Given that `16sin^(5)x=p sin5x+qsin3x+rsinx`.
What is the value of r ?

Options

A. 5
B. 8
C. 10
D. `-10`

Explanation

To find the value of \( r \) in the equation \( 16\sin^5 x = p \sin 5x + q \sin 3x + r \sin x \), we will follow these steps:

### Step 1: Rewrite the Left-Hand Side
We start with the left-hand side of the equation:
\[ 16\sin^5 x \]
We can express \( \sin^5 x \) in terms of \( \sin x \) and \( \sin 3x \) and \( \sin 5x \).

### Step 2: Use the Identity for \( \sin^2 x \)
Recall the identity:
\[ \sin^2 x = 1 - \cos^2 x \]
We can also use the double angle identity to express \( \sin^2 x \) in terms of \( \cos 2x \):
\[ \sin^2 x = \frac{1 - \cos 2x}{2} \]
Thus,
\[ \sin^5 x = \sin x \cdot \sin^4 x = \sin x \cdot (\sin^2 x)^2 = \sin x \cdot \left(\frac{1 - \cos 2x}{2}\right)^2 \]

### Step 3: Expand \( \sin^5 x \)
Now we expand \( \left(\frac{1 - \cos 2x}{2}\right)^2 \):
\[ \sin^5 x = \sin x \cdot \frac{(1 - \cos 2x)^2}{4} = \frac{\sin x (1 - 2\cos 2x + \cos^2 2x)}{4} \]
Using \( \cos^2 2x = \frac{1 + \cos 4x}{2} \):
\[ \sin^5 x = \frac{\sin x}{4} \left( 1 - 2\cos 2x + \frac{1 + \cos 4x}{2} \right) \]
\[ = \frac{\sin x}{4} \left( \frac{2 - 4\cos 2x + 1 + \cos 4x}{2} \right) \]
\[ = \frac{\sin x}{8} (3 - 4\cos 2x + \cos 4x) \]

### Step 4: Distribute \( 16 \)
Now multiply by 16:
\[ 16\sin^5 x = 2\sin x (3 - 4\cos 2x + \cos 4x) \]
\[ = 6\sin x + 2\sin x \cos 4x - 8\sin x \cos 2x \]

### Step 5: Use Product-to-Sum Identities
Using the product-to-sum identities, we can express \( 2\sin x \cos 4x \) and \( 8\sin x \cos 2x \):
\[ 2\sin x \cos 4x = \sin(5x) + \sin(-3x) = \sin(5x) - \sin(3x) \]
\[ 8\sin x \cos 2x = 4\sin(3x) + 4\sin(x) \]

### Step 6: Combine Terms
Putting this all together:
\[ 16\sin^5 x = 6\sin x + \sin(5x) - \sin(3x) - 4\sin(3x) \]
\[ = \sin(5x) - 5\sin(3x) + 10\sin x \]

### Step 7: Compare Coefficients
Now we compare this with the right-hand side \( p \sin 5x + q \sin 3x + r \sin x \):
- Coefficient of \( \sin 5x \) gives \( p = 1 \)
- Coefficient of \( \sin 3x \) gives \( q = -5 \)
- Coefficient of \( \sin x \) gives \( r = 10 \)

### Final Answer
Thus, the value of \( r \) is:
\[ \boxed{10} \]