If `P_(n) = cos^(n) theta + sin^(n) theta`
`6P_(10) -15P_(8) + 10P_(6)` is equal to

Options

A. `-1`
B. 0
C. 1
D. 8

Answer

To solve the problem, we need to evaluate the expression \(6P_{10} - 15P_{8} + 10P_{6}\) where \(P_n = \cos^n \theta + \sin^n \theta\).

### Step-by-Step Solution:

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

2. **Calculate \(P_{10}\), \(P_{8}\), and \(P_{6}\)**:
- \(P_{10} = \cos^{10} \theta + \sin^{10} \theta\)
- \(P_{8} = \cos^{8} \theta + \sin^{8} \theta\)
- \(P_{6} = \cos^{6} \theta + \sin^{6} \theta\)

3. **Use the recurrence relation**:
We can use the recurrence relation derived from \(P_n\):
\[ P_n = P_{n-2} + (\cos^2 \theta + \sin^2 \theta) P_{n-2} = P_{n-2} + P_{n-2} = 2P_{n-2} \]
This means:
\[ P_n = P_{n-2} + \sin^2 \theta P_{n-2} + \cos^2 \theta P_{n-2} \]
Thus,
\[ P_n = P_{n-2} + \sin^2 \theta P_{n-2} + \cos^2 \theta P_{n-2} \]

4. **Express \(P_{10}\), \(P_{8}\), and \(P_{6}\) in terms of \(P_{4}\) and \(P_{2}\)**:
- \(P_{10} = P_{8} + \sin^2 \theta P_{8} + \cos^2 \theta P_{8}\)
- \(P_{8} = P_{6} + \sin^2 \theta P_{6} + \cos^2 \theta P_{6}\)
- \(P_{6} = P_{4} + \sin^2 \theta P_{4} + \cos^2 \theta P_{4}\)

5. **Substituting back into the expression**:
We substitute \(P_{10}\), \(P_{8}\), and \(P_{6}\) into the expression \(6P_{10} - 15P_{8} + 10P_{6}\):
\[ 6P_{10} - 15P_{8} + 10P_{6} = 6(P_{8} + P_{6}) - 15P_{8} + 10P_{6} \]
\[ = 6P_{8} + 6P_{6} - 15P_{8} + 10P_{6} \]
\[ = (6 - 15)P_{8} + (6 + 10)P_{6} \]
\[ = -9P_{8} + 16P_{6} \]

6. **Continue substituting down to \(P_{2}\)**:
Continuing this process down to \(P_{2}\) and \(P_{0}\):
\[ P_{2} = \cos^2 \theta + \sin^2 \theta = 1 \]
\[ P_{0} = 1 \]

7. **Final substitution**:
After substituting all the way down, we will find that:
\[ 6P_{2} - 3P_{0} + 1 = 6(1) - 3(1) + 1 = 6 - 3 + 1 = 4 \]

8. **Final Result**:
Thus, the final result is:
\[ 6P_{10} - 15P_{8} + 10P_{6} = 0 \]