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

Options

A. 1
B. 2
C. `-1`
D. `-2`

Explanation

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

### Step 1: Write the equation
We start with the equation:
\[ 16 \sin^5 x = p \sin 5x + q \sin 3x + r \sin x \]

### Step 2: Express \( \sin^5 x \) in terms of \( \sin 5x \), \( \sin 3x \), and \( \sin x \)
To express \( \sin^5 x \) in terms of \( \sin 5x \), \( \sin 3x \), and \( \sin x \), we can use the identity:
\[ \sin^5 x = \sin x \cdot \sin^4 x \]
We can express \( \sin^4 x \) as:
\[ \sin^4 x = \left( \sin^2 x \right)^2 = \left( 1 - \cos^2 x \right)^2 = 1 - 2\cos^2 x + \cos^4 x \]

### Step 3: Use the double angle and triple angle identities
We know:
\[ \sin 2x = 2 \sin x \cos x \]
\[ \sin 3x = 3 \sin x - 4 \sin^3 x \]
\[ \sin 5x = 16 \sin^5 x - 20 \sin^3 x + 5 \sin x \]

### Step 4: Substitute and simplify
Substituting \( \sin^5 x \) in terms of \( \sin 5x \), \( \sin 3x \), and \( \sin x \):
\[ 16 \sin^5 x = \sin 5x + 5 \sin 3x - 10 \sin x \]

### Step 5: Compare coefficients
Now, we can compare the coefficients of \( \sin 5x \), \( \sin 3x \), and \( \sin x \) on both sides of the equation:
\[ 16 \sin^5 x = p \sin 5x + q \sin 3x + r \sin x \]
From the comparison:
- 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 \( p \) is:
\[ \boxed{1} \]