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Trigonometry 7 Online
OpenStudy (anonymous):

sin(2x)/sin^2x

OpenStudy (psymon):

Simplify I'm guessing?

OpenStudy (anonymous):

yea

OpenStudy (psymon):

Well, you have to know the identity for sin2x here. Theidentity for that is just 2sinxcosx. Once you have that in your numerator you can simplify :3

OpenStudy (anonymous):

you beat me.

OpenStudy (anonymous):

cosx/sinx is either cot or tan. and sinx/cosx is the another one of cot or cos. it may be needed.

OpenStudy (anonymous):

how do you simplify the sin^2x, do you make it 1-cos^2x

OpenStudy (psymon):

No need. Once you simplify sin(2x) you'll have this: \[\frac{ 2\sin(x)\cos(x) }{ \sin ^{2}x }= \frac{ 2\cos(x) }{ \sin(x) } = 2\cot(x)\] Thats all that is needed :3

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