Question

How do you evaluate `3cos((17pi)/6)+2cos((-5pi)/3 )`?

Answer 1

`(-3sqrt3+2)/2=-1.2132,`nearly.

Explanation:

Use `cos (-a) = cos a` and cosine is negative in `Q_2` and positive in `Q_4`.

`3 cos (17/6pi)+ 2 cos (-5/3pi)`

`=3 cos (3pi-pi/6) + 2 cos(-(2pi-pi/3))`

`=-3cos (pi/6)+2cos(2pi-pi/3)`

`=-3 cos (pi/6)+2cos(pi/3)`

`=(-3sqrt3+2)/2`