A possible value of tan left(frac{1}{4} sin ^ | vedclass

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Question

(A) Let $\frac{1}{4} \sin ^{-1} \frac{\sqrt{63}}{8} = 	heta$. Then,$\sin 4	heta = \frac{\sqrt{63}}{8}$. Since $\cos^2 4	heta = 1 - \sin^2 4	heta =

Vedclass · Accepted Answer

$\frac{1}{\sqrt{7}}$

Vedclass · Answer

(A) Let $\frac{1}{4} \sin ^{-1} \frac{\sqrt{63}}{8} = 	heta$. Then,$\sin 4	heta = \frac{\sqrt{63}}{8}$. Since $\cos^2 4	heta = 1 - \sin^2 4	heta = 1 - \frac{63}{64} = \frac{1}{64}$,we have $\cos 4	heta = \frac{1}{8}$. Using the formula $\cos 4	heta = 2\cos^2 2	heta - 1$,we get $2\cos^2 2	heta - 1 = \frac{1}{8}$,which implies $2\cos^2 2	heta = \frac{9}{8}$,so $\cos^2 2	heta = \frac{9}{16}$. Thus,$\cos 2	heta = \frac{3}{4}$. Using the formula $\cos 2	heta = 2\cos^2 	heta - 1$,we get $2\cos^2 	heta - 1 = \frac{3}{4}$,which implies $2\cos^2 	heta = \frac{7}{4}$,so $\cos^2 	heta = \frac{7}{8}$. Then $\sin^2 	heta = 1 - \cos^2 	heta = 1 - \frac{7}{8} = \frac{1}{8}$. Therefore,$	an^2 	heta = \frac{\sin^2 	heta}{\cos^2 	heta} = \frac{1/8}{7/8} = \frac{1}{7}$. Hence,$	an 	heta = \frac{1}{\sqrt{7}}$.

$A$ possible value of $\tan \left(\frac{1}{4} \sin ^{-1} \frac{\sqrt{63}}{8}\right)$ is :

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