If ` sin theta = 1/2 and theta ` is acute then ` (3 cos theta -4 cos^(3) theta)` is equal to

Options

A. 0
B. `1/2`
C. `1/6`
D. `-1`

Explanation

To solve the problem, we will follow these steps:

1. **Identify the value of theta**: We know that \( \sin \theta = \frac{1}{2} \) and that \( \theta \) is acute. The sine function equals \( \frac{1}{2} \) at \( 30^\circ \) (or \( \frac{\pi}{6} \) radians).

2. **Calculate cos theta**: Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can find \( \cos \theta \):
\[ \sin^2 \theta = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{1}{4} = \frac{3}{4} \]
\[ \cos \theta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

3. **Substitute cos theta into the expression**: We need to evaluate \( 3 \cos \theta - 4 \cos^3 \theta \):
\[ 3 \cos \theta - 4 \cos^3 \theta = 3 \left(\frac{\sqrt{3}}{2}\right) - 4 \left(\frac{\sqrt{3}}{2}\right)^3 \]

4. **Calculate \( \cos^3 \theta \)**:
\[ \cos^3 \theta = \left(\frac{\sqrt{3}}{2}\right)^3 = \frac{(\sqrt{3})^3}{2^3} = \frac{3\sqrt{3}}{8} \]

5. **Substitute \( \cos^3 \theta \) back into the expression**:
\[ 3 \cos \theta - 4 \cos^3 \theta = 3 \left(\frac{\sqrt{3}}{2}\right) - 4 \left(\frac{3\sqrt{3}}{8}\right) \]
\[ = \frac{3\sqrt{3}}{2} - \frac{12\sqrt{3}}{8} \]
\[ = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} = 0 \]

6. **Conclusion**: Therefore, the value of \( 3 \cos \theta - 4 \cos^3 \theta \) is \( 0 \).