If `n` be a natural number define polynomial `f_(n)(x)` of `n^(th)` degree as follows `f_(n)(costheta)cosntheta`
i.e. `f_(2)(x)=2x^(2)-1 f_(3)(x)=4x^(3)-3x_(1)`
Then
`f_(6)(x)` is equal to

Options

A. `36x^(6)-48x^(4)+18x^(2)-5`
B. `32x^(6)-48x^(4)+18x^(2)-1`
C. `36x^(6)-45x^(4)+18x^(2)-8`
D. `36x^(6)-48x^(4)+18x^(2)-7`

Explanation

To find the polynomial \( f_6(x) \) defined as \( f_n(\cos \theta) \cos n \theta \), we can use the recursive relationship derived from the properties of Chebyshev polynomials.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We know:
- \( f_2(x) = 2x^2 - 1 \)
- \( f_3(x) = 4x^3 - 3x \)

2. **Using the Recursive Formula**:
The recursive relationship for the polynomials is given by:
\[ f_{n+1}(x) + f_{n-1}(x) = 2x f_n(x) \]
We will use this to find \( f_4(x) \), \( f_5(x) \), and \( f_6(x) \).

3. **Finding \( f_4(x) \)**:
Set \( n = 3 \):
\[ f_4(x) + f_2(x) = 2x f_3(x) \]
Substitute \( f_2(x) \) and \( f_3(x) \):
\[ f_4(x) + (2x^2 - 1) = 2x(4x^3 - 3x) \]
Simplifying the right side:
\[ f_4(x) + 2x^2 - 1 = 8x^4 - 6x^2 \]
Rearranging gives:
\[ f_4(x) = 8x^4 - 6x^2 + 1 \]

4. **Finding \( f_5(x) \)**:
Set \( n = 4 \):
\[ f_5(x) + f_3(x) = 2x f_4(x) \]
Substitute \( f_3(x) \) and \( f_4(x) \):
\[ f_5(x) + (4x^3 - 3x) = 2x(8x^4 - 6x^2 + 1) \]
Simplifying the right side:
\[ f_5(x) + 4x^3 - 3x = 16x^5 - 12x^3 + 2x \]
Rearranging gives:
\[ f_5(x) = 16x^5 - 12x^3 + 2x - 4x^3 + 3x \]
\[ f_5(x) = 16x^5 - 16x^3 + 5x \]

5. **Finding \( f_6(x) \)**:
Set \( n = 5 \):
\[ f_6(x) + f_4(x) = 2x f_5(x) \]
Substitute \( f_4(x) \) and \( f_5(x) \):
\[ f_6(x) + (8x^4 - 6x^2 + 1) = 2x(16x^5 - 16x^3 + 5x) \]
Simplifying the right side:
\[ f_6(x) + 8x^4 - 6x^2 + 1 = 32x^6 - 32x^4 + 10x^2 \]
Rearranging gives:
\[ f_6(x) = 32x^6 - 32x^4 + 10x^2 - 8x^4 + 6x^2 - 1 \]
\[ f_6(x) = 32x^6 - 40x^4 + 16x^2 - 1 \]

### Final Answer:
Thus, the polynomial \( f_6(x) \) is:
\[ f_6(x) = 32x^6 - 40x^4 + 16x^2 - 1 \]