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

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Answer 1 · 2016-12-19T12:32:19

$sqrt13/26(3sqrt3+2)$

$=0.9979$ , nearly.

Let $a = arc sin (-1/2) = -pi/6 and b = arc tan (2/3) in Q_1$ , giving $tan b = 2/3, sin b = 2/sqrt 13 and cos b =3/sqrt 13$ .

The given expression is

$cos (a+b)$

$= cos (-pi/6+b)=cos (-pi/6)cosb-sin(-pi/6)sin b$

$=sqrt 3/2(3/sqrt 13)+1/2(2/sqrt 13)$

$=sqrt13/26(3sqrt3+2)$

$=0.9979$ , nearly.

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