If `8sin^(3)xsin3x=sum_(r=0)^(n)a^(r)cosrx` is an identity in x, then n=

Options

A. 3
B. 4
C. 6
D. 9

Explanation

To solve the given problem, we need to find the value of \( n \) in the identity:

\[ 8 \sin^3 x \sin 3x = \sum_{r=0}^{n} a^r \cos rx \]

### Step 1: Simplify the Left-Hand Side

We start with the left-hand side:

\[ 8 \sin^3 x \sin 3x \]

Using the identity for \( \sin 3x \):

\[ \sin 3x = 3 \sin x - 4 \sin^3 x \]

We can express \( \sin^3 x \) in terms of \( \sin 3x \):

\[ 8 \sin^3 x \sin 3x = 8 \sin^3 x (3 \sin x - 4 \sin^3 x) \]

Distributing \( 8 \sin^3 x \):

\[ = 24 \sin^4 x - 32 \sin^6 x \]

### Step 2: Factor the Expression

We can factor out \( 8 \sin^4 x \):

\[ = 8 \sin^4 x (3 - 4 \sin^2 x) \]

### Step 3: Use the Identity for Sine Products

Now, we can use the product-to-sum identities. We rewrite \( \sin^4 x \) as:

\[ \sin^4 x = \left( \frac{1 - \cos 2x}{2} \right)^2 = \frac{1 - 2\cos 2x + \cos^2 2x}{4} \]

And we know that:

\[ \cos^2 2x = \frac{1 + \cos 4x}{2} \]

Thus, substituting back, we get:

\[ \sin^4 x = \frac{1 - 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} \]

### Step 4: Expand and Combine Terms

Now we can substitute this back into our expression and simplify. However, for our purpose, we need to focus on the maximum cosine term that arises from the left-hand side.

### Step 5: Identify the Maximum Cosine Term

The maximum cosine term from \( 8 \sin^4 x (3 - 4 \sin^2 x) \) will be determined by the highest degree of \( x \) in the expansion. The maximum degree of \( \cos \) that can arise from \( \sin^4 x \) and \( \sin^2 x \) will yield terms involving \( \cos 0x, \cos 2x, \cos 4x, \) and \( \cos 6x \).

### Step 6: Compare with Right-Hand Side

The right-hand side is given by:

\[ \sum_{r=0}^{n} a^r \cos rx \]

The highest term in this sum is \( \cos nx \). For the identity to hold, the maximum term from the left-hand side must match the maximum term from the right-hand side.

### Step 7: Set the Maximum Terms Equal

From our analysis, the maximum term from the left-hand side is \( \cos 6x \). Therefore, we set:

\[ n = 6 \]

### Conclusion

Thus, the value of \( n \) is:

\[ \boxed{6} \]