If `sintheta+costheta=a`, then find the value of `sin^(6)theta+cos^(6)theta`.

Options

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

Answer

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

### Step-by-Step Solution:

1. **Use the identity for \( \sin^6 \theta + \cos^6 \theta \)**:
\[ \sin^6 \theta + \cos^6 \theta = (\sin^2 \theta)^3 + (\cos^2 \theta)^3 \]
We can use the identity for the sum of cubes:
\[ A^3 + B^3 = (A + B)(A^2 - AB + B^2) \]
where \( A = \sin^2 \theta \) and \( B = \cos^2 \theta \).

2. **Calculate \( A + B \)**:
\[ A + B = \sin^2 \theta + \cos^2 \theta = 1 \]

3. **Calculate \( A^2 + B^2 \)**:
\[ A^2 + B^2 = \sin^4 \theta + \cos^4 \theta \]
We can use the identity:
\[ \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \]

4. **Calculate \( AB \)**:
\[ AB = \sin^2 \theta \cos^2 \theta = \left(\frac{1}{2} \sin(2\theta)\right)^2 \]
We can express \( \sin^2 \theta \cos^2 \theta \) in terms of \( a \):
\[ \sin^2 \theta + \cos^2 \theta = 1 \implies \sin^2 \theta \cos^2 \theta = \frac{(a^2 - 1)}{4} \]

5. **Substituting back into the equation**:
Now substituting \( AB \) into the equation for \( A^2 + B^2 \):
\[ A^2 + B^2 = 1 - 2\left(\frac{(a^2 - 1)}{4}\right) = 1 - \frac{(a^2 - 1)}{2} = \frac{2 - (a^2 - 1)}{2} = \frac{3 - a^2}{2} \]

6. **Putting it all together**:
Now substituting back into the sum of cubes formula:
\[ \sin^6 \theta + \cos^6 \theta = (1)\left(\frac{3 - a^2}{2} - \frac{(a^2 - 1)}{4}\right) \]
Simplifying this gives:
\[ = \frac{3 - a^2}{2} - \frac{a^2 - 1}{4} = \frac{6 - 2a^2 - a^2 + 1}{4} = \frac{7 - 3a^2}{4} \]

### Final Answer:
\[ \sin^6 \theta + \cos^6 \theta = \frac{7 - 3a^2}{4} \]