Question

Show how ((cos^2(x)-sin^2(x))/sin(x))-2= sin^2(x)

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

You know you can edit your question right?

Answer 5

I see the latex just fine

Answer 6

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

Answer 7

Aha, hard refresh works!

Answer 8

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

Answer 9

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

Answer 10

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?

Answer 11

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

Answer 12

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

Answer 13

-1-2 does not equal 1

Answer 14

no questions?

Answer 15

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

Answer 16

they are not equal

Answer 17

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