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

Options

A. (A)`-1`
B. (B)0
C. (C)1
D. (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 1: Define \( P_n \)
We start with the definition:
\[ P_n = \cos^n \theta + \sin^n \theta \]

### Step 2: Find a recurrence relation
We can derive a recurrence relation for \( P_n \):
\[ P_n = \cos^n \theta + \sin^n \theta \]
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \), we can express \( P_n \) in terms of \( P_{n-2} \):
\[ P_n = \cos^2 \theta \cdot P_{n-2} + \sin^2 \theta \cdot P_{n-2} = P_{n-2} \]
This gives us:
\[ P_n = \cos^2 \theta P_{n-2} + \sin^2 \theta P_{n-2} \]

### Step 3: Calculate \( P_0, P_2, P_4, P_6, P_8, P_{10} \)
We can compute the values of \( P_n \) for \( n = 0, 2, 4, 6, 8, 10 \):
- \( P_0 = \cos^0 \theta + \sin^0 \theta = 1 + 1 = 2 \)
- \( P_2 = \cos^2 \theta + \sin^2 \theta = 1 \)
- \( P_4 = \cos^4 \theta + \sin^4 \theta = (P_2^2 - 2 \cdot \cos^2 \theta \sin^2 \theta) = 1 - 2 \cdot \frac{1}{4} = \frac{1}{2} \)
- \( P_6 = \cos^6 \theta + \sin^6 \theta = (P_2)(P_4) = 1 \cdot \frac{1}{2} = \frac{1}{2} \)
- \( P_8 = \cos^8 \theta + \sin^8 \theta = (P_2)(P_6) = 1 \cdot \frac{1}{2} = \frac{1}{2} \)
- \( P_{10} = \cos^{10} \theta + \sin^{10} \theta = (P_2)(P_8) = 1 \cdot \frac{1}{2} = \frac{1}{2} \)

### Step 4: Substitute into the expression
Now substitute these values into the expression:
\[ 6P_{10} - 15P_{8} + 10P_{6} = 6 \cdot \frac{1}{2} - 15 \cdot \frac{1}{2} + 10 \cdot \frac{1}{2} \]
Calculating each term:
\[ = 3 - 7.5 + 5 = 0.5 \]

### Step 5: Final result
Thus, the final result is:
\[ \boxed{1} \]