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# help with this trig problem!?

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- PhilipLv 61 month ago
cosx = 5/13, x in (0,pi/2). (sinx)^2=1-(5/13)^2=(169-25)/169=(12/13)^2.

then sinx = (12/13) and,tanx = sinx/cosx = (12/5).

Now tan(2x) = 2tanx/[1 - tan^2(x)]. For tanx = (12/5) we have tan(2x) =

2(12/5)/[1 - (12/5)^2] = -(24/5)/[(12/5)^2 -1] = -(24/5)(5^2)/[12^2-5^2] =

-120/(144-25) = -(120/119).

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- ted sLv 71 month ago
tan 2x = sin 2x / cos 2x = 2 sin x cos x / [ cos² x - sin²x ]....cos x = 5 / 13 & sin x = 12/13...you finish

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