Evaluate: cos ^{-1}left(frac{4}{5}right)+cos | vedclass

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(B) Let $\alpha = \cos ^{-1}\left(\frac{4}{5}\right)$ and $\beta = \cos ^{-1}\left(\frac{12}{13}\right)$.Then $\cos \alpha = \frac{4}{5}$ and $\cos \b

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$\cos ^{-1}\left(\frac{33}{65}\right)$

Vedclass · Answer

(B) Let $\alpha = \cos ^{-1}\left(\frac{4}{5}ight)$ and $\beta = \cos ^{-1}\left(\frac{12}{13}ight)$.Then $\cos \alpha = \frac{4}{5}$ and $\cos \beta = \frac{12}{13}$.Using the identity $\sin 	heta = \sqrt{1 - \cos^2 	heta}$,we get $\sin \alpha = \sqrt{1 - \left(\frac{4}{5}ight)^2} = \frac{3}{5}$ and $\sin \beta = \sqrt{1 - \left(\frac{12}{13}ight)^2} = \frac{5}{13}$.Using the formula $\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$:$\cos(\alpha + \beta) = \left(\frac{4}{5}ight)\left(\frac{12}{13}ight) - \left(\frac{3}{5}ight)\left(\frac{5}{13}ight)$$= \frac{48}{65} - \frac{15}{65} = \frac{33}{65}$.Therefore,$\alpha + \beta = \cos ^{-1}\left(\frac{33}{65}ight)$.

Evaluate: $\cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)$

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