If `sin theta + cos theta = x ` , then the value of `cos^(6) theta + sin^(6) theta` is equal to

Options

A. `(1)/(4)`
B. `(1)/(4) (1 + 6x^(2))`
C. `(1)/(4) ( 1 + 6x^(2) - 3x^(4))`
D. `(1)/(2) (5 - 3x^(2))`

Answer

To solve the problem where \( \sin \theta + \cos \theta = x \), we need to find the value of \( \cos^6 \theta + \sin^6 \theta \).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ \sin \theta + \cos \theta = x \]

2. **Square both sides**:
\[ (\sin \theta + \cos \theta)^2 = x^2 \]
Expanding the left side using the identity \( (A + B)^2 = A^2 + B^2 + 2AB \):
\[ \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = x^2 \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 1 + 2\sin \theta \cos \theta = x^2 \]

3. **Rearranging to find \( \sin \theta \cos \theta \)**:
\[ 2\sin \theta \cos \theta = x^2 - 1 \]
Therefore:
\[ \sin \theta \cos \theta = \frac{x^2 - 1}{2} \]

4. **Now, we need to find \( \sin^2 \theta \cos^2 \theta \)**:
\[ \sin^2 \theta \cos^2 \theta = \left(\sin \theta \cos \theta\right)^2 = \left(\frac{x^2 - 1}{2}\right)^2 = \frac{(x^2 - 1)^2}{4} \]

5. **Using the identity for \( \sin^6 \theta + \cos^6 \theta \)**:
We can use the formula:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Here, let \( a = \sin^2 \theta \) and \( b = \cos^2 \theta \):
\[ \sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \sin^6 \theta + \cos^6 \theta = 1 \cdot (\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) \]

6. **Now, express \( \sin^4 \theta + \cos^4 \theta \)**:
Using the identity:
\[ \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \]
Substitute \( \sin^2 \theta \cos^2 \theta \):
\[ \sin^4 \theta + \cos^4 \theta = 1 - 2 \cdot \frac{(x^2 - 1)^2}{4} = 1 - \frac{(x^2 - 1)^2}{2} \]

7. **Putting it all together**:
\[ \sin^6 \theta + \cos^6 \theta = \sin^4 \theta + \cos^4 \theta - \sin^2 \theta \cos^2 \theta \]
Substitute the values:
\[ = \left(1 - \frac{(x^2 - 1)^2}{2}\right) - \frac{(x^2 - 1)^2}{4} \]
Combine the terms:
\[ = 1 - \frac{2(x^2 - 1)^2}{4} - \frac{(x^2 - 1)^2}{4} = 1 - \frac{3(x^2 - 1)^2}{4} \]

8. **Final expression**:
\[ \sin^6 \theta + \cos^6 \theta = 1 - \frac{3(x^2 - 1)^2}{4} \]

### Final Answer:
\[ \sin^6 \theta + \cos^6 \theta = 1 - \frac{3(x^2 - 1)^2}{4} \]