Double-angle identities?

Question

(Use the identities to find the values of the six trig functions.)
Can someone please explain the process of figuring these kinds of problems out?

1.) x, given cos2x=(-5/12) and pi/2 < x < pi
2.) 2cos^2(67 1/2)-1

EDIT: The answers in the book are...

1.) cosx= √42/12, sinx= √102/12, tanx = -√119/7
2.) -√2/2

Answer 1

cos 2x = 2 cos^2 x - 1 = 1 - 2 sin^2 x.
Since sin x is positive in the given quadrant, I'll use that one. ;-)
So -5/12 = 1 - 2 sin^2 x
=> 2 sin^2 x = 17/12
=> sin^2 x = 17/24
So sin x = √(17/24)
cos x = -√(7/24) (since cos x is negative in the given quadrant).
From here you can work out the others: tan x = -√(17/7), sec x = -√(24/7), cosec x = √(24/17), cot x = -√(7/17).

2) 2 cos^2 (67 1/2°) - 1
Recall: cos 2A = 2cos^2 A - 1
= cos (2 (67 1/2°))
= cos (135°)
= -1/√2.