# How do you calculate cos(tan^-1(4/3)-sin^-1(12/13))?

Oct 13, 2016

$\cos \left({\tan}^{-} 1 \left(\frac{4}{3}\right) - {\sin}^{-} 1 \left(\frac{12}{13}\right)\right) = \frac{63}{65}$

#### Explanation:

$\cos \left({\tan}^{-} 1 \left(\frac{4}{3}\right) - {\sin}^{-} 1 \left(\frac{12}{13}\right)\right)$
The restriction for the range of ${\tan}^{-} 1 x$ is $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ and the restriction for ${\sin}^{-} 1 x$ is$\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$.

We need to draw two triangles. They will both be in quadrant I since both the arguments are positive. Let's call the first one triangle A and the second one triangle B.

For triangle A the opposite is 4 and adjacent is 3 so the hypotenuse is 5. For triangle B the opposite is 12 and the hypotenuse is 13 so the adjacent is 5. Now

Use the formula $\cos \left(A - B\right) = \cos A \cos B + \sin A \sin B$ to evaluate. Note that $A = {\tan}^{-} 1 \left(\frac{4}{3}\right)$ and $B = {\sin}^{-} 1 \left(\frac{12}{13}\right)$. Therefore,

$\cos \left({\tan}^{-} 1 \left(\frac{4}{3}\right) - {\sin}^{-} 1 \left(\frac{12}{13}\right)\right)$

$= \cos \left({\tan}^{-} 1 \left(\frac{4}{3}\right)\right) \cos \left({\sin}^{-} 1 \left(\frac{12}{13}\right)\right) + \sin \left({\tan}^{-} 1 \left(\frac{4}{3}\right)\right) \sin \left({\sin}^{-} 1 \left(\frac{12}{13}\right)\right)$

$= \cos A \cos B + \sin A \sin B$-> Use triangles A and B to find the ratios

$= \frac{3}{5} \cdot \frac{5}{13} + \frac{4}{5} \cdot \frac{12}{13}$

$= \frac{3}{13} + \frac{48}{65} = \frac{63}{65}$