How do you evaluate cos(arctan(-2) + arccos(5/13))?

1 Answer
Jul 30, 2016

=29/(13sqrt 5)

Explanation:

Let a = arc tan (-2) in Q3,

wherein sine and tangent are negative.

Then, tan a = -2, sin a =-2/sqrt 5 and cos a = 1/sqrt 5

Let b = arc cos (5/13) in Q1, wherein sine is positive..

Then, cos b = 5/13 and sin b =12/13.

Now, the given expression is

cos(a+b)

=cos a cos b - sin a sin b

=(1/sqrt 5)(5/13)-(-2/sqrt 5)(12/13)

=29/(13sqrt 5)