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Question 145 of 177

Q.Show that cos⁡−1(45)+cos⁡−1(1213)=cos⁡−1(3365)\cos^{-1}\left(\dfrac{4}{5}\right) + \cos^{-1}\left(\dfrac{12}{13}\right) = \cos^{-1}\left(\dfrac{33}{65}\right)

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2017Subjective· 3mImportance★★★★★
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Let α,β\alpha,\beta be the two inverse-cosine angles, compute cos⁡(α+β)\cos(\alpha+\beta) using the addition formula.

Let α=cos⁡−1(45)\alpha = \cos^{-1}\left(\dfrac45\right), so cos⁡α=45\cos\alpha = \dfrac45, and since α\alpha is acute, sin⁡α=35\sin\alpha = \dfrac35.

Let β=cos⁡−1(1213)\beta = \cos^{-1}\left(\dfrac{12}{13}\right), so cos⁡β=1213\cos\beta = \dfrac{12}{13}, and since β\beta is acute, sin⁡β=513\sin\beta = \dfrac{5}{13}.

cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β=45⋅1213−35⋅513=4865−1565=3365\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta = \dfrac45\cdot\dfrac{12}{13} - \dfrac35\cdot\dfrac{5}{13} = \dfrac{48}{65}-\dfrac{15}{65} = \dfrac{33}{65}

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