## anonymous 4 years ago Show how ((cos^2(x)-sin^2(x))/sin(x))-2= sin^2(x)

1. anonymous

Yeah can someone show me the steps or what trig identities you used.

2. anonymous

$\frac{\cos^2(x)-\sin^2(x)}{\sin(x)}-2=\sin^2(x)$

3. anonymous

it's not you, latex is not rendering properly.

4. anonymous

You know you can edit your question right?

5. myininaya

I see the latex just fine

6. myininaya

$\frac{\cos^2(x)-\sin^2(x)}{\sin(x)-2}=\sin^2(x) ?$

7. anonymous

Aha, hard refresh works!

8. anonymous

I'm doing eigen values and eigen functions so the left expression has to equal the right.

9. myininaya

You said you wrote it wrong? Did I write it right?

10. anonymous

Sorry I deleted the wrong one already. The one I wrote is right. Anyways I should rephrase my questions. I want the left side to equal the right side. How can I change the left side in order to make the equation true so that the left hand side equals the right hand side?

11. myininaya

$\frac{1-\sin^2(x)-\sin^2(x)}{\sin(x)}-2=\frac{1-2\sin^2(x)}{\sin(x)}-2=\frac{1-2\sin^2(x)-2\sin(x)}{\sin(x)}$ hmmm.... I don't think they are the same .... Did you see if you could find a counterexample?

12. myininaya

$x=\frac{\pi}{2}$ $\frac{\cos^2(x)-\sin^2(x)}{\sin(x)}-2=\sin^2(x)$ $\frac{\cos^2(\frac{\pi}{2})-\sin^2(\frac{\pi}{2})}{\sin(\frac{\pi}{2})}=\frac{0-1}{1}=-1$ But the other side is 1 when x=pi/2 Therefore both sides are not the same

13. myininaya

-1-2 does not equal 1

14. myininaya

no questions?

15. anonymous

What do you suggest I do in order to get them equal?

16. myininaya

they are not equal

17. myininaya

this is not an identity I gave you a counterexample above try pluggin' in pi/2 for x we do not get the same thing on both sides unless you meant to type something else