If `sin2 theta=3/4, then sin^(3)theta+cos^(3)theta=`

Options

A. `(sqrt5)/(8)`
B. `(sqrt7)/(8)`
C. 5`(sqrt7)/(16)`
D. none of these

Explanation

To solve the problem, we need to find \( \sin^3 \theta + \cos^3 \theta \) given that \( \sin 2\theta = \frac{3}{4} \).

### Step-by-Step Solution:

1. **Use the identity for \( \sin^3 \theta + \cos^3 \theta \)**:
\[ \sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta) \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \), we can simplify this to:
\[ \sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta) \]

2. **Find \( \sin \theta \cos \theta \)**:
We know that:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
Given \( \sin 2\theta = \frac{3}{4} \), we can find \( \sin \theta \cos \theta \):
\[ 2 \sin \theta \cos \theta = \frac{3}{4} \implies \sin \theta \cos \theta = \frac{3}{8} \]

3. **Now we need to find \( \sin \theta + \cos \theta \)**:
We can use the identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
and the square of the sum:
\[ (\sin \theta + \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 1 + 2 \sin \theta \cos \theta \]
Substituting \( \sin \theta \cos \theta = \frac{3}{8} \):
\[ (\sin \theta + \cos \theta)^2 = 1 + 2 \left(\frac{3}{8}\right) = 1 + \frac{3}{4} = \frac{7}{4} \]
Therefore:
\[ \sin \theta + \cos \theta = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2} \]

4. **Now substitute back into the expression for \( \sin^3 \theta + \cos^3 \theta \)**:
\[ \sin^3 \theta + \cos^3 \theta = \left(\frac{\sqrt{7}}{2}\right) \left(1 - \frac{3}{8}\right) \]
Simplifying the second term:
\[ 1 - \frac{3}{8} = \frac{8}{8} - \frac{3}{8} = \frac{5}{8} \]
Thus:
\[ \sin^3 \theta + \cos^3 \theta = \left(\frac{\sqrt{7}}{2}\right) \left(\frac{5}{8}\right) = \frac{5\sqrt{7}}{16} \]

### Final Answer:
\[ \sin^3 \theta + \cos^3 \theta = \frac{5\sqrt{7}}{16} \]