Trig
posted by Hayden .
given sin(4x)=5/13 in Q II
find Tan(2x) and cos(2x)

sin 4 x = 5/13
then cos 4 x = 12/13 (5,12,13 right triangle)
and
tan 4x = sin 4 x /cos 4x = 5/12
so identities:
sin 4x = 2 sin 2x cos 2x = 5/13
cos 4x = cos^2 2x  sin^2 2x = 12/13
tan 4x = 2 tan 2x / (1tan^2 2x) = 5/12
so
cos^2 2x = 12/13 + sin^2 2x
= 12/13 + [ 5/(26 cos 2x)]^2
= 12/13 + 25 /(114244 cos^2 2x)
114244 cos^4 2x = 105456 cos^2 2x + 25
cos^4 2x  .9231 cos^2 2x  .00021883=0
cos^2 2x = (1/2)[.9231 +/ .96124]
cos^2 2x =.4616 +/ .4806
use + sign
cos^2 2x = .94217
cos 2x = .9707
check my arithmetic then go on and get the sin 2x and tan 2x