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Problem Verifing Identities in Trig

  1. Mar 6, 2008 #1
    Hello this one has stumped me for about a day and I have tried several different approaches. maybe someone could point me in the right direction.
    cot^2x-1/1+cot^2x=2Cos^2x-1

    I startedon the Left side tring to prove the equation.

    cot^2x-1/csc^2x using Pyth. Theory

    (cot^2x/csc^2x)-(1/csc^2x) reciprocal Identites

    ((cos^2x/sin^2x)/csc^2x)-(sin^2x) reciprocal Identites

    (cos^2x/sin^2x)*(sin^2x)-(sin^2x) //sin^2x will cancel outleaving

    cos^2x - sin^2x this does not = 2Cos^2x -1

    Could somone point me in the right direstion .Thank s in advanced Rada3b
     
  2. jcsd
  3. Mar 6, 2008 #2
    Are you sure that

    [tex]\cos^2x - \sin^2x = 2\cos^2x - 1[/tex]

    is not a true statement? What is 1 equal to in terms of trig identities?
     
  4. Mar 7, 2008 #3
    You are correct it is not. I was trying to prove that the equation on the left is = to the equation on the right in the begining of the post. Could I simplify the equation on the right anymore that might = cos^2x -sin^2x??
     
  5. Mar 7, 2008 #4

    Gib Z

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    Yes you can, but taking Tedjins hint =] Ill make it a tiny bit more obvious - Pythagorean identities?
     
  6. Mar 7, 2008 #5

    HallsofIvy

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    "You are correct it is not"? Tedjn was not telling you it is not true! He was suggesting that you check your assumption that it is not true again! Specifically, what do you get if you replace [itex]sin^2(x)[/itex] on the left side by [itex]1- cos^2(x)[/itex]?
     
  7. Mar 7, 2008 #6
    Have you learned about the power reduction formulas yet? It looks easier by using those to me.
     
  8. Mar 8, 2008 #7
    No, we haven't learned the power reduction formula but it sounds alot easier Hydrargyrum. I will look into it and continue trying. I did sub Pi/4 in for x on both side ofthe equation and yes they are =. So back to proving again, Thanks EV1 for the help.
     
  9. Mar 8, 2008 #8

    Gib Z

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    Would you PLEASE listen to any of our hints? The power reduction formulas come from what we are trying to tell you. I don't see why our hint, which has become more so an obvious instruction, is so hard to execute?
     
  10. Mar 8, 2008 #9
    Thank you all. I was on the right track.I must be just overstressed atm not to see something so obvious. I got it now.
     
  11. Mar 9, 2008 #10
    Sorry. I didn't read your posts
     
  12. Mar 9, 2008 #11
    what are the three most basic formulas for proving trig equations?

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  13. Mar 24, 2008 #12
    L= cot^2x-1/csc^2x =cot^2x/csc^2x -1/csc^2x =(cos^2x/sin^2x)/(1/sin^2x)-sin^2x=cos^2x-sin^2x=cos^2x-(1-cos^2x)=cos^2x-1+ cos^2x=2cos^2x-1=R

    Everybody gave you hints. Its been a while now. This should be your answer. Please correct if I am wrong.
     
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