evaluate the expression.\n\n\\( \\cos \\left( 2 \\arcsin \\left( \\frac { 1 } { 4 } \\right) \\right)…

Explanation:

Step1: Let (\theta=\arcsin\left(\frac{1}{4}\right))

By the definition of the inverse - sine function, (\sin\theta=\frac{1}{4}), and (\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]). Then, using the Pythagorean identity (\sin^{2}\theta+\cos^{2}\theta = 1), we can find (\cos\theta). [ \begin{align*} \cos\theta&=\sqrt{1 - \sin^{2}\theta}\ &=\sqrt{1-\left(\frac{1}{4}\right)^{2}}\ &=\sqrt{1-\frac{1}{16}}\ &=\sqrt{\frac{16 - 1}{16}}\ &=\sqrt{\frac{15}{16}}=\frac{\sqrt{15}}{4} \end{align*} ]

Step2: Use the double - angle formula (\cos(2\theta)=1 - 2\sin^{2}\theta)

Since (\theta=\arcsin\left(\frac{1}{4}\right)) and (\sin\theta=\frac{1}{4}), substitute (\sin\theta) into the double - angle formula. [ \begin{align*} \cos\left(2\arcsin\left(\frac{1}{4}\right)\right)&=1-2\sin^{2}\left(\arcsin\left(\frac{1}{4}\right)\right)\ &=1 - 2\times\left(\frac{1}{4}\right)^{2}\ &=1-2\times\frac{1}{16}\ &=1-\frac{2}{16}\ &=\frac{16 - 2}{16}\ &=\frac{14}{16}=\frac{7}{8} \end{align*} ]

Answer:

B. (\frac{7}{8})