# trig

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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 -

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 -

tan x = 13 cot x

• trig -

tan x = 13 cot x
tan x = 13/tan x
tan^2 x = 13
tan x = ±√13
x = 74.5º,105.5º, 254.5º, 285.5º

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