Let cos 67.5° = [√(2(+√2)]/2, find tan 67.5°. Show work and simplify.

Question

I'm not too sure if i'm doing this correct.

I know that the given is cos 67.5° = [√(2(+√2)]/2

sin^2 x + cos^2 x = 1

x=67.5°

sin^2 67.5° + cos^2 67.5° = 1
sin^2 67.5° = 1 - ([√2(+√2)]/2)^2
sin^2 67.5° = 1 - (2+√2)/4
sin^2 67.5° = (2-√2)/4
sin 67.5° = ([√2(+√2)]/2)

and now i needa find tan.
tan 67.5° = ?

Answer 1

well, tan = sin/cos, right?

Also, 67.5° = 3pi/8 = tan(1/2 * 3pi/4)
so you can use the half-angle formula

tan(x/2) = sinx/(1+cosx)
tan 3pi/8 = (1/√2)/(1-1/√2)
= (1/√2) / (√2-1)/√2
= 1/(√2-1)
= √2+1
or, √((2+√2)/(2-√2))