anonymous
  • anonymous
Show how ((cos^2(x)-sin^2(x))/sin(x))-2= sin^2(x)
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Yeah can someone show me the steps or what trig identities you used.
anonymous
  • anonymous
\[\frac{\cos^2(x)-\sin^2(x)}{\sin(x)}-2=\sin^2(x)\]
anonymous
  • anonymous
it's not you, latex is not rendering properly.

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anonymous
  • anonymous
You know you can edit your question right?
myininaya
  • myininaya
I see the latex just fine
myininaya
  • myininaya
\[\frac{\cos^2(x)-\sin^2(x)}{\sin(x)-2}=\sin^2(x) ?\]
anonymous
  • anonymous
Aha, hard refresh works!
anonymous
  • anonymous
I'm doing eigen values and eigen functions so the left expression has to equal the right.
myininaya
  • myininaya
You said you wrote it wrong? Did I write it right?
anonymous
  • 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?
myininaya
  • 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?
myininaya
  • 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
myininaya
  • myininaya
-1-2 does not equal 1
myininaya
  • myininaya
no questions?
anonymous
  • anonymous
What do you suggest I do in order to get them equal?
myininaya
  • myininaya
they are not equal
myininaya
  • 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

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