Posted by **Amandeep** on Thursday, May 1, 2008 at 12:12am.

tanx = 5/12 and sinx<0

Find sin2x and cos2x

How to get the answers of sin2x = 120/169 and cos 2x = 119/169?

- trig -
**Reiny**, Thursday, May 1, 2008 at 8:28am
tanx = 5/12

r^2 = x^2 + y^2

= 12^2 + 5^2

r = 13

tanx >0 and sinx<0, so x must be in quad. III and

and sinx = -5/13 , cosx = -12/13

sin 2x = 2sinxcosx = 2(-5/13)(-12/13)

= 120/169

cos 2x = cos^2 x - sin^2 x

= 144/169-25/169

=119/169

do the others the same way, or the way

drwls showed you in the other post.

- trig -
**Mel**, Thursday, May 1, 2008 at 12:04pm
tan x = 13 cot x

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