`cos^3x.sin2x=sum_(m=1)^n(a_m)sin mx` is an identity in x.

Options

A. `a_(3) = 3/8, a_(2)=0`
B. `n=6, a_(1)=1/2`
C. `n=5, a_(1)=1/4`
D. `suma_(m)=3/4`

Answer

To solve the problem \( \cos^3 x \sin 2x = \sum_{m=1}^n a_m \sin mx \) as an identity in \( x \), we will follow these steps:

### Step 1: Rewrite the left-hand side (LHS)
We start with the left-hand side:
\[ \cos^3 x \sin 2x \]
Using the identity \( \sin 2x = 2 \sin x \cos x \), we can rewrite the expression:
\[ \cos^3 x \sin 2x = \cos^3 x \cdot 2 \sin x \cos x = 2 \cos^4 x \sin x \]

### Step 2: Express \( \cos^4 x \)
Next, we express \( \cos^4 x \) in terms of \( \sin \) and \( \cos \):
\[ \cos^4 x = (\cos^2 x)^2 = \left(\frac{1 + \cos 2x}{2}\right)^2 \]
Expanding this, we get:
\[ \cos^4 x = \frac{(1 + \cos 2x)^2}{4} = \frac{1 + 2\cos 2x + \cos^2 2x}{4} \]
Using \( \cos^2 2x = \frac{1 + \cos 4x}{2} \), we can substitute:
\[ \cos^4 x = \frac{1 + 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} = \frac{2 + 4\cos 2x + 1 + \cos 4x}{8} = \frac{3 + 4\cos 2x + \cos 4x}{8} \]

### Step 3: Substitute back into LHS
Now, we substitute \( \cos^4 x \) back into our expression for LHS:
\[ LHS = 2 \cos^4 x \sin x = 2 \cdot \frac{3 + 4\cos 2x + \cos 4x}{8} \sin x = \frac{3 + 4\cos 2x + \cos 4x}{4} \sin x \]

### Step 4: Expand and simplify
Distributing \( \sin x \):
\[ LHS = \frac{3 \sin x}{4} + \frac{4 \cos 2x \sin x}{4} + \frac{\cos 4x \sin x}{4} \]
This simplifies to:
\[ LHS = \frac{3 \sin x}{4} + \sin 2x + \frac{\sin 4x}{4} \]

### Step 5: Express LHS in terms of the summation
Now we can express the left-hand side as a summation:
\[ LHS = \frac{3}{4} \sin x + 0 \sin 2x + 1 \sin 2x + \frac{1}{4} \sin 4x \]
This means we can identify coefficients:
- \( a_1 = \frac{3}{4} \)
- \( a_2 = 0 \)
- \( a_3 = 1 \)
- \( a_4 = \frac{1}{4} \)
- \( n = 4 \)

### Final Result
Thus, we conclude that:
\[ \cos^3 x \sin 2x = \frac{3}{4} \sin x + 0 \sin 2x + 1 \sin 3x + \frac{1}{4} \sin 4x \]
is an identity in \( x \).