Question

What is the value of `cos[pi/3-pi/4]`?

Answer 1

`(sqrt2 + sqrt6)/4`

Explanation:

Apply the trig identity:
cos (a - b) = cos a.cos b + sin a.sin b
`cos (pi/3 - pi/4) = cos (pi/3).cos (pi/4) + sin (pi/3).sin (pi/4) =`
Trig table of special arcs gives:
`cos (pi/3) = 1/2` , and `cos (pi/4) = sqrt2/2`
`sin (pi/3) = sqrt3/2`, and `sin (pi/4) = sqrt2/2`
There for:
`cos (pi/3 - pi/4) = (1/2)(sqrt2/2) + (sqrt3/2)(sqrt2/2) =`
`= sqrt2/4 + sqrt6/4 = (sqrt2 + sqrt6)/4`