## anonymous 5 years ago Trig-- Solve for 0< x< 2pi, exact values in radian .........cos (2 tan ^-1(5/12))

1. anonymous

let y = tan^-1(5/12) => tan(y) = 5/12 use sin^2y + cos^2y = 1 -> 1 + tan^2y = [1/cos^2y] to find cos(y) in terms of tan(y) Than use double angle formula for cos(2x) to find cos(2y) in terms of cos(y).

2. anonymous

$\sin^2 y + \cos^2y = 1 \implies \tan^2y + 1 = \frac{1}{\cos^2y} \implies \cos^2 = \frac{1}{\tan^2y+1}$ We know tan(y) = 5/12 => tan^2 y = (5/12)^2 -> cos^2 y = 144/169 -> sin^2 y = 1- 144/169 = 25/169 cos(2y) = cos^2(y) - sin^2 y = 144/169 - 25/169 = 119/169.

3. anonymous

thank you very much, its greatly appreciated! :)

4. anonymous

i understand it now!

5. anonymous

no problem.