If `sin theta cos theta = 1//2,` then what is `sin^6 theta +cos^6 theta` equal to?

Options

A. 1
B. 2
C. 3
D. `1/4`

Answer

To solve the problem, we need to find the value of \( \sin^6 \theta + \cos^6 \theta \) given that \( \sin \theta \cos \theta = \frac{1}{2} \).

### Step 1: Use the identity for \( \sin^6 \theta + \cos^6 \theta \)

We can use the identity:
\[ \sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \), we can simplify this to:
\[ \sin^6 \theta + \cos^6 \theta = 1 \cdot (\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) = \sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta \]

### Step 2: Express \( \sin^4 \theta \) and \( \cos^4 \theta \)

We can further simplify \( \sin^4 \theta + \cos^4 \theta \) using the identity:
\[ \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \]
Thus, we have:
\[ \sin^6 \theta + \cos^6 \theta = (1 - 2\sin^2 \theta \cos^2 \theta) - \sin^2 \theta \cos^2 \theta = 1 - 3\sin^2 \theta \cos^2 \theta \]

### Step 3: Substitute \( \sin \theta \cos \theta \)

Given \( \sin \theta \cos \theta = \frac{1}{2} \), we can find \( \sin^2 \theta \cos^2 \theta \):
\[ \sin^2 \theta \cos^2 \theta = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

### Step 4: Substitute back into the equation

Now we substitute \( \sin^2 \theta \cos^2 \theta \) into our expression:
\[ \sin^6 \theta + \cos^6 \theta = 1 - 3\left(\frac{1}{4}\right) = 1 - \frac{3}{4} = \frac{1}{4} \]

### Final Answer

Thus, the value of \( \sin^6 \theta + \cos^6 \theta \) is:
\[ \boxed{\frac{1}{4}} \]