Explanation:
Okay, let's see. The problem says that half the value of 3x minus 9 is equal to x plus 37. I need to find the value of x. Hmm, let me try to translate this into an equation.

So, "half the value of 3x - 9" – that should mean taking (3x - 9) and then dividing it by 2, right? And that's equal to x + 37. Let me write that down:

(3x - 9)/2 = x + 37

Okay, that seems right. Now, I need to solve for x. Let me think about how to do this. Maybe I can get rid of the fraction first. To do that, I can multiply both sides of the equation by 2. Let me try that.

Multiplying both sides by 2:

2 * [(3x - 9)/2] = 2 * (x + 37)

Simplifying the left side, the 2 and the denominator 2 cancel out, so it becomes:

3x - 9 = 2x + 74

Alright, now I have 3x - 9 = 2x + 74. Let me get all the x terms on one side and constants on the other. So, maybe subtract 2x from both sides to get rid of the 2x on the right.

3x - 9 - 2x = 2x + 74 - 2x

Simplifying both sides:

(3x - 2x) - 9 = 74

So that's:

x - 9 = 74

Now, to solve for x, I can add 9 to both sides.

x - 9 + 9 = 74 + 9

Which simplifies to:

x = 83

Wait, let me check if that makes sense. Let me plug x = 83 back into the original equation to verify.

First, compute 3x - 9 when x is 83:

3*83 = 249, then 249 - 9 = 240. Then half of that is 240 / 2 = 120.

Now, compute x + 37 when x is 83:

83 + 37 = 120.

Wow, both sides equal 120. So that checks out. Therefore, x = 83 is correct.

But let me go through the steps again to make sure I didn't skip anything or make a mistake. Starting with the original equation:

(3x - 9)/2 = x + 37

Multiply both sides by 2:

3x - 9 = 2x + 74

Subtract 2x:

x - 9 = 74

Add 9:

x = 83

Yep, same result. Seems solid. Let me think if there's another way to approach this problem, just to confirm.

Alternatively, maybe interpreting the problem differently? Let me check the wording again: "Half the value of 3x - 9 is x + 37." So "half the value" is definitely (3x - 9)/2. There isn't another interpretation, right? Like, maybe "half of (3x - 9)" is the same as multiplying by 1/2. So that's the same as what I did.

Alternatively, could it be that "half the value" refers to something else? Like, maybe half of 3x minus 9? But that would be 3x/2 - 9? Wait, but the wording is "half the value of 3x - 9", so the "value of" applies to the entire expression 3x - 9. So, it's definitely (3x - 9)/2, not 3x/2 - 9.

But just to check, if I took it as 3x/2 - 9, then the equation would be:

3x/2 - 9 = x + 37

Then solving that would be different. Let me try that to see what happens. Maybe I misread the problem?

But that seems less likely. The original problem says "half the value of 3x - 9". The phrase "the value of" modifies the entire expression, so it's half of (3x - 9). So the original equation is correct. But just to explore, if it were 3x/2 - 9, then solving:

3x/2 - 9 = x + 37

Multiply both sides by 2:

3x - 18 = 2x + 74

Subtract 2x:

x - 18 = 74

Add 18:

x = 92

But then check with original equation (if it were that way):

Original equation would be 3x/2 - 9 = x + 37

Plug x = 92:

Left side: 3*92 / 2 - 9 = 276 / 2 - 9 = 138 - 9 = 129

Right side: 92 + 37 = 129. Also works. Wait, so depending on interpretation, there are two possible answers?

But that's a problem. The original wording is crucial here. The problem is in the interpretation of "half the value of 3x - 9". If it's half of (3x - 9), then it's (3x - 9)/2. If it's half of 3x minus 9, then it's (3x)/2 - 9. Which is correct?

In standard English, "half the value of 3x - 9" would mean (3x - 9)/2. Because "the value of 3x - 9" is the entire expression, and then you take half of that. If they meant 3x/2 - 9, they would probably say "half of 3x minus 9" or "half of 3x, minus 9".

But since the problem is from an algebra textbook or similar, they probably intended the first interpretation. Also, when I checked the first answer, x=83 worked, which is correct. If the problem had said "half of 3x minus 9", then x=92 would be the answer. But since the original problem says "half the value of 3x-9", the answer is 83.

But just to be thorough, maybe the problem is from a source where maybe they have different conventions. But since my first interpretation checks out with the problem's original wording, and the answer works, I think 83 is the correct answer.

Therefore, I think I did everything right. So the value of x is 83.

**Final Answer**
The value of $ x $ is \boxed{83}.

To solve the equation:

$$
\frac{1}{2}(3x - 9) = x + 37
$$

we begin by eliminating the fraction. Multiply both sides of the equation by 2:

$$
2 \cdot \frac{1}{2}(3x - 9) = 2(x + 37)
$$

Simplify both sides:

$$
3x - 9 = 2x + 74
$$

Now, isolate $ x $. Subtract $ 2x $ from both sides:

$$
x - 9 = 74
$$

Add 9 to both sides:

$$
x = 83
$$

---

To confirm that this is correct, substitute $ x = 83 $ back into the original equation:

- Compute the left-hand side:
  $$
  \frac{1}{2}(3x - 9) = \frac{1}{2}(3(83) - 9) = \frac{1}{2}(249 - 9) = \frac{1}{2}(240) = 120
  $$

- Compute the right-hand side:
  $$
  x + 37 = 83 + 37 = 120
  $$

Since both sides are equal, the solution is consistent.

---

### Final Answer:

$$
\boxed{83}
$$

[Assistant]
Explanation:
Okay, let's see. The problem says that half the value of 3x minus 9 is equal to x plus 37. I need to find the value of x. Hmm, let me try to translate this into an equation.

So, "half the value of 3x - 9" – that should mean taking (3x - 9) and then dividing it by 2, right? And that's equal to x + 37. Let me write that down:

(3x - 9)/2 = x + 37

Okay, that seems right. Now, I need to solve for x. Let me think about how to do this. Maybe I can get rid of the fraction first. To do that, I can multiply both sides of the equation by 2. Let me try that.

Multiplying both sides by 2:

2 * [(3x - 9)/2] = 2 * (x + 37)

Simplifying the left side, the 2 and the denominator 2 cancel out, so it becomes:

3x - 9 = 2x + 74

Alright, now I have 3x - 9 = 2x + 74. Let me get all the x terms on one side and constants on the other. So, maybe subtract 2x from both sides to get rid of the 2x on the right.

3x - 9 - 2x = 2x + 74 - 2x

Simplifying both sides:

(3x - 2x) - 9 = 74

So that's:

x - 9 = 74

Now, to solve for x, I can add 9 to both sides.

x - 9 + 9 = 74 + 9

Which simplifies to:

x = 83

Wait, let me check if that makes sense. Let me plug x = 83 back into the original equation to verify.

First, compute 3x - 9 when x is 83:

3*83 = 249, then 249 - 9 = 240. Then half of that is 240 / 2 = 120.

Now, compute x + 37 when x is 83:

83 + 37 = 120.

Wow, both sides equal 120. So that checks out. Therefore, x = 83 is correct.

But let me go through the steps again to make sure I didn't skip anything or make a mistake. Starting with the original equation:

(3x - 9)/2 = x + 37

Multiply both sides by 2:

3x - 9 = 2x + 74

Subtract 2x:

x - 9 = 74

Add 9:

x = 83

Yep, same result. Seems solid. Let me think if there's another way to approach this problem, just to confirm.

Alternatively, maybe interpreting the problem differently? Let me check the wording again: "Half the value of 3x - 9 is x + 37." So "half the value" is definitely (3x - 9)/2. There isn't another interpretation, right? Like, maybe "half of (3x - 9)" is the same as multiplying by 1/2. So that's the same as what I did.

Alternatively, could it be that "half the value" refers to something else? Like, maybe half of 3x minus 9? But that would be 3x/2 - 9? Wait, but the wording is "half the value of 3x - 9", so the "value of" applies to the entire expression 3x - 9. So, it's definitely (3x - 9)/2, not 3x/2 - 9.

But just to check, if I took it as 3x/2 - 9, then the equation would be:

3x/2 - 9 = x + 37

Then solving that would be different. Let me try that to see what happens. Maybe I misread the problem?

But that seems less likely. The original problem says "half the value of 3x - 9". The phrase "the value of" modifies the entire expression, so it's half of (3x - 9). So the original equation is correct. But just to explore, if it were 3x/2 - 9, then solving:

3x/2 - 9 = x + 37

Multiply both sides by 2:

3x - 18 = 2x + 74

Subtract 2x:

x - 18 = 74

Add 18:

x = 92

But then check with original equation (if it were that way):

Original equation would be 3x/2 - 9 = x + 37

Plug x = 92:

Left side: 3*92 / 2 - 9 = 276 / 2 - 9 = 138 - 9 = 129

Right side: 92 + 37 = 129. Also works. Wait, so depending on interpretation, there are two possible answers?

But that's a problem. The original wording is crucial here. The problem is in the interpretation of "half the value of 3x - 9". If it's half of (3x - 9), then it's (3x - 9)/2. If it's half of 3x minus 9, then it's (3x)/2 - 9. Which is correct?

In standard English, "half the value of 3x - 9" would mean (3x - 9)/2. Because "the value of 3x - 9" is the entire expression, and then you take half of that. If they meant 3x/2 - 9, they would probably say "half of 3x minus 9" or "half of 3x, minus 9".

But since the problem is from an algebra textbook or similar, they probably intended the first interpretation. Also, when I checked the first answer, x=83 worked, which is correct. If the problem had said "half of 3x minus 9", then x=92 would be the answer. But since the original problem says "half the value of 3x-9", the answer is 83.

But just to be thorough, maybe the problem is from a source where maybe they have different conventions. But since my first interpretation checks out with the problem's original wording, and the answer works, I think 83 is the correct answer.

Therefore, I think I did everything right. So the value of x is 83.

**Final Answer**
The value of $ x $ is \boxed{83}.

To solve the equation:

$$
\frac{1}{2}(3x - 9) = x + 37
$$

we begin by eliminating the fraction. Multiply both sides of the equation by 2:

$$
2 \cdot \frac{1}{2}(3x - 9) = 2(x + 37)
$$

Simplify both sides:

$$
3x - 9 = 2x + 74
$$

Now, isolate $ x $. Subtract $ 2x $ from both sides:

$$
x - 9 = 74
$$

Add 9 to both sides:

$$
x = 83
$$

---

To confirm that this is correct, substitute $ x = 83 $ back into the original equation:

- Compute the left-hand side:
  $$
  \frac{1}{2}(3x - 9) = \frac{1}{2}(3(83) - 9) = \frac{1}{2}(249 - 9) = \frac{1}{2}(240) = 120
  $$

- Compute the right-hand side:
  $$
  x + 37 = 83 + 37 = 120
  $$

Since both sides are equal, the solution is consistent.

---

### Final Answer:

$$
\boxed{83}
$$

Solution: We translate the problem to the equation $\frac{1}{2}(3x-9) = x+37$. Multiplying both sides by 2 gives $3x-9 = 2x+74$. Subtracting $2x$ from both sides yield $x-9 = 74$. Adding $9$ to both sides yields $ x = \boxed{83}$.


Explanation:
Okay, let me try to figure out this problem. So, we have a polynomial f, and we know that when we multiply it by (x - 1), we get this polynomial: 3x⁴ + x³ - 25x² + 38x - 17. The question is asking for the degree of f.

First, I remember that when you multiply two polynomials, the degree of the product is the sum of the degrees of the two polynomials. So, if (x - 1) is multiplied by f(x), the resulting polynomial has degree 4. Let me check that. The given polynomial is 3x⁴ + ... so yeah, it's a fourth-degree polynomial.

Now, (x - 1) is a first-degree polynomial. Let me denote the degree of f(x) as n. Then, according to the rule about the degrees adding up, the degree of (x - 1) times f(x) should be 1 + n. But we know that the product is a degree 4 polynomial. So, 1 + n = 4. Solving for n, that would be n = 4 - 1 = 3. So, does that mean the degree of f is 3? Wait, but maybe I should check if there's something else going on here?

Hold on, maybe there's a trick here. Suppose that f(x) has a factor of (x - 1), but in that case, when we multiply by (x - 1), the degree would be higher. But in this case, the product is a fourth-degree polynomial. However, if f(x) had a factor of (x - 1), then the product would have (x - 1) squared, making the degree 1 + degree of f. But maybe f(x) isn't divisible by (x - 1)? Wait, but if f(x) had a factor of (x - 1), then the product would have (x - 1)^2, so maybe the original problem isn't saying that f(x) is divisible by (x - 1), right? Wait, actually, no. Wait, the equation given is (x - 1)*f(x) equals that polynomial. So, that means that (x - 1) is a factor of the polynomial on the right-hand side, right? Therefore, the polynomial 3x⁴ + x³ - 25x² + 38x - 17 must be divisible by (x - 1). Therefore, maybe we can check if x = 1 is a root of that polynomial? Let me verify that.

Let me plug in x = 1 into the polynomial: 3(1)^4 + (1)^3 - 25(1)^2 + 38(1) - 17. Let's compute that:

3*1 + 1 - 25*1 + 38 - 17 = 3 + 1 - 25 + 38 - 17. Let's compute step by step:

3 + 1 = 4

4 - 25 = -21

-21 + 38 = 17

17 - 17 = 0

Oh, so x = 1 is a root. Therefore, (x - 1) is indeed a factor of the polynomial on the right-hand side, which makes sense because the left-hand side is (x - 1)*f(x). Therefore, that means that f(x) is equal to that polynomial divided by (x - 1). So, if we do the division, we can find f(x) and then check its degree.

But maybe we don't need to do the division? Since we already know that the degree of the product is 4, and since (x - 1) is degree 1, then f(x) must be degree 3. However, I need to be careful here. If the division results in a polynomial with a lower degree? Wait, no. If we divide a degree 4 polynomial by a degree 1 polynomial, we get a degree 3 polynomial. So, that suggests that f(x) is a cubic polynomial. Therefore, the degree of f is 3. So, maybe the answer is 3.

But let me check again. Let me think if there are any cases where dividing by (x - 1) might result in a lower degree? For example, if the original polynomial was of degree 4 and we divide by a degree 1 polynomial, then the quotient is degree 3, unless there is some kind of cancellation. Wait, but (x - 1) is a factor, so the division should result in a polynomial of degree 3. Therefore, the degree of f(x) is 3. Therefore, is that the answer?

But maybe I should check by performing the division to confirm. Let me try polynomial division. Let me divide the polynomial 3x⁴ + x³ - 25x² + 38x - 17 by (x - 1).

Let me recall how polynomial long division works. Dividing 3x⁴ + x³ - 25x² + 38x - 17 by (x - 1).

First, divide the leading term of the dividend by the leading term of the divisor. The leading term of the dividend is 3x⁴, and the leading term of the divisor is x. So, 3x⁴ / x = 3x³. Multiply the entire divisor by 3x³: 3x³*(x - 1) = 3x⁴ - 3x³. Subtract this from the dividend:

(3x⁴ + x³ - 25x² + 38x - 17) - (3x⁴ - 3x³) =

3x⁴ - 3x⁴ + x³ + 3x³ - 25x² + 38x -17 = 4x³ - 25x² + 38x -17.

Now, take the new polynomial 4x³ - 25x² + 38x -17 and divide by (x - 1). The leading term is 4x³ divided by x is 4x². Multiply divisor by 4x²: 4x²*(x -1) = 4x³ - 4x². Subtract that from the current polynomial:

(4x³ - 25x² + 38x -17) - (4x³ - 4x²) = 0x³ + (-25x² + 4x²) + 38x -17 = -21x² + 38x -17.

Next, divide -21x² by x to get -21x. Multiply divisor by -21x: -21x*(x -1) = -21x² + 21x. Subtract that from the current polynomial:

(-21x² + 38x -17) - (-21x² + 21x) = 0x² + (38x - 21x) -17 = 17x -17.

Now, divide 17x by x to get 17. Multiply divisor by 17: 17*(x -1) = 17x -17. Subtract that from the current polynomial:

(17x -17) - (17x -17) = 0.

So, the division comes out exactly with no remainder, as expected, since we already checked that x=1 is a root. Therefore, the quotient is 3x³ + 4x² -21x +17. Therefore, f(x) = 3x³ + 4x² -21x +17. Let me check the degree of this polynomial. The highest degree term is 3x³, so degree 3. Therefore, the answer is 3.

Wait, but just to make sure, let me verify that multiplying (x -1) by this f(x) gives the original polynomial. Let me do that quickly.

Multiply (x - 1) by (3x³ + 4x² -21x +17):

First, x*(3x³ + 4x² -21x +17) = 3x⁴ + 4x³ -21x² +17x

Then, -1*(3x³ + 4x² -21x +17) = -3x³ -4x² +21x -17

Now, add these two results together:

3x⁴ +4x³ -21x² +17x -3x³ -4x² +21x -17

Combine like terms:

3x⁴ + (4x³ - 3x³) + (-21x² -4x²) + (17x +21x) -17

Which is:

3x⁴ + x³ -25x² +38x -17

Which is exactly the polynomial given in the problem. Perfect, that checks out. Therefore, f(x) is indeed a cubic polynomial, so degree 3. Therefore, the answer is 3.

But let me think again if there's another way this could be approached, maybe without polynomial division. For example, considering the degrees. If (x -1) * f(x) is degree 4, then since (x -1) is degree 1, then f(x) must be degree 3. However, suppose that f(x) had a term with (x -1) as a factor? Wait, but even so, when you multiply by (x -1), the degree would still be 1 + degree of f(x). So, even if f(x) had (x -1) as a factor, the resulting product would have degree 1 + n, where n is the degree of f(x). However, in our case, since the product is degree 4, then n must be 3 regardless. Therefore, even if f(x) had (x -1) as a factor, the degree is still 3.

But in reality, when we did the division, we saw that f(x) is 3x³ + 4x² -21x +17, which does not have (x -1) as a factor. Let me check that. Let me plug x = 1 into f(x): 3(1)^3 + 4(1)^2 -21(1) +17 = 3 + 4 -21 +17 = (3 + 4) + (-21 +17) = 7 - 4 = 3 ≠ 0. Therefore, f(1) = 3, so (x -1) is not a factor of f(x). Therefore, even though (x -1) is a factor of the product, f(x) itself isn't necessarily.

So, in conclusion, regardless of whether f(x) has (x -1) as a factor or not, the degree is determined purely by the degree of the product and the degree of (x -1). Therefore, since the product is degree 4 and (x -1) is degree 1, f(x) must be degree 3.

Therefore, the answer is 3.

**Final Answer**
The degree of $ f $ is \boxed{3}.

To determine the degree of the polynomial $ f(x) $, we analyze the equation:

$$
(x - 1) \cdot f(x) = 3x^4 + x^3 - 25x^2 + 38x - 17
$$

---

### Step 1: Analyze the Degree of the Product

The right-hand side of the equation is a **fourth-degree** polynomial, as it has the highest power of $ x^4 $.

We are multiplying $ f(x) $ by $ (x - 1) $, which is a **first-degree** polynomial. When multiplying two polynomials, the degree of the product equals the **sum of the degrees** of the two polynomials. Let the degree of $ f(x) $ be $ n $. Then:

$$
\deg((x - 1) \cdot f(x)) = \deg(x - 1) + \deg(f(x)) = 1 + n
$$

Since the degree of the product is 4, we solve:

$$
1 + n = 4 \Rightarrow n = 3
$$

Thus, the degree of $ f(x) $ must be **3**, unless $ f(x) $ is not a polynomial, which it is not, as we are given that $ f(x) $ is a polynomial.

---

### Step 2: Verification via Polynomial Division

To further verify this result, we can perform polynomial division and find the expression of $ f(x) $. Dividing the polynomial

$$
3x^4 + x^3 - 25x^2 + 38x - 17
$$

by $ (x - 1) $ gives:

$$
f(x) = 3x^3 + 4x^2 - 21x + 17
$$

This is a **cubic (degree 3)** polynomial, confirming the earlier conclusion.

---

### Step 3: Confirming the Result

As a final check, we can multiply $ (x - 1) $ by $ f(x) $ to see if we recover the original polynomial:

$$
(x - 1)(3x^3 + 4x^2 - 21x + 17)
$$

Expanding:

- $ x(3x^3 + 4x^2 - 21x + 17) = 3x^4 + 4x^3 - 21x^2 + 17x $
- $ -1(3x^3 + 4x^2 - 21x + 17) = -3x^3 - 4x^2 + 21x - 17 $

Adding these:

$$
3x^4 + (4x^3 - 3x^3) + (-21x^2 - 4x^2) + (17x + 21x) - 17
= 3x^4 + x^3 - 25x^2 + 38x - 17
$$

This matches the original polynomial, confirming the correctness of $ f(x) $ and the resulting degree.

---

### Final Conclusion

The degree of the polynomial $ f(x) $ is:

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this problem. So, we have a polynomial f, and we know that when we multiply it by (x - 1), we get this polynomial: 3x⁴ + x³ - 25x² + 38x - 17. The question is asking for the degree of f.

First, I remember that when you multiply two polynomials, the degree of the product is the sum of the degrees of the two polynomials. So, if (x - 1) is multiplied by f(x), the resulting polynomial has degree 4. Let me check that. The given polynomial is 3x⁴ + ... so yeah, it's a fourth-degree polynomial.

Now, (x - 1) is a first-degree polynomial. Let me denote the degree of f(x) as n. Then, according to the rule about the degrees adding up, the degree of (x - 1) times f(x) should be 1 + n. But we know that the product is a degree 4 polynomial. So, 1 + n = 4. Solving for n, that would be n = 4 - 1 = 3. So, does that mean the degree of f is 3? Wait, but maybe I should check if there's something else going on here?

Hold on, maybe there's a trick here. Suppose that f(x) has a factor of (x - 1), but in that case, when we multiply by (x - 1), the degree would be higher. But in this case, the product is a fourth-degree polynomial. However, if f(x) had a factor of (x - 1), then the product would have (x - 1) squared, making the degree 1 + degree of f. But maybe f(x) isn't divisible by (x - 1)? Wait, but if f(x) had a factor of (x - 1), then the product would have (x - 1)^2, so maybe the original problem isn't saying that f(x) is divisible by (x - 1), right? Wait, actually, no. Wait, the equation given is (x - 1)*f(x) equals that polynomial. So, that means that (x - 1) is a factor of the polynomial on the right-hand side, right? Therefore, the polynomial 3x⁴ + x³ - 25x² + 38x - 17 must be divisible by (x - 1). Therefore, maybe we can check if x = 1 is a root of that polynomial? Let me verify that.

Let me plug in x = 1 into the polynomial: 3(1)^4 + (1)^3 - 25(1)^2 + 38(1) - 17. Let's compute that:

3*1 + 1 - 25*1 + 38 - 17 = 3 + 1 - 25 + 38 - 17. Let's compute step by step:

3 + 1 = 4

4 - 25 = -21

-21 + 38 = 17

17 - 17 = 0

Oh, so x = 1 is a root. Therefore, (x - 1) is indeed a factor of the polynomial on the right-hand side, which makes sense because the left-hand side is (x - 1)*f(x). Therefore, that means that f(x) is equal to that polynomial divided by (x - 1). So, if we do the division, we can find f(x) and then check its degree.

But maybe we don't need to do the division? Since we already know that the degree of the product is 4, and since (x - 1) is degree 1, then f(x) must be degree 3. However, I need to be careful here. If the division results in a polynomial with a lower degree? Wait, no. If we divide a degree 4 polynomial by a degree 1 polynomial, we get a degree 3 polynomial. So, that suggests that f(x) is a cubic polynomial. Therefore, the degree of f is 3. So, maybe the answer is 3.

But let me check again. Let me think if there are any cases where dividing by (x - 1) might result in a lower degree? For example, if the original polynomial was of degree 4 and we divide by a degree 1 polynomial, then the quotient is degree 3, unless there is some kind of cancellation. Wait, but (x - 1) is a factor, so the division should result in a polynomial of degree 3. Therefore, the degree of f(x) is 3. Therefore, is that the answer?

But maybe I should check by performing the division to confirm. Let me try polynomial division. Let me divide the polynomial 3x⁴ + x³ - 25x² + 38x - 17 by (x - 1).

Let me recall how polynomial long division works. Dividing 3x⁴ + x³ - 25x² + 38x - 17 by (x - 1).

First, divide the leading term of the dividend by the leading term of the divisor. The leading term of the dividend is 3x⁴, and the leading term of the divisor is x. So, 3x⁴ / x = 3x³. Multiply the entire divisor by 3x³: 3x³*(x - 1) = 3x⁴ - 3x³. Subtract this from the dividend:

(3x⁴ + x³ - 25x² + 38x - 17) - (3x⁴ - 3x³) =

3x⁴ - 3x⁴ + x³ + 3x³ - 25x² + 38x -17 = 4x³ - 25x² + 38x -17.

Now, take the new polynomial 4x³ - 25x² + 38x -17 and divide by (x - 1). The leading term is 4x³ divided by x is 4x². Multiply divisor by 4x²: 4x²*(x -1) = 4x³ - 4x². Subtract that from the current polynomial:

(4x³ - 25x² + 38x -17) - (4x³ - 4x²) = 0x³ + (-25x² + 4x²) + 38x -17 = -21x² + 38x -17.

Next, divide -21x² by x to get -21x. Multiply divisor by -21x: -21x*(x -1) = -21x² + 21x. Subtract that from the current polynomial:

(-21x² + 38x -17) - (-21x² + 21x) = 0x² + (38x - 21x) -17 = 17x -17.

Now, divide 17x by x to get 17. Multiply divisor by 17: 17*(x -1) = 17x -17. Subtract that from the current polynomial:

(17x -17) - (17x -17) = 0.

So, the division comes out exactly with no remainder, as expected, since we already checked that x=1 is a root. Therefore, the quotient is 3x³ + 4x² -21x +17. Therefore, f(x) = 3x³ + 4x² -21x +17. Let me check the degree of this polynomial. The highest degree term is 3x³, so degree 3. Therefore, the answer is 3.

Wait, but just to make sure, let me verify that multiplying (x -1) by this f(x) gives the original polynomial. Let me do that quickly.

Multiply (x - 1) by (3x³ + 4x² -21x +17):

First, x*(3x³ + 4x² -21x +17) = 3x⁴ + 4x³ -21x² +17x

Then, -1*(3x³ + 4x² -21x +17) = -3x³ -4x² +21x -17

Now, add these two results together:

3x⁴ +4x³ -21x² +17x -3x³ -4x² +21x -17

Combine like terms:

3x⁴ + (4x³ - 3x³) + (-21x² -4x²) + (17x +21x) -17

Which is:

3x⁴ + x³ -25x² +38x -17

Which is exactly the polynomial given in the problem. Perfect, that checks out. Therefore, f(x) is indeed a cubic polynomial, so degree 3. Therefore, the answer is 3.

But let me think again if there's another way this could be approached, maybe without polynomial division. For example, considering the degrees. If (x -1) * f(x) is degree 4, then since (x -1) is degree 1, then f(x) must be degree 3. However, suppose that f(x) had a term with (x -1) as a factor? Wait, but even so, when you multiply by (x -1), the degree would still be 1 + degree of f(x). So, even if f(x) had (x -1) as a factor, the resulting product would have degree 1 + n, where n is the degree of f(x). However, in our case, since the product is degree 4, then n must be 3 regardless. Therefore, even if f(x) had (x -1) as a factor, the degree is still 3.

But in reality, when we did the division, we saw that f(x) is 3x³ + 4x² -21x +17, which does not have (x -1) as a factor. Let me check that. Let me plug x = 1 into f(x): 3(1)^3 + 4(1)^2 -21(1) +17 = 3 + 4 -21 +17 = (3 + 4) + (-21 +17) = 7 - 4 = 3 ≠ 0. Therefore, f(1) = 3, so (x -1) is not a factor of f(x). Therefore, even though (x -1) is a factor of the product, f(x) itself isn't necessarily.

So, in conclusion, regardless of whether f(x) has (x -1) as a factor or not, the degree is determined purely by the degree of the product and the degree of (x -1). Therefore, since the product is degree 4 and (x -1) is degree 1, f(x) must be degree 3.

Therefore, the answer is 3.

**Final Answer**
The degree of $ f $ is \boxed{3}.

To determine the degree of the polynomial $ f(x) $, we analyze the equation:

$$
(x - 1) \cdot f(x) = 3x^4 + x^3 - 25x^2 + 38x - 17
$$

---

### Step 1: Analyze the Degree of the Product

The right-hand side of the equation is a **fourth-degree** polynomial, as it has the highest power of $ x^4 $.

We are multiplying $ f(x) $ by $ (x - 1) $, which is a **first-degree** polynomial. When multiplying two polynomials, the degree of the product equals the **sum of the degrees** of the two polynomials. Let the degree of $ f(x) $ be $ n $. Then:

$$
\deg((x - 1) \cdot f(x)) = \deg(x - 1) + \deg(f(x)) = 1 + n
$$

Since the degree of the product is 4, we solve:

$$
1 + n = 4 \Rightarrow n = 3
$$

Thus, the degree of $ f(x) $ must be **3**, unless $ f(x) $ is not a polynomial, which it is not, as we are given that $ f(x) $ is a polynomial.

---

### Step 2: Verification via Polynomial Division

To further verify this result, we can perform polynomial division and find the expression of $ f(x) $. Dividing the polynomial

$$
3x^4 + x^3 - 25x^2 + 38x - 17
$$

by $ (x - 1) $ gives:

$$
f(x) = 3x^3 + 4x^2 - 21x + 17
$$

This is a **cubic (degree 3)** polynomial, confirming the earlier conclusion.

---

### Step 3: Confirming the Result

As a final check, we can multiply $ (x - 1) $ by $ f(x) $ to see if we recover the original polynomial:

$$
(x - 1)(3x^3 + 4x^2 - 21x + 17)
$$

Expanding:

- $ x(3x^3 + 4x^2 - 21x + 17) = 3x^4 + 4x^3 - 21x^2 + 17x $
- $ -1(3x^3 + 4x^2 - 21x + 17) = -3x^3 - 4x^2 + 21x - 17 $

Adding these:

$$
3x^4 + (4x^3 - 3x^3) + (-21x^2 - 4x^2) + (17x + 21x) - 17
= 3x^4 + x^3 - 25x^2 + 38x - 17
$$

This matches the original polynomial, confirming the correctness of $ f(x) $ and the resulting degree.

---

### Final Conclusion

The degree of the polynomial $ f(x) $ is:

$$
\boxed{3}
$$

Solution: Since the product of $f$ and a polynomial with degree 1 equals a polynomial with degree 4, we know that $f$ is a polynomial with degree $4-1=\boxed{3}$.


Explanation:
Okay, so I need to find the value of $ d $ in the equation:

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

for all angles $ x $. The problem states that constants $ a $, $ b $, $ c $, and $ d $ exist such that this equation holds. My task is specifically to find $ d $.

First, I recall that powers of sine functions can often be expressed as linear combinations of multiple angle sine functions. This is probably related to Fourier series or using trigonometric identities to expand higher powers. Since the left side is $ (\sin x)^7 $, which is an odd power, maybe I can use some multiple-angle identities or perhaps recursion relations.

I remember that there are formulas for expressing $ \sin^n x $ in terms of multiple angles. For example, for $ \sin^2 x $, we have the identity:

$$
\sin^2 x = \frac{1 - \cos 2x}{2}
$$

But here we have $ \sin^7 x $, which is a higher power. I need a way to express this as a combination of sines of multiple angles. Maybe I can use complex numbers? Euler's formula? Let me think.

Alternatively, I remember that using the binomial theorem might help, but since it's sine to the seventh power, perhaps expanding $ (\sin x)^7 $ using multiple-angle identities step by step. Let me recall that there are formulas for reducing powers of sine to multiple angles.

Another approach might be to use the identity for $ \sin^n x $ in terms of multiple angles. For example, for odd powers, we can express them as a sum of sine functions with different frequencies. Since the exponent is 7, which is odd, and we're dealing with multiple angles up to 7x, which is present on the right-hand side.

Alternatively, maybe I can use the method of expressing $ \sin^7 x $ in terms of complex exponentials. Let me try that.

Recall that $ \sin x = \frac{e^{ix} - e^{-ix}}{2i} $. Therefore, $ (\sin x)^7 $ would be:

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

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:

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

So, this is a sum of complex exponentials. Since $ \sin x $ is the imaginary part of $ e^{ix} $, perhaps we can take the imaginary part of this expression to get back to the sine functions. Wait, but we already started with the sine function. Maybe I need to consider that.

Alternatively, since $ (\sin x)^7 $ is real, the expression above must be real. Therefore, when we take the imaginary part of $ e^{i(7 - 2k)x} $, we might get terms that can be combined into sine functions. Let me think.

Let me write the expression again:

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

But $ (2i)^7 = 2^7 i^7 = 128 \cdot i^7 = 128 \cdot i^{4+3} = 128 \cdot (i^4 \cdot i^3) = 128 \cdot (1 \cdot (-i)) = -128i $.

Therefore, the entire expression becomes:

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

Multiplying numerator and denominator by $ i $ to make it easier:

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

Now, let's consider the sum $ \sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} $. Let me note that $ e^{i(7 - 2k)x} = e^{i7x} e^{-i2k x} $. Therefore, we can write:

$$
e^{i7x} \sum_{k=0}^7 \binom{7}{k} (-1)^k e^{-i2k x}
$$

So, this is $ e^{i7x} \times \sum_{k=0}^7 \binom{7}{k} (-1)^k (e^{-i2x})^k $

Which is $ e^{i7x} \times (1 - e^{-i2x})^7 $. Wait, because the binomial expansion of $ (a + b)^n $ is $ \sum_{k=0}^n \binom{n}{k} a^{n - k} b^k $. In this case, $ a = 1 $, $ b = -e^{-i2x} $, so the sum becomes $ (1 - e^{-i2x})^7 $.

Therefore, our entire expression becomes:

$$
\frac{i}{128} e^{i7x} (1 - e^{-i2x})^7
$$

Let me compute $ (1 - e^{-i2x})^7 $. Let me first compute $ 1 - e^{-i2x} $. Recall that $ 1 - e^{-i2x} = e^{-ix} (e^{ix} - e^{-ix}) = e^{-ix} \cdot 2i \sin x $. Therefore:

$$
1 - e^{-i2x} = 2i \sin x e^{-ix}
$$

Wait, let me verify:

$ e^{ix} - e^{-ix} = 2i \sin x $. So, $ 1 - e^{-i2x} = 1 - e^{-i2x} $. Alternatively, perhaps I should write:

Let me compute $ 1 - e^{-i2x} $:

Using Euler's formula, $ e^{-i2x} = \cos 2x - i \sin 2x $. Therefore,

$$
1 - e^{-i2x} = 1 - \cos 2x + i \sin 2x
$$

But that seems more complicated. However, if I think of $ 1 - e^{-i2x} $ as $ e^{-ix} \cdot 2i \sin x $? Let me check with another approach.

Let me write $ 1 - e^{-i2x} = e^{-ix} \cdot e^{ix} - e^{-i2x} = e^{-ix} (e^{ix} - e^{-ix}) $, which is $ e^{-ix} \cdot 2i \sin x $. Yes! That's correct. So:

$$
1 - e^{-i2x} = 2i \sin x \cdot e^{-ix}
$$

Therefore, substituting back into our expression:

$$
(1 - e^{-i2x})^7 = (2i \sin x \cdot e^{-ix})^7 = (2i)^7 \sin^7 x \cdot e^{-i7x}
$$

Therefore, substituting back into our earlier expression:

$$
\frac{i}{128} e^{i7x} \times (2i)^7 \sin^7 x \cdot e^{-i7x}
$$

Wait, let's compute that:

First, $ e^{i7x} \times e^{-i7x} = e^{0} = 1 $. So they cancel out. Then we have:

$$
\frac{i}{128} \times (2i)^7 \sin^7 x
$$

Compute $ (2i)^7 $:

$ 2^7 \times i^7 = 128 \times (-i) $, since $ i^7 = i^{4+3} = i^3 = -i $.

Therefore, the entire expression becomes:

$$
\frac{i}{128} \times 128 \times (-i) \sin^7 x = \frac{i \times (-i)}{1} \sin^7 x = (-i^2) \sin^7 x = -(-1) \sin^7 x = \sin^7 x
$$

Wait, so we just got back $ \sin^7 x $, which is the original expression. So this approach seems to bring us back to where we started. Therefore, not helpful in expressing $ \sin^7 x $ in terms of multiple sine functions.

Hmm. So perhaps I need a different approach. Let me think.

Alternatively, maybe using the identities for multiple angles. Let me recall that there is an identity for $ \sin^n x $ expressed as a sum of multiple angles. For example, for $ \sin^3 x $, we have:

$$
\sin^3 x = \frac{3 \sin x - \sin 3x}{4}
$$

Similarly, for higher powers. Maybe there's a general formula.

Alternatively, perhaps using recurrence relations. For example, using the identity involving $ \sin^{2n+1} x $ expressed as a combination of sines of multiple angles.

Alternatively, I can use the formula for $ \sin^7 x $ in terms of multiple angles. Let me try using the power-reduction formulas step by step.

First, note that $ \sin^7 x $ can be written as $ \sin x \cdot \sin^6 x $. Then, $ \sin^6 x $ can be expressed as $ (\sin^2 x)^3 $, which can be written using power-reduction formulas.

Let me try that. Start with $ \sin^2 x = \frac{1 - \cos 2x}{2} $. Therefore:

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

Expanding that:

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

Now, we need to express $ \cos^2 2x $ and $ \cos^3 2x $ in terms of multiple angles.

Starting with $ \cos^2 2x $:

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

For $ \cos^3 2x $, perhaps I can use the identity:

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

So replacing $ \theta $ with $ 2x $, we have:

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

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

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

Simplify term by term:

First term: $ 1 $

Second term: $ -3 \cos 2x $

Third term: $ 3 \cdot \frac{1 + \cos 4x}{2} = \frac{3}{2} + \frac{3}{2} \cos 4x $

Fourth term: $ - \frac{3 \cos 2x + \cos 6x}{4} = -\frac{3}{4} \cos 2x - \frac{1}{4} \cos 6x $

Now combine all terms inside the brackets:

$$
1 - 3 \cos 2x + \frac{3}{2} + \frac{3}{2} \cos 4x - \frac{3}{4} \cos 2x - \frac{1}{4} \cos 6x
$$

Combine like terms:

Constants: $ 1 + \frac{3}{2} = \frac{5}{2} $

Cos 2x terms: $ -3 \cos 2x - \frac{3}{4} \cos 2x = -\frac{15}{4} \cos 2x $

Cos 4x term: $ \frac{3}{2} \cos 4x $

Cos 6x term: $ - \frac{1}{4} \cos 6x $

Therefore, the entire expression becomes:

$$
\frac{1}{8} \left[ \frac{5}{2} - \frac{15}{4} \cos 2x + \frac{3}{2} \cos 4x - \frac{1}{4} \cos 6x \right]
$$

Multiply through by $ \frac{1}{8} $:

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

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

Now, going back to $ \sin^7 x $, which is $ \sin x \cdot \sin^6 x $. Therefore, we can write:

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

Now, we need to multiply this by $ \sin x $ and express the result in terms of sine functions. Remember that $ \sin x \cos nx $ can be written using the identity $ \sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)] $.

Let me compute each term separately.

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 $

Let me handle each term:

First term: $ \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 ] $

So, putting all these 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. Let's collect all the coefficients for each sine term:

For $ \sin x $:

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

Second term: $ + \frac{15}{64} $

So total: $ \frac{5}{16} + \frac{15}{64} = \frac{20}{64} + \frac{15}{64} = \frac{35}{64} $

For $ \sin 3x $:

Second term: $ -\frac{15}{64} $

Third term: $ -\frac{3}{32} $

Convert to 64 denominators:

$ -\frac{15}{64} - \frac{6}{64} = -\frac{21}{64} $

For $ \sin 5x $:

Third term: $ + \frac{3}{32} $

Fourth term: $ + \frac{1}{64} $

Convert to 64 denominators:

$ \frac{6}{64} + \frac{1}{64} = \frac{7}{64} $

For $ \sin 7x $:

Fourth term: $ -\frac{1}{64} $

So putting it all together:

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

Therefore, comparing with the original equation:

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

We can see that:

- 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 verify this result to make sure I didn't make any calculation errors. Let me check each step again.

Starting with expanding $ \sin^6 x $:

Original steps:

1. $ \sin^2 x = \frac{1 - \cos 2x}{2} $

So $ \sin^6 x = (\sin^2 x)^3 = \left( \frac{1 - \cos 2x}{2} \right)^3 $. Then expanding that cube:

Yes, the expansion is $ \frac{1}{8}(1 - 3\cos 2x + 3\cos^2 2x - \cos^3 2x) $. That seems correct.

Then, for $ \cos^2 2x $, using $ \frac{1 + \cos 4x}{2} $, correct.

For $ \cos^3 2x $, using $ \frac{3 \cos 2x + \cos 6x}{4} $, correct. Then substituting back:

$$
\frac{1}{8} [ 1 - 3 \cos 2x + 3*(1 + \cos 4x)/2 - (3 \cos 2x + \cos 6x)/4 ]

Yes, that seems correct. Then combining terms step by step:

Constant term: 1 + 3/2 = 5/2, yes.

Coefficient for cos 2x: -3 - 3/4 = -15/4, yes.

Coefficient for cos 4x: 3/2, yes.

Coefficient for cos 6x: -1/4, yes. Then multiplying by 1/8 gives 5/16, -15/32, 3/16, and -1/32. That seems correct.

Then multiplying by sin x term by term:

First term: 5/16 sinx, okay.

Second term: -15/32 sinx cos2x, which expanded as per identity gives -15/64 [sin3x - sinx], correct.

Third term: 3/16 sinx cos4x gives 3/32 [sin5x - sin3x], correct.

Fourth term: -1/32 sinx cos6x gives -1/64 [sin7x - sin5x], correct.

Then combining coefficients:

For sinx: 5/16 + 15/64 = 35/64, yes.

For sin3x: -15/64 - 3/32 = -15/64 -6/64 = -21/64, correct.

For sin5x: 3/32 + 1/64 = 7/64, yes.

For sin7x: -1/64, correct.

Therefore, the coefficients are correct. Therefore, d is 35/64. Therefore, the answer is 35/64.

But just to make sure, maybe I can check with a specific value of x. Let's pick x = 0. Wait, but sin(0) = 0, so both sides would be 0. Not helpful. Let me try x = π/2.

At x = π/2:

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

So substituting:

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

According to our coefficients:

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

So:

-(-1/64) + (7/64) - (-21/64) + (35/64) = 1/64 + 7/64 + 21/64 + 35/64 = (1 + 7 + 21 + 35)/64 = 64/64 = 1, which matches the left side. Good.

Another check: x = π/4. Let me compute both sides.

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/(sqrt(2^7)) = 1/(sqrt(128)) = 1/(8*sqrt(2)) ≈ 0.088388...

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:

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

Compute [ -a - b + c + d ] with a = -1/64, b =7/64, c = -21/64, d =35/64:

-(-1/64) - (7/64) + (-21/64) + 35/64 = 1/64 -7/64 -21/64 +35/64 = (1 -7 -21 +35)/64 = (8)/64 = 1/8

Multiply by sqrt(2)/2: 1/8 * sqrt(2)/2 = sqrt(2)/16 ≈ 0.088388...

Which matches the left side. So that seems correct.

Therefore, my answer for d is 35/64. Therefore, I think that is correct. However, since the problem says "Find d", and I need to present it in boxed notation. So \boxed{\dfrac{35}{64}}.

But let me check one more value just to be safe. Let me take x = π/6.

Left side: sin(π/6)^7 = (1/2)^7 = 1/128 ≈ 0.0078125

Right side: a sin(7π/6) + b sin(5π/6) + c sin(3π/6) + d sin(π/6)

Compute each term:

sin(7π/6) = -1/2

sin(5π/6) = 1/2

sin(3π/6) = sin(π/2) = 1

sin(π/6) = 1/2

Therefore:

a*(-1/2) + b*(1/2) + c*(1) + d*(1/2) = (-a/2 + b/2 + c + d/2)

Plug in the values:

-(-1/64)/2 + (7/64)/2 + (-21/64) + (35/64)/2

Compute each term:

First term: (1/64)/2 = 1/128

Second term: (7/64)/2 = 7/128

Third term: -21/64 = -42/128

Fourth term: (35/64)/2 = 35/128

Adding them up:

1/128 + 7/128 - 42/128 + 35/128 = (1 + 7 - 42 + 35)/128 = (1 + 7 + 35 - 42)/128 = (43 - 42)/128 = 1/128. Which matches the left side. Perfect.

Therefore, my value of d is indeed 35/64. So I can be confident that this is the correct answer.

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

To determine the value of $ d $ in the equation

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

we aim to express $ (\sin x)^7 $ as a linear combination of sines of different multiples of $ x $.

---

### Step 1: Using a Known Expansion

One powerful method to achieve this is to use a known identity or expansion of $ \sin^n x $ in terms of multiple sine functions. A well-known technique involves breaking down $ \sin^7 x $ through a step-by-step application of trigonometric identities, starting from $ \sin^2 x $, then $ \sin^4 x $, and so on.

Following the method of expressing powers of sine in terms of multiple angles, we find the identity:

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

This expresses $ \sin^7 x $ as a sum of sine functions with coefficients:

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

---

### Step 2: Comparing Coefficients

Now, comparing this with the original equation:

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

we see that the coefficients of the sines on both sides correspond directly. Therefore, the coefficient of $ \sin x $, which is $ d $, is equal to:

$$
d = \frac{35}{64}.
$$

---

### Final Answer

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

[Assistant]
Explanation:
Okay, so I need to find the value of $ d $ in the equation:

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

for all angles $ x $. The problem states that constants $ a $, $ b $, $ c $, and $ d $ exist such that this equation holds. My task is specifically to find $ d $.

First, I recall that powers of sine functions can often be expressed as linear combinations of multiple angle sine functions. This is probably related to Fourier series or using trigonometric identities to expand higher powers. Since the left side is $ (\sin x)^7 $, which is an odd power, maybe I can use some multiple-angle identities or perhaps recursion relations.

I remember that there are formulas for expressing $ \sin^n x $ in terms of multiple angles. For example, for $ \sin^2 x $, we have the identity:

$$
\sin^2 x = \frac{1 - \cos 2x}{2}
$$

But here we have $ \sin^7 x $, which is a higher power. I need a way to express this as a combination of sines of multiple angles. Maybe I can use complex numbers? Euler's formula? Let me think.

Alternatively, I remember that using the binomial theorem might help, but since it's sine to the seventh power, perhaps expanding $ (\sin x)^7 $ using multiple-angle identities step by step. Let me recall that there are formulas for reducing powers of sine to multiple angles.

Another approach might be to use the identity for $ \sin^n x $ in terms of multiple angles. For example, for odd powers, we can express them as a sum of sine functions with different frequencies. Since the exponent is 7, which is odd, and we're dealing with multiple angles up to 7x, which is present on the right-hand side.

Alternatively, maybe I can use the method of expressing $ \sin^7 x $ in terms of complex exponentials. Let me try that.

Recall that $ \sin x = \frac{e^{ix} - e^{-ix}}{2i} $. Therefore, $ (\sin x)^7 $ would be:

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

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:

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

So, this is a sum of complex exponentials. Since $ \sin x $ is the imaginary part of $ e^{ix} $, perhaps we can take the imaginary part of this expression to get back to the sine functions. Wait, but we already started with the sine function. Maybe I need to consider that.

Alternatively, since $ (\sin x)^7 $ is real, the expression above must be real. Therefore, when we take the imaginary part of $ e^{i(7 - 2k)x} $, we might get terms that can be combined into sine functions. Let me think.

Let me write the expression again:

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

But $ (2i)^7 = 2^7 i^7 = 128 \cdot i^7 = 128 \cdot i^{4+3} = 128 \cdot (i^4 \cdot i^3) = 128 \cdot (1 \cdot (-i)) = -128i $.

Therefore, the entire expression becomes:

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

Multiplying numerator and denominator by $ i $ to make it easier:

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

Now, let's consider the sum $ \sum_{k=0}^7 \binom{7}{k} (-1)^k e^{i(7 - 2k)x} $. Let me note that $ e^{i(7 - 2k)x} = e^{i7x} e^{-i2k x} $. Therefore, we can write:

$$
e^{i7x} \sum_{k=0}^7 \binom{7}{k} (-1)^k e^{-i2k x}
$$

So, this is $ e^{i7x} \times \sum_{k=0}^7 \binom{7}{k} (-1)^k (e^{-i2x})^k $

Which is $ e^{i7x} \times (1 - e^{-i2x})^7 $. Wait, because the binomial expansion of $ (a + b)^n $ is $ \sum_{k=0}^n \binom{n}{k} a^{n - k} b^k $. In this case, $ a = 1 $, $ b = -e^{-i2x} $, so the sum becomes $ (1 - e^{-i2x})^7 $.

Therefore, our entire expression becomes:

$$
\frac{i}{128} e^{i7x} (1 - e^{-i2x})^7
$$

Let me compute $ (1 - e^{-i2x})^7 $. Let me first compute $ 1 - e^{-i2x} $. Recall that $ 1 - e^{-i2x} = e^{-ix} (e^{ix} - e^{-ix}) = e^{-ix} \cdot 2i \sin x $. Therefore:

$$
1 - e^{-i2x} = 2i \sin x e^{-ix}
$$

Wait, let me verify:

$ e^{ix} - e^{-ix} = 2i \sin x $. So, $ 1 - e^{-i2x} = 1 - e^{-i2x} $. Alternatively, perhaps I should write:

Let me compute $ 1 - e^{-i2x} $:

Using Euler's formula, $ e^{-i2x} = \cos 2x - i \sin 2x $. Therefore,

$$
1 - e^{-i2x} = 1 - \cos 2x + i \sin 2x
$$

But that seems more complicated. However, if I think of $ 1 - e^{-i2x} $ as $ e^{-ix} \cdot 2i \sin x $? Let me check with another approach.

Let me write $ 1 - e^{-i2x} = e^{-ix} \cdot e^{ix} - e^{-i2x} = e^{-ix} (e^{ix} - e^{-ix}) $, which is $ e^{-ix} \cdot 2i \sin x $. Yes! That's correct. So:

$$
1 - e^{-i2x} = 2i \sin x \cdot e^{-ix}
$$

Therefore, substituting back into our expression:

$$
(1 - e^{-i2x})^7 = (2i \sin x \cdot e^{-ix})^7 = (2i)^7 \sin^7 x \cdot e^{-i7x}
$$

Therefore, substituting back into our earlier expression:

$$
\frac{i}{128} e^{i7x} \times (2i)^7 \sin^7 x \cdot e^{-i7x}
$$

Wait, let's compute that:

First, $ e^{i7x} \times e^{-i7x} = e^{0} = 1 $. So they cancel out. Then we have:

$$
\frac{i}{128} \times (2i)^7 \sin^7 x
$$

Compute $ (2i)^7 $:

$ 2^7 \times i^7 = 128 \times (-i) $, since $ i^7 = i^{4+3} = i^3 = -i $.

Therefore, the entire expression becomes:

$$
\frac{i}{128} \times 128 \times (-i) \sin^7 x = \frac{i \times (-i)}{1} \sin^7 x = (-i^2) \sin^7 x = -(-1) \sin^7 x = \sin^7 x
$$

Wait, so we just got back $ \sin^7 x $, which is the original expression. So this approach seems to bring us back to where we started. Therefore, not helpful in expressing $ \sin^7 x $ in terms of multiple sine functions.

Hmm. So perhaps I need a different approach. Let me think.

Alternatively, maybe using the identities for multiple angles. Let me recall that there is an identity for $ \sin^n x $ expressed as a sum of multiple angles. For example, for $ \sin^3 x $, we have:

$$
\sin^3 x = \frac{3 \sin x - \sin 3x}{4}
$$

Similarly, for higher powers. Maybe there's a general formula.

Alternatively, perhaps using recurrence relations. For example, using the identity involving $ \sin^{2n+1} x $ expressed as a combination of sines of multiple angles.

Alternatively, I can use the formula for $ \sin^7 x $ in terms of multiple angles. Let me try using the power-reduction formulas step by step.

First, note that $ \sin^7 x $ can be written as $ \sin x \cdot \sin^6 x $. Then, $ \sin^6 x $ can be expressed as $ (\sin^2 x)^3 $, which can be written using power-reduction formulas.

Let me try that. Start with $ \sin^2 x = \frac{1 - \cos 2x}{2} $. Therefore:

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

Expanding that:

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

Now, we need to express $ \cos^2 2x $ and $ \cos^3 2x $ in terms of multiple angles.

Starting with $ \cos^2 2x $:

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

For $ \cos^3 2x $, perhaps I can use the identity:

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

So replacing $ \theta $ with $ 2x $, we have:

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

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

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

Simplify term by term:

First term: $ 1 $

Second term: $ -3 \cos 2x $

Third term: $ 3 \cdot \frac{1 + \cos 4x}{2} = \frac{3}{2} + \frac{3}{2} \cos 4x $

Fourth term: $ - \frac{3 \cos 2x + \cos 6x}{4} = -\frac{3}{4} \cos 2x - \frac{1}{4} \cos 6x $

Now combine all terms inside the brackets:

$$
1 - 3 \cos 2x + \frac{3}{2} + \frac{3}{2} \cos 4x - \frac{3}{4} \cos 2x - \frac{1}{4} \cos 6x
$$

Combine like terms:

Constants: $ 1 + \frac{3}{2} = \frac{5}{2} $

Cos 2x terms: $ -3 \cos 2x - \frac{3}{4} \cos 2x = -\frac{15}{4} \cos 2x $

Cos 4x term: $ \frac{3}{2} \cos 4x $

Cos 6x term: $ - \frac{1}{4} \cos 6x $

Therefore, the entire expression becomes:

$$
\frac{1}{8} \left[ \frac{5}{2} - \frac{15}{4} \cos 2x + \frac{3}{2} \cos 4x - \frac{1}{4} \cos 6x \right]
$$

Multiply through by $ \frac{1}{8} $:

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

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

Now, going back to $ \sin^7 x $, which is $ \sin x \cdot \sin^6 x $. Therefore, we can write:

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

Now, we need to multiply this by $ \sin x $ and express the result in terms of sine functions. Remember that $ \sin x \cos nx $ can be written using the identity $ \sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)] $.

Let me compute each term separately.

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 $

Let me handle each term:

First term: $ \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 ] $

So, putting all these 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. Let's collect all the coefficients for each sine term:

For $ \sin x $:

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

Second term: $ + \frac{15}{64} $

So total: $ \frac{5}{16} + \frac{15}{64} = \frac{20}{64} + \frac{15}{64} = \frac{35}{64} $

For $ \sin 3x $:

Second term: $ -\frac{15}{64} $

Third term: $ -\frac{3}{32} $

Convert to 64 denominators:

$ -\frac{15}{64} - \frac{6}{64} = -\frac{21}{64} $

For $ \sin 5x $:

Third term: $ + \frac{3}{32} $

Fourth term: $ + \frac{1}{64} $

Convert to 64 denominators:

$ \frac{6}{64} + \frac{1}{64} = \frac{7}{64} $

For $ \sin 7x $:

Fourth term: $ -\frac{1}{64} $

So putting it all together:

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

Therefore, comparing with the original equation:

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

We can see that:

- 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 verify this result to make sure I didn't make any calculation errors. Let me check each step again.

Starting with expanding $ \sin^6 x $:

Original steps:

1. $ \sin^2 x = \frac{1 - \cos 2x}{2} $

So $ \sin^6 x = (\sin^2 x)^3 = \left( \frac{1 - \cos 2x}{2} \right)^3 $. Then expanding that cube:

Yes, the expansion is $ \frac{1}{8}(1 - 3\cos 2x + 3\cos^2 2x - \cos^3 2x) $. That seems correct.

Then, for $ \cos^2 2x $, using $ \frac{1 + \cos 4x}{2} $, correct.

For $ \cos^3 2x $, using $ \frac{3 \cos 2x + \cos 6x}{4} $, correct. Then substituting back:

$$
\frac{1}{8} [ 1 - 3 \cos 2x + 3*(1 + \cos 4x)/2 - (3 \cos 2x + \cos 6x)/4 ]

Yes, that seems correct. Then combining terms step by step:

Constant term: 1 + 3/2 = 5/2, yes.

Coefficient for cos 2x: -3 - 3/4 = -15/4, yes.

Coefficient for cos 4x: 3/2, yes.

Coefficient for cos 6x: -1/4, yes. Then multiplying by 1/8 gives 5/16, -15/32, 3/16, and -1/32. That seems correct.

Then multiplying by sin x term by term:

First term: 5/16 sinx, okay.

Second term: -15/32 sinx cos2x, which expanded as per identity gives -15/64 [sin3x - sinx], correct.

Third term: 3/16 sinx cos4x gives 3/32 [sin5x - sin3x], correct.

Fourth term: -1/32 sinx cos6x gives -1/64 [sin7x - sin5x], correct.

Then combining coefficients:

For sinx: 5/16 + 15/64 = 35/64, yes.

For sin3x: -15/64 - 3/32 = -15/64 -6/64 = -21/64, correct.

For sin5x: 3/32 + 1/64 = 7/64, yes.

For sin7x: -1/64, correct.

Therefore, the coefficients are correct. Therefore, d is 35/64. Therefore, the answer is 35/64.

But just to make sure, maybe I can check with a specific value of x. Let's pick x = 0. Wait, but sin(0) = 0, so both sides would be 0. Not helpful. Let me try x = π/2.

At x = π/2:

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

So substituting:

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

According to our coefficients:

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

So:

-(-1/64) + (7/64) - (-21/64) + (35/64) = 1/64 + 7/64 + 21/64 + 35/64 = (1 + 7 + 21 + 35)/64 = 64/64 = 1, which matches the left side. Good.

Another check: x = π/4. Let me compute both sides.

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/(sqrt(2^7)) = 1/(sqrt(128)) = 1/(8*sqrt(2)) ≈ 0.088388...

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:

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

Compute [ -a - b + c + d ] with a = -1/64, b =7/64, c = -21/64, d =35/64:

-(-1/64) - (7/64) + (-21/64) + 35/64 = 1/64 -7/64 -21/64 +35/64 = (1 -7 -21 +35)/64 = (8)/64 = 1/8

Multiply by sqrt(2)/2: 1/8 * sqrt(2)/2 = sqrt(2)/16 ≈ 0.088388...

Which matches the left side. So that seems correct.

Therefore, my answer for d is 35/64. Therefore, I think that is correct. However, since the problem says "Find d", and I need to present it in boxed notation. So \boxed{\dfrac{35}{64}}.

But let me check one more value just to be safe. Let me take x = π/6.

Left side: sin(π/6)^7 = (1/2)^7 = 1/128 ≈ 0.0078125

Right side: a sin(7π/6) + b sin(5π/6) + c sin(3π/6) + d sin(π/6)

Compute each term:

sin(7π/6) = -1/2

sin(5π/6) = 1/2

sin(3π/6) = sin(π/2) = 1

sin(π/6) = 1/2

Therefore:

a*(-1/2) + b*(1/2) + c*(1) + d*(1/2) = (-a/2 + b/2 + c + d/2)

Plug in the values:

-(-1/64)/2 + (7/64)/2 + (-21/64) + (35/64)/2

Compute each term:

First term: (1/64)/2 = 1/128

Second term: (7/64)/2 = 7/128

Third term: -21/64 = -42/128

Fourth term: (35/64)/2 = 35/128

Adding them up:

1/128 + 7/128 - 42/128 + 35/128 = (1 + 7 - 42 + 35)/128 = (1 + 7 + 35 - 42)/128 = (43 - 42)/128 = 1/128. Which matches the left side. Perfect.

Therefore, my value of d is indeed 35/64. So I can be confident that this is the correct answer.

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

To determine the value of $ d $ in the equation

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

we aim to express $ (\sin x)^7 $ as a linear combination of sines of different multiples of $ x $.

---

### Step 1: Using a Known Expansion

One powerful method to achieve this is to use a known identity or expansion of $ \sin^n x $ in terms of multiple sine functions. A well-known technique involves breaking down $ \sin^7 x $ through a step-by-step application of trigonometric identities, starting from $ \sin^2 x $, then $ \sin^4 x $, and so on.

Following the method of expressing powers of sine in terms of multiple angles, we find the identity:

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

This expresses $ \sin^7 x $ as a sum of sine functions with coefficients:

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

---

### Step 2: Comparing Coefficients

Now, comparing this with the original equation:

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

we see that the coefficients of the sines on both sides correspond directly. Therefore, the coefficient of $ \sin x $, which is $ d $, is equal to:

$$
d = \frac{35}{64}.
$$

---

### 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 percent the area of a square decreases when two opposite sides are decreased by 40% and the other two sides are increased by 50%, turning it into a rectangle. Let me think through this step by step.

First, let me recall that 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 multiplied by s, which is s². Got that down.

Now, the problem says that two opposite sides are decreased by 40%. Let me visualize the square. If I imagine a square, there are two pairs of opposite sides. Let me say that the original square has sides of length 's' horizontally and vertically. If I decrease two opposite sides by 40%, I need to decide which sides those are. But since it's a square, all sides are equal, so decreasing two opposite sides by 40% would mean that those sides become shorter, while the other two sides are increased by 50%.

Wait, but hold on. If it's a square, all sides are the same. So if two opposite sides are decreased by 40%, then the other two sides are increased by 50%? But the problem says "the lengths of two opposite sides... are decreased by 40% while the lengths of the other two sides are increased by 50%". So, the resulting figure is a rectangle. So, if you take a square, and stretch two opposite sides and shrink the other two, you get a rectangle.

But let me make sure. Let me think of a square with side length 's'. Let me say that the original square has sides of length 's'. So, if I take two opposite sides (let's say the top and bottom sides) and decrease their lengths by 40%, then their new lengths would be s - 40% of s. Which is s * (1 - 0.4) = 0.6s. Similarly, the other two sides (left and right sides) are increased by 50%, so their new lengths would be s + 50% of s = s * 1.5 = 1.5s. Wait, but in a rectangle, opposite sides are equal. So, if the top and bottom sides are now 0.6s, then the left and right sides are 1.5s. Therefore, the rectangle has length 1.5s and width 0.6s? Wait, but depending on which sides we are talking about. Wait, actually, if the original square had all sides as s, then when we decrease two opposite sides by 40%, perhaps they become 0.6s, and the other two sides are increased by 50%, becoming 1.5s. Therefore, the resulting rectangle would have sides of 0.6s and 1.5s? Wait, but in a rectangle, there are two different lengths. Let me make sure.

Alternatively, maybe when they say "the lengths of two opposite sides are decreased by 40%", they mean that for example, the length is decreased by 40% and the width is increased by 50%? Wait, but in a square, length and width are the same. So maybe we need to think about changing two sides. Let me think again.

Let me take the square with side length 's'. Let me consider that the square has sides of length 's' horizontally and vertically. If I decrease two opposite sides (say, the horizontal sides) by 40%, then the horizontal sides become 0.6s each, and the vertical sides are increased by 50%, so they become 1.5s each. Therefore, the resulting figure is a rectangle with length 1.5s and width 0.6s? Wait, but in that case, the sides are 1.5s and 0.6s. But maybe I need to clarify which sides are being changed. However, since it's a square, all sides are the same. So changing two opposite sides (either horizontal or vertical) by 40% and the other two by 50% will result in a rectangle where the sides are 0.6s and 1.5s. Therefore, the area of the rectangle would be (0.6s) * (1.5s) = 0.9s². Wait, original area was s², so the new area is 0.9s²? So the area decreased by 10%? Wait, that seems straightforward, but let me verify.

Alternatively, maybe I need to check again. Let me think again. Suppose original square has side length s. So original area is s². Now, two opposite sides are decreased by 40%. Let me think of the square as having length and width both equal to s. If we decrease the length by 40%, then the new length is 0.6s. If we increase the width by 50%, the new width is 1.5s. Therefore, the area is 0.6s * 1.5s = 0.9s². So the area decreased by 10%? So the answer would be 10%? But wait, maybe there's a mistake here. Because when they say "the lengths of two opposite sides are decreased by 40% while the lengths of the other two sides are increased by 50%", so perhaps it's not decreasing length and increasing width, but maybe decreasing both horizontal sides and increasing both vertical sides? But since the square is symmetric, the result is the same. Therefore, the area is 0.6s * 1.5s = 0.9s². So the area is 90% of the original, meaning a 10% decrease. Therefore, the answer is 10%? Wait, but let me check again.

Alternatively, maybe I need to think of changing two sides in a different way? For example, suppose that changing the sides affects the same dimension? Wait, but in a square, all sides are the same. So for example, if you take two opposite sides (say, top and bottom) and decrease them by 40%, then those are horizontal sides, so the length is decreased. Then the other two sides (left and right) are vertical sides, which are increased by 50%. Therefore, the horizontal sides are now shorter, and vertical sides are longer. Therefore, the rectangle would have length 0.6s and width 1.5s? Or is it length 1.5s and width 0.6s? Doesn't matter, because area is length times width. So 0.6s * 1.5s = 0.9s². Thus, the area is decreased by 10%? That seems straightforward. But maybe I need to check again.

Alternatively, perhaps I made an error in assuming which sides are being changed? Let me try with numbers. Let me take a square with side length 10 units. Original area is 10*10=100. If two opposite sides are decreased by 40%, so 40% of 10 is 4, so the new length is 10 - 4 = 6. The other two sides are increased by 50%, so 50% of 10 is 5, so new length is 10 + 5 = 15. Therefore, the new rectangle has sides 6 and 15. Area is 6*15=90. Original area 100, so 90 is 10% less. Therefore, 10% decrease. So that seems to check out. So the answer is 10%? Therefore, maybe the answer is 10%? But the problem says "by what percent does the square's area decrease?" So 10% decrease.

Wait, but let me think again. Maybe there's a different way that the problem is interpreted? For example, maybe the problem is not changing the sides in the way I thought? Let me read the problem again:

"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. By what percent does the square's area decrease?"

So, yes, two opposite sides are decreased by 40%, other two increased by 50%. Therefore, as we saw, the area becomes 0.9s², which is 10% less. But maybe I should approach this algebraically to make sure.

Let me define the original square with side length s. Original area = 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 rectangle has sides of 0.6s and 1.5s. Therefore, area is 0.6s * 1.5s = 0.9s². So the area is decreased by 10%.

Alternatively, let me compute the percentage decrease. The formula for percentage decrease is [(Original Area - New Area)/Original Area] * 100%.

So that is [(s² - 0.9s²)/s²] * 100% = (0.1s² / s²) * 100% = 10%. So that's 10% decrease. Therefore, the answer is 10%.

But just to be thorough, let me consider another approach. Suppose instead of taking the sides as length and width, maybe there's a different way to interpret the problem? For example, maybe decreasing two opposite sides by 40% refers to decreasing each of those sides by 40% of the original length? Which is how I interpreted it. So, for example, if each side is decreased by 40%, so each of the two sides becomes 60% of original. Similarly, other two sides become 150% of original. Therefore, the rectangle sides are 0.6s and 1.5s. Therefore, area is 0.9s². So same result.

Alternatively, suppose someone thought that decreasing two opposite sides by 40% meant that the total length of those sides is decreased by 40%? Wait, but that would be different. For example, if you have two sides, each of length s, and you decrease the total length by 40%, then total length of those two sides would be 2s - 0.4*(2s) = 2s - 0.8s = 1.2s. But that would mean that each side is decreased by 0.4s total? But that would mean that each side is decreased by 0.2s, so 20% decrease. But this is not what the problem says. The problem says "the lengths of two opposite sides are decreased by 40%", which is more naturally interpreted as each of those sides is decreased by 40%, not the total length of both sides. So, for example, if you have two sides, each of length s, and you decrease each by 40%, meaning each becomes 0.6s. So that seems correct.

Therefore, the answer is 10%. Wait, but let me check if maybe the problem is referring to something else. For example, sometimes percentage changes can be confusing. Let me check with another example. Suppose original square has side length 1, area 1. Decrease two opposite sides by 40%: 1 - 0.4 = 0.6. Increase other two sides by 50%: 1 + 0.5 = 1.5. Area is 0.6 * 1.5 = 0.9. So decrease by 0.1, which is 10%. So yes. So 10% decrease. So answer is 10%.

But wait, maybe I need to check the problem again? Wait, 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%". So, the square has four sides. Two opposite sides are decreased by 40%: so each of those sides becomes 60% of original. The other two sides are increased by 50%, so each of those sides becomes 150% of original. Therefore, the rectangle formed has sides of 0.6s and 1.5s, hence area is 0.6*1.5*s² = 0.9s², which is 10% less than original. So that's correct.

But maybe I should check if there is a different interpretation. For example, if the problem meant that the total length of two opposite sides is decreased by 40%, but that seems less likely. For example, if you have two opposite sides, each of length s. If you decrease their total length by 40%, you have 2s - 0.4*2s = 1.2s. If that's spread over the two sides, then each side would be 0.6s, which is same as before. So same result. Therefore, either way, you get the same result. Therefore, answer is 10%.

Alternatively, maybe someone might think that decreasing two opposite sides by 40% means that each side is decreased by 40% of the original length, which is what I did. So the answer is 10%.

Therefore, I think the answer is 10% decrease. Therefore, I should put \boxed{10} as the final answer.

Wait, but just to be 100% sure, let me think again. Let me take another example with numbers. Let me take a square with side length 20. Original area is 400. Decrease two opposite sides by 40%: 40% of 20 is 8, so new length is 20 - 8 = 12. Increase the other two sides by 50%: 50% of 20 is 10, so new length is 20 + 10 = 30. Therefore, the area is 12 * 30 = 360. Original area 400, new area 360. The decrease is 40. Percentage decrease is (40 / 400) * 100 = 10%. Yep, same result. So 10% decrease. So I think this is correct.

Therefore, I think the answer is 10%. So the square's area decreases by 10%.

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

To determine the percentage decrease in the area of a square when two opposite sides are each decreased by 40% and the other two sides are increased by 50%, we follow a structured approach:

---

### Step 1: Define the Original Square

Let the side length of the original square be $ s $.
The area of the original square is:

$$
\text{Original Area} = s \times s = s^2
$$

---

### Step 2: Apply the Changes to the Sides

- Two opposite sides (say, the top and bottom) are decreased by 40%.
  The new length of these sides becomes:

  $$
  s - 0.4s = 0.6s
  $$

- The other two opposite sides (say, the left and right) are increased by 50%.
  The new length of these sides becomes:

  $$
  s + 0.5s = 1.5s
  $$

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

---

### Step 3: Calculate the New Area

$$
\text{New Area} = (0.6s)(1.5s) = 0.9s^2
$$

---

### Step 4: Determine the Percentage Change in Area

$$
\text{Percentage Decrease} = \left( \frac{\text{Original Area} - \text{New Area}}{\text{Original Area}} \right) \times 100\%
$$

$$
= \left( \frac{s^2 - 0.9s^2}{s^2} \right) \times 100\% = \left( \frac{0.1s^2}{s^2} \right) \times 100\% = 10\%
$$

---

### Final Answer

The area of the square decreases by:

$$
\boxed{10}
$$

percent.

[Assistant]
Explanation:
Okay, so I need to figure out by what percent the area of a square decreases when two opposite sides are decreased by 40% and the other two sides are increased by 50%, turning it into a rectangle. Let me think through this step by step.

First, let me recall that 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 multiplied by s, which is s². Got that down.

Now, the problem says that two opposite sides are decreased by 40%. Let me visualize the square. If I imagine a square, there are two pairs of opposite sides. Let me say that the original square has sides of length 's' horizontally and vertically. If I decrease two opposite sides by 40%, I need to decide which sides those are. But since it's a square, all sides are equal, so decreasing two opposite sides by 40% would mean that those sides become shorter, while the other two sides are increased by 50%.

Wait, but hold on. If it's a square, all sides are the same. So if two opposite sides are decreased by 40%, then the other two sides are increased by 50%? But the problem says "the lengths of two opposite sides... are decreased by 40% while the lengths of the other two sides are increased by 50%". So, the resulting figure is a rectangle. So, if you take a square, and stretch two opposite sides and shrink the other two, you get a rectangle.

But let me make sure. Let me think of a square with side length 's'. Let me say that the original square has sides of length 's'. So, if I take two opposite sides (let's say the top and bottom sides) and decrease their lengths by 40%, then their new lengths would be s - 40% of s. Which is s * (1 - 0.4) = 0.6s. Similarly, the other two sides (left and right sides) are increased by 50%, so their new lengths would be s + 50% of s = s * 1.5 = 1.5s. Wait, but in a rectangle, opposite sides are equal. So, if the top and bottom sides are now 0.6s, then the left and right sides are 1.5s. Therefore, the rectangle has length 1.5s and width 0.6s? Wait, but depending on which sides we are talking about. Wait, actually, if the original square had all sides as s, then when we decrease two opposite sides by 40%, perhaps they become 0.6s, and the other two sides are increased by 50%, becoming 1.5s. Therefore, the resulting rectangle would have sides of 0.6s and 1.5s? Wait, but in a rectangle, there are two different lengths. Let me make sure.

Alternatively, maybe when they say "the lengths of two opposite sides are decreased by 40%", they mean that for example, the length is decreased by 40% and the width is increased by 50%? Wait, but in a square, length and width are the same. So maybe we need to think about changing two sides. Let me think again.

Let me take the square with side length 's'. Let me consider that the square has sides of length 's' horizontally and vertically. If I decrease two opposite sides (say, the horizontal sides) by 40%, then the horizontal sides become 0.6s each, and the vertical sides are increased by 50%, so they become 1.5s each. Therefore, the resulting figure is a rectangle with length 1.5s and width 0.6s? Wait, but in that case, the sides are 1.5s and 0.6s. But maybe I need to clarify which sides are being changed. However, since it's a square, all sides are the same. So changing two opposite sides (either horizontal or vertical) by 40% and the other two by 50% will result in a rectangle where the sides are 0.6s and 1.5s. Therefore, the area of the rectangle would be (0.6s) * (1.5s) = 0.9s². Wait, original area was s², so the new area is 0.9s²? So the area decreased by 10%? Wait, that seems straightforward, but let me verify.

Alternatively, maybe I need to check again. Let me think again. Suppose original square has side length s. So original area is s². Now, two opposite sides are decreased by 40%. Let me think of the square as having length and width both equal to s. If we decrease the length by 40%, then the new length is 0.6s. If we increase the width by 50%, the new width is 1.5s. Therefore, the area is 0.6s * 1.5s = 0.9s². So the area decreased by 10%? So the answer would be 10%? But wait, maybe there's a mistake here. Because when they say "the lengths of two opposite sides are decreased by 40% while the lengths of the other two sides are increased by 50%", so perhaps it's not decreasing length and increasing width, but maybe decreasing both horizontal sides and increasing both vertical sides? But since the square is symmetric, the result is the same. Therefore, the area is 0.6s * 1.5s = 0.9s². So the area is 90% of the original, meaning a 10% decrease. Therefore, the answer is 10%? Wait, but let me check again.

Alternatively, maybe I need to think of changing two sides in a different way? For example, suppose that changing the sides affects the same dimension? Wait, but in a square, all sides are the same. So for example, if you take two opposite sides (say, top and bottom) and decrease them by 40%, then those are horizontal sides, so the length is decreased. Then the other two sides (left and right) are vertical sides, which are increased by 50%. Therefore, the horizontal sides are now shorter, and vertical sides are longer. Therefore, the rectangle would have length 0.6s and width 1.5s? Or is it length 1.5s and width 0.6s? Doesn't matter, because area is length times width. So 0.6s * 1.5s = 0.9s². Thus, the area is decreased by 10%? That seems straightforward. But maybe I need to check again.

Alternatively, perhaps I made an error in assuming which sides are being changed? Let me try with numbers. Let me take a square with side length 10 units. Original area is 10*10=100. If two opposite sides are decreased by 40%, so 40% of 10 is 4, so the new length is 10 - 4 = 6. The other two sides are increased by 50%, so 50% of 10 is 5, so new length is 10 + 5 = 15. Therefore, the new rectangle has sides 6 and 15. Area is 6*15=90. Original area 100, so 90 is 10% less. Therefore, 10% decrease. So that seems to check out. So the answer is 10%? Therefore, maybe the answer is 10%? But the problem says "by what percent does the square's area decrease?" So 10% decrease.

Wait, but let me think again. Maybe there's a different way that the problem is interpreted? For example, maybe the problem is not changing the sides in the way I thought? Let me read the problem again:

"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. By what percent does the square's area decrease?"

So, yes, two opposite sides are decreased by 40%, other two increased by 50%. Therefore, as we saw, the area becomes 0.9s², which is 10% less. But maybe I should approach this algebraically to make sure.

Let me define the original square with side length s. Original area = 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 rectangle has sides of 0.6s and 1.5s. Therefore, area is 0.6s * 1.5s = 0.9s². So the area is decreased by 10%.

Alternatively, let me compute the percentage decrease. The formula for percentage decrease is [(Original Area - New Area)/Original Area] * 100%.

So that is [(s² - 0.9s²)/s²] * 100% = (0.1s² / s²) * 100% = 10%. So that's 10% decrease. Therefore, the answer is 10%.

But just to be thorough, let me consider another approach. Suppose instead of taking the sides as length and width, maybe there's a different way to interpret the problem? For example, maybe decreasing two opposite sides by 40% refers to decreasing each of those sides by 40% of the original length? Which is how I interpreted it. So, for example, if each side is decreased by 40%, so each of the two sides becomes 60% of original. Similarly, other two sides become 150% of original. Therefore, the rectangle sides are 0.6s and 1.5s. Therefore, area is 0.9s². So same result.

Alternatively, suppose someone thought that decreasing two opposite sides by 40% meant that the total length of those sides is decreased by 40%? Wait, but that would be different. For example, if you have two sides, each of length s, and you decrease the total length by 40%, then total length of those two sides would be 2s - 0.4*(2s) = 2s - 0.8s = 1.2s. But that would mean that each side is decreased by 0.4s total? But that would mean that each side is decreased by 0.2s, so 20% decrease. But this is not what the problem says. The problem says "the lengths of two opposite sides are decreased by 40%", which is more naturally interpreted as each of those sides is decreased by 40%, not the total length of both sides. So, for example, if you have two sides, each of length s, and you decrease each by 40%, meaning each becomes 0.6s. So that seems correct.

Therefore, the answer is 10%. Wait, but let me check if maybe the problem is referring to something else. For example, sometimes percentage changes can be confusing. Let me check with another example. Suppose original square has side length 1, area 1. Decrease two opposite sides by 40%: 1 - 0.4 = 0.6. Increase other two sides by 50%: 1 + 0.5 = 1.5. Area is 0.6 * 1.5 = 0.9. So decrease by 0.1, which is 10%. So yes. So 10% decrease. So answer is 10%.

But wait, maybe I need to check the problem again? Wait, 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%". So, the square has four sides. Two opposite sides are decreased by 40%: so each of those sides becomes 60% of original. The other two sides are increased by 50%, so each of those sides becomes 150% of original. Therefore, the rectangle formed has sides of 0.6s and 1.5s, hence area is 0.6*1.5*s² = 0.9s², which is 10% less than original. So that's correct.

But maybe I should check if there is a different interpretation. For example, if the problem meant that the total length of two opposite sides is decreased by 40%, but that seems less likely. For example, if you have two opposite sides, each of length s. If you decrease their total length by 40%, you have 2s - 0.4*2s = 1.2s. If that's spread over the two sides, then each side would be 0.6s, which is same as before. So same result. Therefore, either way, you get the same result. Therefore, answer is 10%.

Alternatively, maybe someone might think that decreasing two opposite sides by 40% means that each side is decreased by 40% of the original length, which is what I did. So the answer is 10%.

Therefore, I think the answer is 10% decrease. Therefore, I should put \boxed{10} as the final answer.

Wait, but just to be 100% sure, let me think again. Let me take another example with numbers. Let me take a square with side length 20. Original area is 400. Decrease two opposite sides by 40%: 40% of 20 is 8, so new length is 20 - 8 = 12. Increase the other two sides by 50%: 50% of 20 is 10, so new length is 20 + 10 = 30. Therefore, the area is 12 * 30 = 360. Original area 400, new area 360. The decrease is 40. Percentage decrease is (40 / 400) * 100 = 10%. Yep, same result. So 10% decrease. So I think this is correct.

Therefore, I think the answer is 10%. So the square's area decreases by 10%.

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

To determine the percentage decrease in the area of a square when two opposite sides are each decreased by 40% and the other two sides are increased by 50%, we follow a structured approach:

---

### Step 1: Define the Original Square

Let the side length of the original square be $ s $.
The area of the original square is:

$$
\text{Original Area} = s \times s = s^2
$$

---

### Step 2: Apply the Changes to the Sides

- Two opposite sides (say, the top and bottom) are decreased by 40%.
  The new length of these sides becomes:

  $$
  s - 0.4s = 0.6s
  $$

- The other two opposite sides (say, the left and right) are increased by 50%.
  The new length of these sides becomes:

  $$
  s + 0.5s = 1.5s
  $$

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

---

### Step 3: Calculate the New Area

$$
\text{New Area} = (0.6s)(1.5s) = 0.9s^2
$$

---

### Step 4: Determine the Percentage Change in Area

$$
\text{Percentage Decrease} = \left( \frac{\text{Original Area} - \text{New Area}}{\text{Original Area}} \right) \times 100\%
$$

$$
= \left( \frac{s^2 - 0.9s^2}{s^2} \right) \times 100\% = \left( \frac{0.1s^2}{s^2} \right) \times 100\% = 10\%
$$

---

### Final Answer

The area of the square decreases by:

$$
\boxed{10}
$$

percent.

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.