## Romero Group Title Show how ((cos^2(x)-sin^2(x))/sin(x))-2= sin^2(x) 2 years ago 2 years ago

1. Romero Group Title

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

2. Romero Group Title

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

3. FoolForMath Group Title

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

4. FoolForMath Group Title

You know you can edit your question right?

5. myininaya Group Title

I see the latex just fine

6. myininaya Group Title

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

7. FoolForMath Group Title

Aha, hard refresh works!

8. Romero Group Title

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

9. myininaya Group Title

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

10. Romero Group Title

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 Group Title

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

12. myininaya Group Title

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

13. myininaya Group Title

-1-2 does not equal 1

14. myininaya Group Title

no questions?

15. Romero Group Title

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

16. myininaya Group Title

they are not equal

17. myininaya Group Title

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