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# How do you evaluate cos(arcsin(-1/2)+arctan(2/3))?

Dec 19, 2016

$\frac{\sqrt{13}}{26} \left(3 \sqrt{3} + 2\right)$

$= 0.9979$, nearly.

#### Explanation:

Let $a = a r c \sin \left(- \frac{1}{2}\right) = - \frac{\pi}{6} \mathmr{and} b = a r c \tan \left(\frac{2}{3}\right) \in {Q}_{1}$, giving $\tan b = \frac{2}{3} , \sin b = \frac{2}{\sqrt{13}} \mathmr{and} \cos b = \frac{3}{\sqrt{13}}$.

The given expression is

$\cos \left(a + b\right)$

$= \cos \left(- \frac{\pi}{6} + b\right) = \cos \left(- \frac{\pi}{6}\right) \cos b - \sin \left(- \frac{\pi}{6}\right) \sin b$

$= \frac{\sqrt{3}}{2} \left(\frac{3}{\sqrt{13}}\right) + \frac{1}{2} \left(\frac{2}{\sqrt{13}}\right)$

$= \frac{\sqrt{13}}{26} \left(3 \sqrt{3} + 2\right)$

$= 0.9979$, nearly.

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