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Romero Group Title

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

  • 2 years ago
  • 2 years ago

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  1. Romero Group Title
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    Yeah can someone show me the steps or what trig identities you used.

    • 2 years ago
  2. Romero Group Title
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    \[\frac{\cos^2(x)-\sin^2(x)}{\sin(x)}-2=\sin^2(x)\]

    • 2 years ago
  3. FoolForMath Group Title
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    it's not you, latex is not rendering properly.

    • 2 years ago
  4. FoolForMath Group Title
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    You know you can edit your question right?

    • 2 years ago
  5. myininaya Group Title
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    I see the latex just fine

    • 2 years ago
  6. myininaya Group Title
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    \[\frac{\cos^2(x)-\sin^2(x)}{\sin(x)-2}=\sin^2(x) ?\]

    • 2 years ago
  7. FoolForMath Group Title
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    Aha, hard refresh works!

    • 2 years ago
  8. Romero Group Title
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    I'm doing eigen values and eigen functions so the left expression has to equal the right.

    • 2 years ago
  9. myininaya Group Title
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    You said you wrote it wrong? Did I write it right?

    • 2 years ago
  10. Romero Group Title
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    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?

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

    • 2 years ago
  12. myininaya Group Title
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    \[x=\frac{\pi}{2}\] \[\frac{\cos^2(x)-\sin^2(x)}{\sin(x)}-2=\sin^2(x)\] \[\frac{\cos^2(\frac{\pi}{2})-\sin^2(\frac{\pi}{2})}{\sin(\frac{\pi}{2})}=\frac{0-1}{1}=-1\] But the other side is 1 when x=pi/2 Therefore both sides are not the same

    • 2 years ago
  13. myininaya Group Title
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    -1-2 does not equal 1

    • 2 years ago
  14. myininaya Group Title
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    no questions?

    • 2 years ago
  15. Romero Group Title
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    What do you suggest I do in order to get them equal?

    • 2 years ago
  16. myininaya Group Title
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    they are not equal

    • 2 years ago
  17. myininaya Group Title
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    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

    • 2 years ago
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