If `sin theta cos theta = sqrt3//4`,then the value of `sin^4theta + cos^4 theta` is

Options

A. `7//8`
B. `5//8`
C. `3//8`
D. `1//8`

Answer

To solve the problem, we need to find the value of \( \sin^4 \theta + \cos^4 \theta \) given that \( \sin \theta \cos \theta = \frac{\sqrt{3}}{4} \).

### Step 1: Use the identity for \( \sin^2 \theta + \cos^2 \theta \)
We know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]

### Step 2: Square both sides
Now, we square both sides:
\[ (\sin^2 \theta + \cos^2 \theta)^2 = 1^2 \]
This expands to:
\[ \sin^4 \theta + \cos^4 \theta + 2\sin^2 \theta \cos^2 \theta = 1 \]

### Step 3: Rearrange the equation
We can rearrange this equation to isolate \( \sin^4 \theta + \cos^4 \theta \):
\[ \sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \]

### Step 4: Substitute \( \sin \theta \cos \theta \)
We know \( \sin \theta \cos \theta = \frac{\sqrt{3}}{4} \). Thus, we can find \( \sin^2 \theta \cos^2 \theta \):
\[ \sin^2 \theta \cos^2 \theta = \left(\sin \theta \cos \theta\right)^2 = \left(\frac{\sqrt{3}}{4}\right)^2 = \frac{3}{16} \]

### Step 5: Substitute back into the equation
Now, substitute \( \sin^2 \theta \cos^2 \theta \) back into the equation:
\[ \sin^4 \theta + \cos^4 \theta = 1 - 2 \cdot \frac{3}{16} \]
Calculating \( 2 \cdot \frac{3}{16} \):
\[ 2 \cdot \frac{3}{16} = \frac{6}{16} = \frac{3}{8} \]
Thus, we have:
\[ \sin^4 \theta + \cos^4 \theta = 1 - \frac{3}{8} \]

### Step 6: Simplify the expression
Now, simplify \( 1 - \frac{3}{8} \):
\[ 1 = \frac{8}{8} \quad \text{so} \quad \sin^4 \theta + \cos^4 \theta = \frac{8}{8} - \frac{3}{8} = \frac{5}{8} \]

### Final Answer
Thus, the value of \( \sin^4 \theta + \cos^4 \theta \) is:
\[ \boxed{\frac{5}{8}} \]