Question

How do you find the value of `cos((17pi)/12)`?

Answer 1

`- sqrt(2 - sqrt3)/2`

Explanation:

Trig unit circle -->
`cos ((17pi)/12) = cos ((5pi)/12 + pi) = - cos ((5pi)/12)` (1)
Find `cos ((5pi)/12` by using the trig identity:
`2cos^2 t = 1 + cos 2t`
Call `cos ((5pi)/12) = cos t`
Trig table -->
`2cos^2 t = 1 + cos ((10pi)/12) = 1 + cos ((5pi)/6) = 1 - sqrt3/2`
`cos ^2 t = (2 - sqrt3)/4`
`cos t = cos ((5pi)/12) = sqrt(2 - sqrt3)/2`
Take the positive answer because `cos ((5pi)/12)` is positive.
Back to (1)
`cos ((17pi)/12) = - cos t = - sqrt(2 - sqrt3)/2`