Posted by **Hayden** on Wednesday, February 18, 2009 at 3:35pm.

given sin(4x)=5/13 in Q II

find Tan(2x) and cos(2x)

- Trig -
**Damon**, Wednesday, February 18, 2009 at 4:00pm
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 / (1-tan^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

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