If `cos^3xsin2x=sum_(r=0)^n a_xsin(r x),AAx in R` then

Options

A. `n=5,a_(1)=1//2`
B. `n=5,alpha_(1)=1//4`
C. `n=5,a_(2)=1//8`
D. `n=5,a_(2)=1//4`

Explanation

To solve the problem \( \cos(3x) \sin(2x) = \sum_{r=0}^{n} a_r \sin(rx) \), we will use trigonometric identities and properties to express the left-hand side in the desired summation form.

### Step-by-Step Solution:

1. **Use the Product-to-Sum Formulas**:
We start with the expression \( \cos(3x) \sin(2x) \). We can use the product-to-sum identities:
\[ \cos(A) \sin(B) = \frac{1}{2} [\sin(A + B) - \sin(A - B)] \]
Here, let \( A = 3x \) and \( B = 2x \):
\[ \cos(3x) \sin(2x) = \frac{1}{2} [\sin(3x + 2x) - \sin(3x - 2x)] = \frac{1}{2} [\sin(5x) - \sin(x)] \]

2. **Rewrite the Expression**:
Thus, we can rewrite the left-hand side:
\[ \cos(3x) \sin(2x) = \frac{1}{2} \sin(5x) - \frac{1}{2} \sin(x) \]

3. **Express in Summation Form**:
Now we can express this in the form of a summation:
\[ \cos(3x) \sin(2x) = -\frac{1}{2} \sin(x) + \frac{1}{2} \sin(5x) \]
This can be represented as:
\[ = \sum_{r=0}^{5} a_r \sin(rx) \]
where:
- \( a_1 = -\frac{1}{2} \) for \( r=1 \) (coefficient of \( \sin(x) \)),
- \( a_5 = \frac{1}{2} \) for \( r=5 \) (coefficient of \( \sin(5x) \)),
- \( a_r = 0 \) for all other \( r \).

4. **Identify \( n \) and \( a_r \)**:
From the expression, we see that the highest value of \( r \) is 5, thus \( n = 5 \). The coefficients are:
- \( a_1 = -\frac{1}{2} \)
- \( a_5 = \frac{1}{2} \)
- \( a_r = 0 \) for \( r = 0, 2, 3, 4 \).

### Conclusion:
Thus, the values are:
- \( n = 5 \)
- \( a_1 = -\frac{1}{2} \)
- \( a_5 = \frac{1}{2} \)