Question

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

Answer 1

`- sqrt(2 - sqrt3)/2`

Explanation:

`sin (13pi)/12 = sin (pi/12 + pi) = - sin (pi/12).`
Evaluate sin (pi/12) by the trig identity:
`cos 2a = 1 - 2sin^2 a`
Call `pi/12) = t`, we get `cos (2t) = cos (pi/6) = sqrt3/2`
The equation becomes:
`sqrt3/2 = 1 - 2 sin^2 t`
`2sin^2 t = 1 - sqrt3/2 = (2 - sqrt3)/2`
`sin^2 t = (2 - sqrt3)/4`
`sin t = sin (pi/12) = +- sqrt(2 - sqrt3)/2`
Because `sin (pi/12)` is positive, therefor,
`sin ((13pi)/12) = - sin (pi/12)` is negative.
Answer: `sin ((13pi)/12) = - (sqrt(2 - sqrt3))/2`