Question

How do you evaluate `-sin((13pi)/12)`?

Answer 1

`sqrt(2 - sqrt3)/2`

Explanation:

Trig unit circle and property of supplementary arc-->
`- sin ((13pi)/12) = - sin (pi/12 + pi) = sin (pi/12)`
Evaluate `sin (pi/12)` by applying the trig identity:
`cos 2a = 1 - 2sin^2 a`
`cos (pi/6) = sqrt3/2 = 1 - 2sin^2 (pi/12)`
`2sin^2 (pi/12) = 1 - sqrt3/2 = (2 - sqrt3)/2`
`sin^2 (pi/12) = (2 - sqrt3)/4`
`sin (pi/12) = +- sqrt(2 - sqrt3)/2` -->
Since `sin (pi/12)` is positive, therefor:
`- sin ((13pi)/12) = sin (pi/12) = sqrt(2 - sqrt3)/2`