $ \sin^n x $ as a sum of multiple angles?

Alternatively, I recall that complex exponentials can be useful for such expansions. Euler's formula says that $ \sin x = \frac{e^{ix} - e^{-ix}}{2i} $. So, perhaps I can express $ (\sin x)^7 $ in terms of exponentials and then convert back to sine terms?

Let me try that approach. Let me write $ \sin x $ as $ \frac{e^{ix} - e^{-ix}}{2i} $. Then $ (\sin x)^7 $ would be $ \left( \frac{e^{ix} - e^{-ix}}{2i} \right)^7 $. Expanding this using the binomial theorem might give me terms involving $ e^{i k x} $ for various integers $ k $, which I can then convert back to sine functions.

Let me write that out:

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

Wait, let me check that. The binomial expansion of $ (a + b)^n $ is $ \sum_{k=0}^n \binom{n}{k} a^{n - k} b^k $. In this case, $ a = e^{ix} $, $ b = -e^{-ix} $, and $ n = 7 $. So:

$ (e^{ix} - e^{-ix})^7 = \sum_{k=0}^7 \binom{7}{k} (e^{ix})^{7 - k} (-e^{-ix})^k $.

Simplify the exponents:

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

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

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

Simplify the constants: $ \frac{1}{-128i} = \frac{i}{128} $, since multiplying numerator and denominator by $ i $ gives $ \frac{i}{-128i^2} = \frac{i}{-128(-1)} = \frac{i}{128} $.

So, $ (\sin x)^7 = \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 must cancel out, leaving only sine terms.

Alternatively, perhaps I can pair terms to get sine functions. Let me consider that each term $ e^{i(7 - 2k)x} $ can be written as $ \cos((7 - 2k)x) + i \sin((7 - 2k)x) $. Therefore, when we take the sum over $ k $, and multiply by $ \frac{i}{128} $, the result should be purely imaginary, since the original expression is real. Therefore, the imaginary parts will combine to give sine terms.

Alternatively, maybe I can write the entire expression as a sum of sine terms. Let me think.

Alternatively, since $ e^{i\theta} + e^{-i\theta} = 2\cos \theta $, and $ e^{i\theta} - e^{-i\theta} = 2i \sin \theta $. However, in our case, the sum is over different exponents. Let me try to compute the sum.

Let me write out the sum:

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

Let me make a substitution. Let me let $ m = 7 - 2k $. Then, when $ k = 0 $, $ m = 7 $; when $ k = 1 $, $ m = 5 $; $ k = 2 $, $ m = 3 $; $ k = 3 $, $ m = 1 $; $ k = 4 $, $ m = -1 $; $ k = 5 $, $ m = -3 $; $ k = 6 $, $ m = -5 $; $ k = 7 $, $ m = -7 $.

So, the exponents go from 7 down to -7 in steps of -2. Therefore, the sum can be rewritten as:

$ \sum_{m = -7, -5, -3, -1, 1, 3, 5, 7} \binom{7}{(7 - m)/2} (-1)^{(7 - m)/2} e^{i m x} $.

Wait, maybe that's more complicated. Alternatively, note that the terms for $ k $ and $ 7 - k $ might be related. Let me check for symmetry.

Let me consider that when $ k $ ranges from 0 to 7, $ 7 - 2k $ ranges from 7 to -7. Let me note that for each term with exponent $ m $, there is a corresponding term with exponent $ -m $, except when $ m = 0 $. However, since 7 is odd, $ 7 - 2k = 0 $ would require $ 2k = 7 $, which is not possible for integer $ k $. Therefore, all exponents are non-zero and come in pairs $ m $ and $ -m $.

Therefore, perhaps I can pair terms. Let me consider that:

For each $ k $, there is a term with exponent $ m = 7 - 2k $, and another term with exponent $ -m $ when $ k' $ satisfies $ 7 - 2k' = -m $. Let me check for $ k' $:

$ 7 - 2k' = -m \Rightarrow 2k' = 7 + m \Rightarrow k' = (7 + m)/2 $.

But since $ m = 7 - 2k $, substituting:

$ k' = (7 + 7 - 2k)/2 = (14 - 2k)/2 = 7 - k $.

Therefore, each term with index $ k $ has a corresponding term with index $ 7 - k $. Therefore, the sum can be written as:

$ \sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} = \sum_{k=0}^3 \left[ \binom{7}{k} (-1)^k e^{i(7 - 2k)x} + \binom{7}{7 - k} (-1)^{7 - k} e^{i(7 - 2(7 - k))x} \right] $.

Wait, since when $ k $ goes from 0 to 7, pairing $ k $ and $ 7 - k $, but since 7 is odd, the middle term would be when $ k = 3.5 $, which isn't an integer. Therefore, the sum is symmetric around $ k = 3.5 $, so pairing terms from $ k = 0 $ and $ k = 7 $, $ k = 1 $ and $ k = 6 $, $ k = 2 $ and $ k = 5 $, $ k = 3 $ and $ k = 4 $.

Let me compute each pair:

First pair: $ k = 0 $ and $ k = 7 $:

Term for $ k = 0 $: $ \binom{7}{0} (-1)^0 e^{i(7 - 0)x} = 1 \cdot 1 \cdot e^{i7x} $.

Term for $ k = 7 $: $ \binom{7}{7} (-1)^7 e^{i(7 - 14)x} = 1 \cdot (-1)^7 e^{-i7x} = -e^{-i7x} $.

Adding them together: $ e^{i7x} - e^{-i7x} = 2i \sin 7x $.

Second pair: $ k = 1 $ and $ k = 6 $:

Term for $ k = 1 $: $ \binom{7}{1} (-1)^1 e^{i(7 - 2)x} = 7 \cdot (-1) \cdot e^{i5x} = -7 e^{i5x} $.

Term for $ k = 6 $: $ \binom{7}{6} (-1)^6 e^{i(7 - 12)x} = 7 \cdot 1 \cdot e^{-i5x} = 7 e^{-i5x} $.

Adding them together: $ -7 e^{i5x} + 7 e^{-i5x} = 7(-e^{i5x} + e^{-i5x}) = -14i \sin 5x $.

Third pair: $ k = 2 $ and $ k = 5 $:

Term for $ k = 2 $: $ \binom{7}{2} (-1)^2 e^{i(7 - 4)x} = 21 \cdot 1 \cdot e^{i3x} = 21 e^{i3x} $.

Term for $ k = 5 $: $ \binom{7}{5} (-1)^5 e^{i(7 - 10)x} = 21 \cdot (-1)^5 e^{-i3x} = -21 e^{-i3x} $.

Adding them together: $ 21 e^{i3x} - 21 e^{-i3x} = 42i \sin 3x $.

Fourth pair: $ k = 3 $ and $ k = 4 $:

Term for $ k = 3 $: $ \binom{7}{3} (-1)^3 e^{i(7 - 6)x} = 35 \cdot (-1)^3 e^{i1x} = -35 e^{ix} $.

Term for $ k = 4 $: $ \binom{7}{4} (-1)^4 e^{i(7 - 8)x} = 35 \cdot 1 \cdot e^{-ix} = 35 e^{-ix} $.

Adding them together: $ -35 e^{ix} + 35 e^{-ix} = 70i \sin x $.

Therefore, the entire sum is:

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

Therefore, going back to the original expression:

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

Let me compute this:

Multiply $ \frac{i}{128} $ by each term:

First term: $ \frac{i}{128} \times 2i \sin 7x = \frac{2i^2}{128} \sin 7x = \frac{2(-1)}{128} \sin 7x = -\frac{1}{64} \sin 7x $.

Second term: $ \frac{i}{128} \times (-14i) \sin 5x = \frac{-14i^2}{128} \sin 5x = \frac{-14(-1)}{128} \sin 5x = \frac{14}{128} \sin 5x = \frac{7}{64} \sin 5x $.

Third term: $ \frac{i}{128} \times 42i \sin 3x = \frac{42i^2}{128} \sin 3x = \frac{42(-1)}{128} \sin 3x = -\frac{21}{64} \sin 3x $.

Fourth term: $ \frac{i}{128} \times 70i \sin x = \frac{70i^2}{128} \sin x = \frac{70(-1)}{128} \sin x = -\frac{35}{64} \sin x $.

Therefore, combining all these terms:

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

But according to the problem statement, this should be equal to $ a \sin 7x + b \sin 5x + c \sin 3x + d \sin x $. Comparing 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} $.

Wait, but let me check if I did the calculations correctly. Let me verify each step again.

First, when I expanded the binomial, I think that was correct. Then, pairing the terms, for each pair, I calculated the sum. Let me check the coefficients again:

First pair: $ e^{i7x} - e^{-i7x} = 2i \sin 7x $. Then multiplied by $ \frac{i}{128} $ gives $ \frac{i}{128} \times 2i \sin 7x = \frac{2i^2}{128} \sin 7x = -2/128 \sin 7x = -1/64 \sin 7x $. Correct.

Second pair: $ -7 e^{i5x} + 7 e^{-i5x} = -14i \sin 5x $. Then multiplied by $ \frac{i}{128} $: $ \frac{i}{128} \times (-14i) \sin 5x = \frac{-14i^2}{128} \sin 5x = \frac{14}{128} \sin 5x = 7/64 \sin 5x $. Correct.

Third pair: $ 21 e^{i3x} - 21 e^{-i3x} = 42i \sin 3x $. Multiply by $ \frac{i}{128} $: $ \frac{i}{128} \times 42i \sin 3x = \frac{42i^2}{128} \sin 3x = -42/128 \sin 3x = -21/64 \sin 3x $. Correct.

Fourth pair: $ -35 e^{ix} + 35 e^{-ix} = 70i \sin x $. Multiply by $ \frac{i}{128} $: $ \frac{i}{128} \times 70i \sin x = \frac{70i^2}{128} \sin x = -70/128 \sin x = -35/64 \sin x $. Correct.

Therefore, the coefficients are as above. Therefore, $ d = -35/64 $.

But let me check if there's another way to verify this. Maybe using known identities or another method?

Alternatively, I can use the formula for $ \sin^n x $ in terms of multiple angles. For example, I know that for $ \sin^7 x $, there is a formula. Alternatively, I can use recurrence relations or Chebyshev polynomials?

Alternatively, I can use the identity for $ \sin^7 x $ by writing it as $ \sin x \cdot \sin^6 x $, and then using the identity for $ \sin^6 x $, which can be expressed in terms of lower powers. But that might take longer. Let me try.

First, recall that $ \sin^2 x = \frac{1 - \cos 2x}{2} $.

Then $ \sin^4 x = (\sin^2 x)^2 = \left( \frac{1 - \cos 2x}{2} \right)^2 = \frac{1 - 2\cos 2x + \cos^2 2x}{4} $.

Then $ \cos^2 2x = \frac{1 + \cos 4x}{2} $, so:

$ \sin^4 x = \frac{1 - 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} = \frac{2 - 4\cos 2x + 1 + \cos 4x}{8} = \frac{3 - 4\cos 2x + \cos 4x}{8} $.

Then $ \sin^6 x = (\sin^2 x)^3 = \left( \frac{1 - \cos 2x}{2} \right)^3 = \frac{1 - 3\cos 2x + 3\cos^2 2x - \cos^3 2x}{8} $.

Now, $ \cos^2 2x = \frac{1 + \cos 4x}{2} $, and $ \cos^3 2x $ can be expressed using the identity $ \cos^3 \theta = \frac{3\cos \theta + \cos 3\theta}{4} $.

Therefore:

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

Therefore, substituting back into $ \sin^6 x $:

$ \sin^6 x = \frac{1 - 3\cos 2x + 3 \cdot \frac{1 + \cos 4x}{2} - \frac{3\cos 2x + \cos 6x}{4}}{8} $.

Let me compute numerator step by step:

First term: 1

Second term: -3 cos 2x

Third term: 3*(1 + cos 4x)/2 = 3/2 + (3/2) cos 4x

Fourth term: - [3 cos 2x + cos 6x]/4 = -3/4 cos 2x - 1/4 cos 6x

So combining all terms:

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

Combine constants: 1 + 3/2 = 5/2.

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

Then, (3/2) cos 4x and -1/4 cos 6x.

Therefore, numerator is:

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

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

Simplify:

Multiply numerator and denominator:

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

Therefore, $ \sin^6 x = \frac{5}{16} - \frac{15}{32} \cos 2x + \frac{3}{16} \cos 4x - \frac{1}{32} \cos 6x $.

Now, $ \sin^7 x = \sin x \cdot \sin^6 x $. Therefore, multiply this expression by $ \sin x $:

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

Now, distribute $ \sin x $:

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

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

Let me apply this to each term:

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

Second term: $ -\frac{15}{32} \sin x \cos 2x = -\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} \sin x \cos 4x = \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} \sin x \cos 6x = -\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, let me write all terms together:

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 7x $: only the fourth term has it: $ -\frac{1}{64} \sin 7x $.

For $ \sin 5x $: third term $ \frac{3}{32} \sin 5x $ and fourth term $ \frac{1}{64} \sin 5x $. Let me convert to 64 denominator:

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

For $ \sin 3x $: second term $ -\frac{15}{64} \sin 3x $ and third term $ -\frac{3}{32} \sin 3x $. Convert $ -\frac{3}{32} = -\frac{6}{64} $, so total $ -\frac{15}{64} - \frac{6}{64} = -\frac{21}{64} \sin 3x $.

For $ \sin x $: first term $ \frac{5}{16} \sin x $ and second term $ \frac{15}{64} \sin x $. Convert $ \frac{5}{16} = \frac{20}{64} $, so total $ \frac{20}{64} + \frac{15}{64} = \frac{35}{64} \sin x $.

Therefore, combining all:

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

Wait, but earlier with the complex exponential method, I had $ -\frac{35}{64} \sin x $, but here it's $ +\frac{35}{64} \sin x $. There's a discrepancy here. Which one is correct?

Wait, let me check the fourth term again. The fourth term was:

Fourth term: $ -\frac{1}{64} [\sin 7x - \sin 5x] $, which is $ -\frac{1}{64} \sin 7x + \frac{1}{64} \sin 5x $. That's correct.

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

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

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

Then combining $ \sin x $ terms: $ \frac{5}{16} + \frac{15}{64} = \frac{20 + 15}{64} = \frac{35}{64} $. So positive 35/64. But in the complex exponential method, I had negative 35/64. Which is conflicting. Therefore, there must be a mistake in one of the methods.

Wait, let me check the complex exponential approach again. Let me go back.

Earlier, I had:

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

But in the second method, I have:

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

So the difference is in the sign of the $ \sin x $ term. Which one is correct?

Let me test with a specific value of $ x $. Let me take $ x = 0 $. Then $ \sin 0 = 0 $, so $ (\sin 0)^7 = 0 $. Let me compute both expressions.

First, complex exponential method:

$ -\frac{1}{64} \sin 0 + \frac{7}{64} \sin 0 - \frac{21}{64} \sin 0 - \frac{35}{64} \sin 0 = 0 $. Correct.

Second method:

$ -\frac{1}{64} \sin 0 + \frac{7}{64} \sin 0 - \frac{21}{64} \sin 0 + \frac{35}{64} \sin 0 = 0 $. Also correct. So both expressions give 0. Not helpful.

Let me try $ x = \pi/2 $. Then $ \sin(\pi/2) = 1 $, so $ (\sin x)^7 = 1 $.

Compute using complex exponential method:

$ -\frac{1}{64} \sin(7\pi/2) + \frac{7}{64} \sin(5\pi/2) - \frac{21}{64} \sin(3\pi/2) - \frac{35}{64} \sin(\pi/2) $.

Compute each term:

- $ \sin(7\pi/2) = \sin(3\pi + \pi/2) = -\sin(\pi/2) = -1 $
- $ \sin(5\pi/2) = \sin(2\pi + \pi/2) = \sin(\pi/2) = 1 $
- $ \sin(3\pi/2) = -1 $
- $ \sin(\pi/2) = 1 $

Therefore:

$ -\frac{1}{64}(-1) + \frac{7}{64}(1) - \frac{21}{64}(-1) - \frac{35}{64}(1) $

= $ \frac{1}{64} + \frac{7}{64} + \frac{21}{64} - \frac{35}{64} $

= $ (1 + 7 + 21 - 35)/64 = (-6)/64 = -3/32 $. But the actual value is 1. So this is wrong. Therefore, the complex exponential method must have an error.

Wait, that's a problem. So according to this, the complex exponential method gives -3/32 instead of 1. Therefore, there must be a mistake in that approach. Let me check.

Wait, let me check the complex exponential approach again. Let me go back.

Original expression:

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

Wait, let me recompute that multiplication:

Multiply $ \frac{i}{128} $ by each term:

First term: $ \frac{i}{128} \times 2i \sin 7x = \frac{2i^2}{128} \sin 7x = \frac{-2}{128} \sin 7x = -\frac{1}{64} \sin 7x $. Correct.

Second term: $ \frac{i}{128} \times (-14i) \sin 5x = \frac{-14i^2}{128} \sin 5x = \frac{14}{128} \sin 5x = \frac{7}{64} \sin 5x $. Correct.

Third term: $ \frac{i}{128} \times 42i \sin 3x = \frac{42i^2}{128} \sin 3x = \frac{-42}{128} \sin 3x = -\frac{21}{64} \sin 3x $. Correct.

Fourth term: $ \frac{i}{128} \times 70i \sin x = \frac{70i^2}{128} \sin x = \frac{-70}{128} \sin x = -\frac{35}{64} \sin x $. Correct.

But when I plug in $ x = \pi/2 $, I get -3/32 instead of 1. Therefore, there must be an error in the complex exponential approach. Let me check the earlier steps.

Wait, when I expanded $ (\sin x)^7 $, I had:

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

But let me check the initial expansion. The original expression was:

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

Which is equal to $ \frac{1}{(2i)^7} (e^{ix} - e^{-ix})^7 $.

But $ (2i)^7 = 128 i^7 = 128 (-i) $, so $ \frac{1}{(2i)^7} = \frac{1}{-128i} = \frac{i}{128} $. Correct.

Then expanding $ (e^{ix} - e^{-ix})^7 $, which is $ \sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} $. Correct.

Then pairing terms and getting the sum as $ 2i \sin 7x - 14i \sin 5x + 42i \sin 3x + 70i \sin x $. Then multiplying by $ \frac{i}{128} $ gives the expression. But when I plug in $ x = \pi/2 $, it doesn't match. Therefore, there must be an error in the pairing step.

Wait, let me check the pairing again. Let me recompute the sum:

Original sum after expansion:

For k from 0 to 7:

k=0: binom(7,0)*(-1)^0*e^{i7x} = 1*1*e^{i7x}

k=1: binom(7,1)*(-1)^1*e^{i5x} = 7*(-1)*e^{i5x}

k=2: binom(7,2)*(-1)^2*e^{i3x} = 21*1*e^{i3x}

k=3: binom(7,3)*(-1)^3*e^{ix} = 35*(-1)*e^{ix}

k=4: binom(7,4)*(-1)^4*e^{-ix} = 35*1*e^{-ix}

k=5: binom(7,5)*(-1)^5*e^{-i3x} = 21*(-1)*e^{-i3x}

k=6: binom(7,6)*(-1)^6*e^{-i5x} = 7*1*e^{-i5x}

k=7: binom(7,7)*(-1)^7*e^{-i7x} = 1*(-1)*e^{-i7x}

Therefore, the entire sum is:

e^{i7x} -7e^{i5x} +21e^{i3x} -35e^{ix} +35e^{-ix} -21e^{-i3x} +7e^{-i5x} -e^{-i7x}

Now, grouping terms:

(e^{i7x} - e^{-i7x}) + (-7e^{i5x} +7e^{-i5x}) + (21e^{i3x} -21e^{-i3x}) + (-35e^{ix} +35e^{-ix})

Which is:

2i sin7x + (-14i sin5x) + 42i sin3x + 70i sinx.

Wait, that's what I had before. Then multiplying by i/128:

i/128*(2i sin7x -14i sin5x +42i sin3x +70i sinx) = i/128*(2i sin7x) + ...

But let me compute this again:

First term: 2i sin7x * i/128 = 2i^2 sin7x /128 = -2 sin7x /128 = -sin7x /64

Second term: -14i sin5x * i/128 = -14i^2 sin5x /128 = 14 sin5x /128 = 7 sin5x /64

Third term: 42i sin3x * i/128 = 42i^2 sin3x /128 = -42 sin3x /128 = -21 sin3x /64

Fourth term: 70i sinx * i/128 = 70i^2 sinx /128 = -70 sinx /128 = -35 sinx /64

So the expression is:

- (1/64) sin7x + (7/64) sin5x - (21/64) sin3x - (35/64) sinx.

But when I plug in x = π/2, this gives:

- (1/64) sin(7π/2) + (7/64) sin(5π/2) - (21/64) sin(3π/2) - (35/64) 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

So substituting:

- (1/64)(-1) + (7/64)(1) - (21/64)(-1) - (35/64)(1)

= 1/64 + 7/64 + 21/64 - 35/64

= (1 + 7 + 21 - 35)/64

= (-6)/64 = -3/32

But the actual value of (sin(π/2))^7 is 1. So clearly, there is a mistake here. Therefore, my complex exponential approach must have an error.

But where?

Wait, let me check the original expansion. Let me compute (sinx)^7 at x = π/2. It's 1. Let me compute the expression from the second method:

From the second method, we had:

sin^7x = -1/64 sin7x + 7/64 sin5x -21/64 sin3x + 35/64 sinx.

At x = π/2:

sin7x = sin(7π/2) = -1

sin5x = sin(5π/2) = 1

sin3x = sin(3π/2) = -1

sinx = sin(π/2) = 1

Therefore:

-1/64*(-1) + 7/64*(1) -21/64*(-1) + 35/64*(1)

= 1/64 + 7/64 + 21/64 + 35/64

= (1 + 7 + 21 + 35)/64 = 64/64 = 1. Correct!

So the second method gives the correct result, while the first method gives -3/32. Therefore, there must be a mistake in the first method.

Wait, so why is there a discrepancy? Let me check the first method again.

Wait, in the first method, I had:

(\sin x)^7 = \frac{i}{128} [2i \sin 7x - 14i \sin 5x + 42i \sin 3x + 70i \sin x]

But when I compute this expression, it's:

\frac{i}{128} * [2i sin7x -14i sin5x +42i sin3x +70i sinx]

= \frac{i}{128} * [i(2 sin7x -14 sin5x +42 sin3x +70 sinx)]

= \frac{i^2}{128} [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= \frac{-1}{128} [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= -\frac{2}{128} sin7x + \frac{14}{128} sin5x - \frac{42}{128} sin3x - \frac{70}{128} sinx

Simplify:

= -\frac{1}{64} sin7x + \frac{7}{64} sin5x - \frac{21}{64} sin3x - \frac{35}{64} sinx

But according to the second method, it should be:

- \frac{1}{64} sin7x + \frac{7}{64} sin5x - \frac{21}{64} sin3x + \frac{35}{64} sinx

So the difference is the sign of the last term. Therefore, there must be a mistake in the first method's calculation of the sum.

Wait, let me check the pairing again. When I paired the terms, I had:

For k=0 and k=7: e^{i7x} - e^{-i7x} = 2i sin7x. Correct.

For k=1 and k=6: -7e^{i5x} +7e^{-i5x} = -14i sin5x. Correct.

For k=2 and k=5: 21e^{i3x} -21e^{-i3x} = 42i sin3x. Correct.

For k=3 and k=4: -35e^{ix} +35e^{-ix} = 70i sinx. Correct.

Therefore, the sum is 2i sin7x -14i sin5x +42i sin3x +70i sinx.

But then multiplying by i/128 gives:

i/128*(2i sin7x -14i sin5x +42i sin3x +70i sinx)

= i/128 * [i(2 sin7x -14 sin5x +42 sin3x +70 sinx)]

= i^2 / 128 * [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= -1/128 * [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= -2/128 sin7x +14/128 sin5x -42/128 sin3x -70/128 sinx

= -1/64 sin7x +7/64 sin5x -21/64 sin3x -35/64 sinx

But according to the second method, it should be +35/64 sinx. So why is there a discrepancy?

Wait, but in the second method, the expression was:

sin^7x = -1/64 sin7x +7/64 sin5x -21/64 sin3x +35/64 sinx

So the difference is the sign of the last term. Therefore, there must be an error in the first method's calculation.

Wait, let me check the original expansion again. The original expansion was:

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

But (2i)^7 = 128 i^7 = 128*(-i). Therefore, 1/(2i)^7 = 1/( -128i ) = i/128. Correct.

But then the sum is:

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

But when I expanded this sum, I got:

e^{i7x} -7e^{i5x} +21e^{i3x} -35e^{ix} +35e^{-ix} -21e^{-i3x} +7e^{-i5x} -e^{-i7x}

But when I group terms:

(e^{i7x} - e^{-i7x}) + (-7e^{i5x} +7e^{-i5x}) + (21e^{i3x} -21e^{-i3x}) + (-35e^{ix} +35e^{-ix})

Which is:

2i sin7x -14i sin5x +42i sin3x +70i sinx

But if I instead consider that the original expansion is:

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

But when I expand this, the terms are:

For k=0: binom(7,0)*(-1)^0*e^{i7x} = 1*1*e^{i7x}

k=1: binom(7,1)*(-1)^1*e^{i(7 - 2)x} = 7*(-1)*e^{i5x}

k=2: binom(7,2)*(-1)^2*e^{i(7 - 4)x} = 21*1*e^{i3x}

k=3: binom(7,3)*(-1)^3*e^{i(7 - 6)x} = 35*(-1)*e^{i1x}

k=4: binom(7,4)*(-1)^4*e^{i(7 - 8)x} = 35*1*e^{-i1x}

k=5: binom(7,5)*(-1)^5*e^{i(7 - 10)x} = 21*(-1)*e^{-i3x}

k=6: binom(7,6)*(-1)^6*e^{i(7 - 12)x} = 7*1*e^{-i5x}

k=7: binom(7,7)*(-1)^7*e^{i(7 - 14)x} = 1*(-1)*e^{-i7x}

So the sum is:

e^{i7x} -7e^{i5x} +21e^{i3x} -35e^{ix} +35e^{-ix} -21e^{-i3x} +7e^{-i5x} -e^{-i7x}

Now, grouping terms:

(e^{i7x} - e^{-i7x}) + (-7e^{i5x} +7e^{-i5x}) + (21e^{i3x} -21e^{-i3x}) + (-35e^{ix} +35e^{-ix})

Which is:

2i sin7x -14i sin5x +42i sin3x +70i sinx

But when I multiply by i/128, I get:

i/128*(2i sin7x -14i sin5x +42i sin3x +70i sinx) = -1/64 sin7x +7/64 sin5x -21/64 sin3x -35/64 sinx

But according to the second method, it should be +35/64 sinx. So why is there a discrepancy?

Wait, but in the second method, we had:

sin^7x = -1/64 sin7x +7/64 sin5x -21/64 sin3x +35/64 sinx

So the difference is the sign of the last term. Let me check the second method again.

In the second method, when I calculated the fourth term, I had:

Fourth term: -1/32 sinx cos6x = -1/32 * [sin(x + 6x) + sin(x - 6x)] / 2 = -1/64 [sin7x + sin(-5x)] = -1/64 [sin7x - sin5x].

Wait, but in the expansion, this term is -1/64 sin7x + 1/64 sin5x.

Then, when combining all terms, the sinx terms were:

First term: 5/16 sinx

Second term: +15/64 sinx

Total: 5/16 + 15/64 = 20/64 + 15/64 = 35/64 sinx.

But in the complex exponential method, it's -35/64 sinx.

So why is there a sign difference?

Wait, let me check the fourth term in the second method again.

Fourth term: -1/32 sinx cos6x.

Using identity: sinA cosB = [sin(A+B) + sin(A-B)] / 2.

So sinx cos6x = [sin(x + 6x) + sin(x - 6x)] / 2 = [sin7x + sin(-5x)] / 2 = [sin7x - sin5x]/2.

Therefore, -1/32 sinx cos6x = -1/32 * [sin7x - sin5x]/2 = -1/64 sin7x + 1/64 sin5x.

Yes, that's correct.

Then, when combining all terms, the sinx terms are from first term (5/16 sinx) and second term (15/64 sinx).

But where does the 15/64 sinx come from?

Second term was: -15/32 sinx cos2x = -15/32 * [sin3x - sinx]/2 = -15/64 sin3x + 15/64 sinx.

Yes, so that gives +15/64 sinx.

Therefore, total sinx terms: 5/16 sinx + 15/64 sinx = 35/64 sinx.

Therefore, the second method gives +35/64 sinx.

But the complex exponential method gives -35/64 sinx.

So there must be a sign error in the complex exponential method.

Wait, let me check the original expansion again.

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

But (2i)^7 = 128 i^7 = 128 * (-i). Therefore, 1/(2i)^7 = 1/( -128i ) = i/128.

But let me check this calculation again.

i^7 = i^(4 + 3) = i^3 = -i.

Yes, so (2i)^7 = 2^7 * i^7 = 128 * (-i).

Therefore, 1/(2i)^7 = 1/( -128i ) = multiply numerator and denominator by i: i / (-128i^2) = i / (-128 * -1) = i / 128.

Yes, correct.

Then, the sum is 2i sin7x -14i sin5x +42i sin3x +70i sinx.

Multiply by i/128:

i/128 * [2i sin7x -14i sin5x +42i sin3x +70i sinx]

= i/128 * i [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= i^2 / 128 [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= -1/128 [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= -2/128 sin7x +14/128 sin5x -42/128 sin3x -70/128 sinx

= -1/64 sin7x +7/64 sin5x -21/64 sin3x -35/64 sinx

But according to the second method, it should be +35/64 sinx.

Wait, but if I take the negative of this expression, I get the second method's result.

Wait, what if there was a sign error in the initial expansion? Let me check the original expansion of (\sin x)^7.

(\sin x)^7 = [ (e^{ix} - e^{-ix}) / (2i) ]^7

= (e^{ix} - e^{-ix})^7 / (2i)^7

= [sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} ] / (2i)^7

Yes, that's correct.

But if I instead had [ (e^{-ix} - e^{ix}) / (2i) ]^7, which is [ - (e^{ix} - e^{-ix}) / (2i) ]^7 = (-1)^7 [ (e^{ix} - e^{-ix}) / (2i) ]^7 = - (\sin x)^7.

But no, the original expression is correct.

Wait, but maybe there is a mistake in the sign when expanding (e^{ix} - e^{-ix})^7? Let me check.

The binomial expansion is (a + b)^n = sum_{k=0}^n \binom{n}{k} a^{n - k} b^k.

Here, a = e^{ix}, b = -e^{-ix}.

So (e^{ix} - e^{-ix})^7 = sum_{k=0}^7 \binom{7}{k} (e^{ix})^{7 - k} (-e^{-ix})^k

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

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

Yes, correct.

So the expansion is correct.

But then why is there a discrepancy?

Wait, let me check with x = π/2.

From the complex exponential method, we have:

-1/64 sin7x +7/64 sin5x -21/64 sin3x -35/64 sinx

At x = π/2, this is -1/64*(-1) +7/64*(1) -21/64*(-1) -35/64*(1) = 1/64 +7/64 +21/64 -35/64 = (1 +7 +21 -35)/64 = -6/64 = -3/32.

But the correct value is 1.

From the second method, it's -1/64 sin7x +7/64 sin5x -21/64 sin3x +35/64 sinx, which at x = π/2 is 1/64 +7/64 +21/64 +35/64 = 64/64 = 1.

So the second method is correct, and the first method is wrong.

But why? Where is the error in the first method?

Wait, let me check the initial step.

(\sin x)^7 = [ (e^{ix} - e^{-ix}) / (2i) ]^7

= (e^{ix} - e^{-ix})^7 / (2i)^7

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

But if I instead write this as [ - (e^{-ix} - e^{ix}) ]^7 = (-1)^7 (e^{-ix} - e^{ix})^7 = - (e^{-ix} - e^{ix})^7

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

Wait, this seems complicated.

Alternatively, perhaps there is a mistake in the sign when converting the exponential terms to sine.

Wait, let me recall that e^{iθ} - e^{-iθ} = 2i sinθ.

But in the complex exponential method, we have:

The sum is 2i sin7x -14i sin5x +42i sin3x +70i sinx.

But if I multiply this by i/128, I get:

i/128 * [2i sin7x -14i sin5x +42i sin3x +70i sinx]

= i/128 * i [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= i^2 / 128 [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= -1/128 [2 sin7x -14 sin5x +42 sin3x +70 sinx]

= -2/128 sin7x +14/128 sin5x -42/128 sin3x -70/128 sinx

= -1/64 sin7x +7/64 sin5x -21/64 sin3x -35/64 sinx

But according to the second method, it should be +35/64 sinx.

Wait, but if I instead had multiplied by -i/128 instead of i/128, would that fix it? Let me check.

If I had a sign error in the initial step, like if the original expression was [ (e^{ix} - e^{-ix}) / (2i) ]^7 = [ - (e^{-ix} - e^{ix}) / (2i) ]^7 = (-1)^7 [ (e^{-ix} - e^{ix}) / (2i) ]^7 = - [ (e^{-ix} - e^{ix}) / (2i) ]^7.

But (e^{-ix} - e^{ix}) / (2i) = [ - (e^{ix} - e^{-ix}) ] / (2i) = - [ (e^{ix} - e^{-ix}) / (2i) ] = -sinx.

So [ (e^{-ix} - e^{ix}) / (2i) ]^7 = (-sinx)^7 = -sin^7x.

Therefore, the original expression is - [ (e^{-ix} - e^{ix}) / (2i) ]^7 = -(-sin^7x) = sin^7x.

Wait, this seems to suggest that the original expansion is correct.

But why is there a discrepancy?

Wait, let me try to compute the complex exponential expression at x = π/2.

If I use the second method's result, which is correct, and compare it to the complex exponential method's result, which is wrong, there must be a miscalculation in the complex exponential method.

Wait, let me try to compute the complex exponential expression again.

Let me compute the complex exponential expression:

-1/64 sin7x +7/64 sin5x -21/64 sin3x -35/64 sinx

At x = π/2, this is:

-1/64 * sin(7π/2) +7/64 * sin(5π/2) -21/64 * sin(3π/2) -35/64 * sin(π/2)

= -1/64 * (-1) +7/64 * 1 -21/64 * (-1) -35/64 * 1

= 1/64 +7/64 +21/64 -35/64

= (1 +7 +21 -35)/64 = -6/64 = -3/32

But the correct value is 1.

However, if I take the negative of this expression, I get 3/32, which is still not 1.

Wait, but if I had the expression as -1/64 sin7x +7/64 sin5x -21/64 sin3x +35/64 sinx, then at x = π/2, it would be 1/64 +7/64 +21/64 +35/64 = 64/64 = 1.

So the error is in the sign of the last term in the complex exponential method.

But why? Where is the mistake?

Wait, let me check the original expansion again.

The sum was:

e^{i7x} -7e^{i5x} +21e^{i3x} -35e^{ix} +35e^{-ix} -21e^{-i3x} +7e^{-i5x} -e^{-i7x}

When I grouped terms, I said this is:

(e^{i7x} - e^{-i7x}) + (-7e^{i5x} +7e^{-i5x}) + (21e^{i3x} -21e^{-i3x}) + (-35e^{ix} +35e^{-ix})

But let me check the last term: -35e^{ix} +35e^{-ix} = 35(e^{-ix} - e^{ix}) = -35(e^{ix} - e^{-ix}) = -35 * 2i sinx = -70i sinx.

Wait, but earlier I said this is 70i sinx.

Wait, this is the mistake!

Wait, -35e^{ix} +35e^{-ix} = 35(e^{-ix} - e^{ix}) = -35(e^{ix} - e^{-ix}) = -35 * 2i sinx = -70i sinx.

But earlier, I said this is 70i sinx.

Ah! Here is the error!

Previously, I said that -35e^{ix} +35e^{-ix} = 70i sinx, but it should be -70i sinx.

Let me recheck:

-35e^{ix} +35e^{-ix} = 35(e^{-ix} - e^{ix}) = 35 * (- (e^{ix} - e^{-ix})) = -35 * 2i sinx = -70i sinx.

Yes! So previously, I had said this equals 70i sinx, but it should be -70i sinx.

Therefore, the sum is:

(e^{i7x} - e^{-i7x}) + (-7e^{i5x} +7e^{-i5x}) + (21e^{i3x} -21e^{-i3x}) + (-35e^{ix} +35e^{-ix})

= 2i sin7x -14i sin5x +42i sin3x -70i sinx

Therefore, the sum is 2i sin7x -14i sin5x +42i sin3x -70i sinx.

Previously, I had +70i sinx, but it should be -70i sinx.

Therefore, the correct sum is 2i sin7x -14i sin5x +42i sin3x -70i sinx.

Then, multiplying by i/128:

i/128 * [2i sin7x -14i sin5x +42i sin3x -70i sinx]

= i/128 * i [2 sin7x -14 sin5x +42 sin3x -70 sinx]

= i^2 / 128 [2 sin7x -14 sin5x +42 sin3x -70 sinx]

= -1/128 [2 sin7x -14 sin5x +42 sin3x -70 sinx]

= -2/128 sin7x +14/128 sin5x -42/128 sin3x +70/128 sinx

= -1/64 sin7x +7/64 sin5x -21/64 sin3x +35/64 sinx

Which matches the second method!

So the error was in the sign of the last term in the pairing step. I had incorrectly said that -35e^{ix} +35e^{-ix} = 70i sinx, but it should be -70i sinx.

Therefore, the correct expression is:

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

Therefore, the coefficient d is 35/64.

But wait, in the second method, we had the same result. So the correct value of d is 35/64.

But initially, I thought it was -35/64, but that was due to a sign error in the complex exponential method.

Therefore, the correct value of d is 35/64.

But let me confirm with x = π/2.

Using the corrected complex exponential expression:

-1/64 sin7x +7/64 sin5x -21/64 sin3x +35/64 sinx

At x = π/2:

-1/64*(-1) +7/64*(1) -21/64*(-1) +35/64*(1) = 1/64 +7/64 +21/64 +35/64 = (1 +7 +21 +35)/64 = 64/64 = 1. Correct.

Therefore, the correct value of d is 35/64.

But in the initial complex exponential calculation, I had mistakenly taken the last term as +70i sinx instead of -70i sinx, leading to the wrong sign.

Therefore, the correct answer is d = 35/64.

But let me check with the second method's result. In the second method, we had:

sin^7x = -1/64 sin7x +7/64 sin5x -21/64 sin3x +35/64 sinx

So yes, d = 35/64.

Therefore, the correct value of d is 35/64.

But the problem says "Find d." So the answer is 35/64.

But let me check if this matches with any known formulas.

Alternatively, I can recall that the general formula for sin^n x can be expressed as a combination of multiple angles. For example, for odd powers, it's a combination of sines of odd multiples.

Alternatively, I can use the formula from the second method, which is reliable, and it gives d = 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 find the constant $ d $.

---

### Step 1: Use Complex Exponentials

We begin by expressing $ \sin x $ using Euler's formula:

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

Then,

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

Now, compute $ (2i)^7 $:

$$
(2i)^7 = 2^7 \cdot i^7 = 128 \cdot i^7 = 128 \cdot (-i) = -128i
$$

So,

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

Multiply numerator and denominator by $ i $ to simplify:

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

---

### Step 2: Group Terms and Convert to Sine

We now evaluate the sum:

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

This sum can be grouped into pairs of terms that are complex conjugates, and each pair can be expressed as a sine function. After careful pairing and simplification, we find:

$$
\sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} = 2i \sin 7x - 14i \sin 5x + 42i \sin 3x - 70i \sin x
$$

Now multiply by $ \frac{i}{128} $:

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

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

$$
= \frac{-1}{128} \left( 2 \sin 7x - 14 \sin 5x + 42 \sin 3x - 70 \sin x \right)
$$

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

---

### Step 3: Match with Given Form

We are given:

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

Comparing 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 $\boxed{\frac{35}{64}}$.