Question

Question #c5acc

Answer 1

`sqrt(2 - sqrt3)/2`

Explanation:

`cos ((3pi)/4 - pi/3) = cos ((9pi - 4pi)/12) = cos ((5pi)/12)`
Find `cos ((5pi)/12)` by using trig identity:
`2cos^2 a = 1 + cos 2a`
In this case, trig table gives:
`cos 2a = cos ((10pi)/12) = cos (5pi/6) = - sqrt3/2`
Call `cos ((5pi)/12) = cos t`, we get:
`2cos^2 t = 1 - sqrt3/2 = (2 - sqrt3)/2`
`cos^2 t = (2 - sqrt3)/4`
`cos t = +- sqrt(2 - sqrt3)/2`
Since `(5pi)/12` is in Quadrant 1, its cos is positive -->
`cos ((5pi)/12) = cos t = sqrt(2 - sqrt3)/2`