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

1 Answer
Dec 19, 2016

sqrt13/26(3sqrt3+2)

=0.9979, nearly.

Explanation:

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.