sin left(2 sin ^{-1} sqrt{frac{63}{65}}right) | vedclass

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Question

(A) Let $	heta = \sin^{-1} \sqrt{\frac{63}{65}}$. Then $\sin 	heta = \sqrt{\frac{63}{65}}$.We need to find $\sin(2	heta)$.Using the identity $\sin(

Vedclass · Accepted Answer

$\frac{2 \sqrt{126}}{65}$

Vedclass · Answer

(A) Let $	heta = \sin^{-1} \sqrt{\frac{63}{65}}$. Then $\sin 	heta = \sqrt{\frac{63}{65}}$.We need to find $\sin(2	heta)$.Using the identity $\sin(2	heta) = 2 \sin 	heta \cos 	heta$.First,find $\cos 	heta = \sqrt{1 - \sin^2 	heta} = \sqrt{1 - \frac{63}{65}} = \sqrt{\frac{2}{65}}$.Now,substitute the values into the identity:$\sin(2	heta) = 2 	imes \sqrt{\frac{63}{65}} 	imes \sqrt{\frac{2}{65}}$$\sin(2	heta) = 2 	imes \frac{\sqrt{63 	imes 2}}{65} = 2 	imes \frac{\sqrt{126}}{65} = \frac{2 \sqrt{126}}{65}$.

$\sin \left(2 \sin ^{-1} \sqrt{\frac{63}{65}}\right)$ is equal to

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