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.