How do you evaluate $cos (arcsin (- \frac{1}{2}) + arctan (\frac{2}{3}))$ $cos(arcsin(-1/2)+arctan(2/3))$ ?

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How do you evaluate $cos (arcsin (- \frac{1}{2}) + arctan (\frac{2}{3}))$ $cos(arcsin(-1/2)+arctan(2/3))$ ?

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Answer 1 · 2016-12-19T12:32:19

$\frac{\sqrt{13}}{26} (3 \sqrt{3} + 2)$ $sqrt13/26(3sqrt3+2)$

$= 0.9979$ $=0.9979$ , nearly.

Explanation:

Let $a = a r c sin (- \frac{1}{2}) = - \frac{π}{6} and b = a r c tan (\frac{2}{3}) \in Q_{1}$ $a = arc sin (-1/2) = -pi/6 and b = arc tan (2/3) in Q_1$ , giving $tan b = \frac{2}{3}, sin b = \frac{2}{\sqrt{13}} and cos b = \frac{3}{\sqrt{13}}$ $tan b = 2/3, sin b = 2/sqrt 13 and cos b =3/sqrt 13$ .

The given expression is

$cos (a + b)$ $cos (a+b)$

$= cos (- \frac{π}{6} + b) = cos (- \frac{π}{6}) cos b - sin (- \frac{π}{6}) sin b$ $= cos (-pi/6+b)=cos (-pi/6)cosb-sin(-pi/6)sin b$

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

$= \frac{\sqrt{13}}{26} (3 \sqrt{3} + 2)$ $=sqrt13/26(3sqrt3+2)$

$= 0.9979$ $=0.9979$ , nearly.

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