If `mu_(n) = cos^(n)alpha + sin^(n)alpha`, then the value of `2mu_(6)- 3mu_(4) +1` is:

Options

A. a)1
B. b)4
C. c)6
D. d)0

Answer

To solve the problem where \( \mu_n = \cos^n \alpha + \sin^n \alpha \), we need to find the value of \( 2\mu_6 - 3\mu_4 + 1 \).

### Step-by-Step Solution:

1. **Define \( \mu_n \)**:
\[ \mu_n = \cos^n \alpha + \sin^n \alpha \]

2. **Calculate \( \mu_6 \)**:
\[ \mu_6 = \cos^6 \alpha + \sin^6 \alpha \]
We can use the identity:
\[ \cos^6 \alpha + \sin^6 \alpha = 1 - 3 \sin^2 \alpha \cos^2 \alpha \]
Thus,
\[ \mu_6 = 1 - 3 \sin^2 \alpha \cos^2 \alpha \]

3. **Calculate \( \mu_4 \)**:
\[ \mu_4 = \cos^4 \alpha + \sin^4 \alpha \]
We can use the identity:
\[ \cos^4 \alpha + \sin^4 \alpha = 1 - 2 \sin^2 \alpha \cos^2 \alpha \]
Thus,
\[ \mu_4 = 1 - 2 \sin^2 \alpha \cos^2 \alpha \]

4. **Substitute \( \mu_6 \) and \( \mu_4 \) into the expression**:
\[ 2\mu_6 - 3\mu_4 + 1 = 2(1 - 3 \sin^2 \alpha \cos^2 \alpha) - 3(1 - 2 \sin^2 \alpha \cos^2 \alpha) + 1 \]

5. **Expand the expression**:
\[ = 2 - 6 \sin^2 \alpha \cos^2 \alpha - 3 + 6 \sin^2 \alpha \cos^2 \alpha + 1 \]

6. **Combine like terms**:
\[ = (2 - 3 + 1) + (-6 \sin^2 \alpha \cos^2 \alpha + 6 \sin^2 \alpha \cos^2 \alpha) \]
\[ = 0 + 0 \]
\[ = 0 \]

### Final Answer:
The value of \( 2\mu_6 - 3\mu_4 + 1 \) is \( 0 \).