Question

How do you evaluate `1-2cos^2(pi/8)`?

Answer 1

`- sqrt2/2`

Explanation:

Call `cos (pi/8) = cos t`. Then `cos 2t = cos (pi/4)`
Replace cos^2 t by (1 - sin^2 t), we get:
`f(t) = 1 - 2(cos^2 t) = 1 - 2(1 - sin^2 t) = - 1 + 2sin^2 t`
Use trig identity:
`cos 2t = 1 - 2sin^2 t` --> `- cos 2t = - 1 + sin^2 t`
There for:
`f(t) = - 1 + 2sin^2 t = - cos 2t = - cos (pi/4) = - sqrt2/2`
Check by calculator.
`cos (pi/8) = cos (22.5) = 0.924` --> cos^2 (pi/8) = 0.854
`1 - 2cos^@ (pi/8) = 1 - 1.708 = - 0.707`
`-sqrt2/2 = -1.414/2 = - 0.707`. OK