trig identity help...

**katiethegreat** · May 17th 2011, 09:07 AM

I can't solve this :S

cosxsinx/tanx = 1 - sin^2x

thanks

**TheEmptySet** · May 17th 2011, 09:10 AM

Originally Posted by katiethegreat

I can't solve this :S

cosxsinx/tanx = 1 - sin^2x

thanks

Hint:

$\frac{\cos(x)\sin(x)}{\tan(x)}=\frac{\cos(x)\sin(x )}{1}\cdot \frac{\cos(x)}{\sin(x)}=...$

**katiethegreat** · May 17th 2011, 09:20 AM

I don't know why....but i still don't get it :/

**TheEmptySet** · May 17th 2011, 09:21 AM

Originally Posted by katiethegreat

I don't know why....but i still don't get it :/

what is "it" you don't understand how I change what you had into this

$\frac{\cos(x)\sin(x)}{\tan(x)}=\frac{\cos(x)\sin(x )}{1}\cdot \frac{\cos(x)}{\sin(x)}=...$

or you don't know how to finish the problem. Where are you stuck?

**katiethegreat** · May 17th 2011, 09:33 AM

both :/

**masters** · May 17th 2011, 09:43 AM

Hi katiethegreat,

Maybe if you see it like this, it would help:

$\dfrac{\cos (x) \sin(x)}{\tan (x)}=$

$\dfrac{\cos(x) \sin(x)}{\frac{\sin (x)}{\cos (x)}}=$

$\dfrac{\cos (x) \sin (x)}{1} \cdot \frac{\cos (x)}{\sin (x)}=$

He just used the identity: $\tan (x) = \frac{\sin (x)}{\cos (x)}$

You should be able to finish now if you know $\sin^2(x)+\cos^2(x)=1$

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