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

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Question

(B) Let $\alpha = \sin^{-1}\left(\frac{3}{5}\right)$ and $\beta = \sin^{-1}\left(\frac{8}{17}\right)$.Then $\sin \alpha = \frac{3}{5}$ and $\sin \beta

Vedclass · Accepted Answer

$\cos ^{-1} \frac{84}{85}$

Vedclass · Answer

(B) Let $\alpha = \sin^{-1}\left(\frac{3}{5}ight)$ and $\beta = \sin^{-1}\left(\frac{8}{17}ight)$.Then $\sin \alpha = \frac{3}{5}$ and $\sin \beta = \frac{8}{17}$.Using the identity $\cos 	heta = \sqrt{1 - \sin^2 	heta}$,we get $\cos \alpha = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{\frac{16}{25}} = \frac{4}{5}$ and $\cos \beta = \sqrt{1 - (\frac{8}{17})^2} = \sqrt{\frac{225}{289}} = \frac{15}{17}$.We use the formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$.$\cos(\alpha - \beta) = \left(\frac{4}{5}ight)\left(\frac{15}{17}ight) + \left(\frac{3}{5}ight)\left(\frac{8}{17}ight) = \frac{60}{85} + \frac{24}{85} = \frac{84}{85}$.Therefore,$\alpha - \beta = \cos^{-1}\left(\frac{84}{85}ight)$.

$\sin ^{-1}\left(\frac{3}{5}\right)-\sin ^{-1}\left(\frac{8}{17}\right)=$ . . . . . .

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