Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 260: 84

$cos~(2~sin^{-1}~\frac{1}{4}) = \frac{7}{8}$

Let $~~\theta = sin^{-1}~\frac{1}{4}$ Then: $~~sin~\theta = \frac{1}{4}$ We need to find $cos~2\theta$: $cos~2\theta = 1-2~sin^2~\theta$ $cos~2\theta = 1-2~(\frac{1}{4})^2$ $cos~2\theta = \frac{7}{8}$ Therefore, $~~cos~(2~sin^{-1}~\frac{1}{4}) = \frac{7}{8}$