Question

Value of $4 \cos \left[\frac{1}{2} \cos^{-1}(\frac{1}{8})\right]$ is

• A. 3
• B. -3
• C. 1
• D. -1

Answer 1

Answer: (A)

3

Answer 2

Let $\theta = \cos^{-1}(\frac{1}{8})$. Then $\cos \theta = \frac{1}{8}$. Using the half-angle identity $2 \cos^2(\frac{\theta}{2}) - 1 = \cos \theta$, we get: $2 \cos^2(\frac{\theta}{2}) - 1 = \frac{1}{8}$ $2 \cos^2(\frac{\theta}{2}) = \frac{9}{8}$ $\cos^2(\frac{\theta}{2}) = \frac{9}{16}$ Since $0 \le \theta \le \pi$, it follows that $0 \le \frac{\theta}{2} \le \frac{\pi}{2}$. In this interval, $\cos(\frac{\theta}{2})$ is non-negative. Thus, $\cos(\frac{\theta}{2}) = \sqrt{\frac{9}{16}} = \frac{3}{4}$. The required value is $4 \cos(\frac{\theta}{2}) = 4 \times \frac{3}{4} = 3$.