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Question: $\cos \left[\sin^{-1}\left(\frac{3}{5}\right) + \cos^{-1}\left(\frac{12}{13}\right)\right] =$...

cos⁡[sin⁡−1(35)+cos⁡−1(1213)]=\cos \left[\sin^{-1}\left(\frac{3}{5}\right) + \cos^{-1}\left(\frac{12}{13}\right)\right] =

A

3665\frac{36}{65}

B

1265\frac{12}{65}

C

3365\frac{33}{65}

D

365\frac{3}{65}

Answer

3365\frac{33}{65}

Explanation

Solution

Let

α=sin⁡−1(35)\alpha = \sin^{-1}\left(\frac{3}{5}\right) and β=cos⁡−1(1213)\beta = \cos^{-1}\left(\frac{12}{13}\right).

Then:

sin⁡α=35\sin\alpha = \frac{3}{5} and cos⁡α=1−(35)2=45\cos\alpha = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5};

cos⁡β=1213\cos\beta = \frac{12}{13} and sin⁡β=1−(1213)2=513\sin\beta = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \frac{5}{13}.

Using the cosine addition formula:

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 = \frac{4}{5}\cdot\frac{12}{13} - \frac{3}{5}\cdot\frac{5}{13} = \frac{48}{65} - \frac{15}{65} = \frac{33}{65}.