sin left[ 3 sin^{-1} left( frac{1}{5} right) | vedclass

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Question

(A) We use the identity $3 \sin^{-1}(x) = \sin^{-1}(3x - 4x^3)$ for $|x| \leq \frac{1}{2}$.Here,$x = \frac{1}{5}$,which satisfies the condition.$\sin

Vedclass · Accepted Answer

$71/125$

Vedclass · Answer

(A) We use the identity $3 \sin^{-1}(x) = \sin^{-1}(3x - 4x^3)$ for $|x| \leq \frac{1}{2}$.Here,$x = \frac{1}{5}$,which satisfies the condition.$\sin \left[ 3 \sin^{-1} \left( \frac{1}{5} \right) \right] = \sin \left[ \sin^{-1} \left( 3 \left( \frac{1}{5} \right) - 4 \left( \frac{1}{5} \right)^3 \right) \right]$$= \sin \left[ \sin^{-1} \left( \frac{3}{5} - \frac{4}{125} \right) \right]$$= \sin \left[ \sin^{-1} \left( \frac{75 - 4}{125} \right) \right]$$= \sin \left[ \sin^{-1} \left( \frac{71}{125} \right) \right] = \frac{71}{125}$.

$\sin \left[ 3 \sin^{-1} \left( \frac{1}{5} \right) \right] = $

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