Question

How do you evaluate `cos ((7pi)/12)`?

Answer 1

`- sqrt(2 - sqrt3)/2`

Explanation:

Use the trig identity: `cos 2a = 2cos^2 a - 1`
`cos 2a = cos ((14pi)/12) = cos ((7pi)/6) = cos (pi/6 + pi) =`
`-cos (pi/6) = - sqrt3/2`
`cos ((7pi)/6) = -sqrt3/2 = 2cos^2 ((7pi)/12) - 1`
`2cos^2 ((7pi)/12) = 1 - sqrt3/2 = (2 - sqrt3)/2`
`cos^2 ((7pi)/12) = (2 - sqrt3)/4`
`cos ((7pi)/12) = +- sqrt(2 - sqrt3)/2`
Since `(7pi)/12` is in Quadrant II, its cos is negative -->
`cos ((7pi)/12) = - sqrt(2 - sqrt3)/2`