`sin^(6) theta + cos^(6) theta ` is equal to

Options

A. `3 sin^(2) theta cos^(2) theta `
B. `(sin^(3) theta + cos^(3) theta)^(2)`
C. `(3 sin^(3) theta cos^(3) theta)/(cosec theta sec theta)`
D. `1-3 sin^(2) theta cos^(2) theta`

Explanation

To solve the expression \( \sin^6 \theta + \cos^6 \theta \), we can use the identity for the sum of cubes. The expression can be rewritten as follows:

1. **Recognize the sum of cubes**:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Here, let \( a = \sin^2 \theta \) and \( b = \cos^2 \theta \). Thus, we can express \( \sin^6 \theta + \cos^6 \theta \) as:
\[ \sin^6 \theta + \cos^6 \theta = (\sin^2 \theta)^3 + (\cos^2 \theta)^3 \]

2. **Apply the sum of cubes formula**:
Using the formula:
\[ \sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)((\sin^2 \theta)^2 - \sin^2 \theta \cos^2 \theta + (\cos^2 \theta)^2) \]

3. **Simplify using the Pythagorean identity**:
We know that \( \sin^2 \theta + \cos^2 \theta = 1 \). Therefore, the expression simplifies to:
\[ \sin^6 \theta + \cos^6 \theta = 1 \cdot ((\sin^2 \theta)^2 - \sin^2 \theta \cos^2 \theta + (\cos^2 \theta)^2) \]
Which simplifies to:
\[ (\sin^2 \theta)^2 + (\cos^2 \theta)^2 - \sin^2 \theta \cos^2 \theta \]

4. **Further simplification**:
Now, we can express \( (\sin^2 \theta)^2 + (\cos^2 \theta)^2 \) using the identity \( a^2 + b^2 = (a + b)^2 - 2ab \):
\[ (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ = 1^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \]

5. **Final expression**:
Therefore, we have:
\[ \sin^6 \theta + \cos^6 \theta = 1 - 3\sin^2 \theta \cos^2 \theta \]

Thus, the final answer is:
\[ \sin^6 \theta + \cos^6 \theta = 1 - 3\sin^2 \theta \cos^2 \theta \]