Question

How do you use half-angle formulas to find exact values of `sin((-17pi)/12)` and `cos((-17pi)/12)`?

Answer 1

Call `t = (-17pi)/12`
Use the trig identity: `2sin^2 t = 1 - cos 2t`

Explanation:

`sin t = sin (2pi - (17pi)/12) = sin ((7pi)/12)` =

`cos ((14pi)/12) = cos ((2pi)/12 + pi) = cos ((2pi)/12) = cos (pi/6) = 1/2`

`2sin^2 t = 1 - cos 2t = 1 - 1/2 = 1/2`

`sin^2 t = 1/4` --> `sin t = sin ((-17pi)/12) = +- 1/2`

`2cos^2 t = cos 2t + 1 = 1/2 + 1 = 3/2`

`cos^2 t = 3/4 -> cos t = cos ((-17pi)/12) = +- sqrt3/2`