If `sin^2A=x` , then `sinAsin2Asin3Asin4A` is a polynomial in x , the sum of whose cofficients is :

Options

A. 0
B. 40
C. 168
D. 336

Answer

To solve the problem, we need to express \( \sin A \sin 2A \sin 3A \sin 4A \) in terms of \( x \), where \( \sin^2 A = x \). We will follow these steps:

### Step 1: Express \( \sin A \) and \( \cos A \) in terms of \( x \)
Given that \( \sin^2 A = x \), we can write:
\[ \sin A = \sqrt{x} \quad \text{or} \quad \sin A = -\sqrt{x} \]
Also, using the Pythagorean identity, we have:
\[ \cos^2 A = 1 - \sin^2 A = 1 - x \quad \Rightarrow \quad \cos A = \sqrt{1 - x} \quad \text{or} \quad \cos A = -\sqrt{1 - x} \]

### Step 2: Write \( \sin 2A \), \( \sin 3A \), and \( \sin 4A \) in terms of \( \sin A \) and \( \cos A \)
Using the double angle and triple angle formulas:
\[ \sin 2A = 2 \sin A \cos A \]
\[ \sin 3A = 3 \sin A - 4 \sin^3 A = 3 \sin A - 4 (\sin^2 A) \sin A = 3 \sin A - 4x \sin A = \sin A (3 - 4x) \]
\[ \sin 4A = 2 \sin 2A \cos 2A = 2 (2 \sin A \cos A) (1 - 2 \sin^2 A) = 4 \sin A \cos A (1 - 2x) \]

### Step 3: Substitute these expressions into \( \sin A \sin 2A \sin 3A \sin 4A \)
Now we can write:
\[ \sin A \sin 2A \sin 3A \sin 4A = \sin A (2 \sin A \cos A) (\sin A (3 - 4x)) (4 \sin A \cos A (1 - 2x)) \]
This simplifies to:
\[ = 8 \sin^4 A \cos^2 A (3 - 4x)(1 - 2x) \]

### Step 4: Substitute \( \sin^2 A \) and \( \cos^2 A \)
Substituting \( \sin^2 A = x \) and \( \cos^2 A = 1 - x \):
\[ = 8 x^2 (1 - x) (3 - 4x)(1 - 2x) \]

### Step 5: Expand the polynomial
Now we will expand \( 8 x^2 (1 - x) (3 - 4x)(1 - 2x) \):
1. First, expand \( (3 - 4x)(1 - 2x) \):
\[ = 3 - 6x - 4x + 8x^2 = 3 - 10x + 8x^2 \]
2. Now multiply by \( (1 - x) \):
\[ (1 - x)(3 - 10x + 8x^2) = 3 - 10x + 8x^2 - 3x + 10x^2 - 8x^3 = 3 - 13x + 18x^2 - 8x^3 \]
3. Finally, multiply by \( 8x^2 \):
\[ 8x^2(3 - 13x + 18x^2 - 8x^3) = 24x^2 - 104x^3 + 144x^4 - 64x^5 \]

### Step 6: Find the sum of coefficients
The polynomial is:
\[ -64x^5 + 144x^4 - 104x^3 + 24x^2 \]
To find the sum of the coefficients, substitute \( x = 1 \):
\[ -64(1)^5 + 144(1)^4 - 104(1)^3 + 24(1)^2 = -64 + 144 - 104 + 24 \]
Calculating this gives:
\[ = 144 + 24 - 64 - 104 = 0 \]

### Final Answer
The sum of the coefficients is \( 0 \).