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Trig Identities

  1. Nov 18, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove the following identity:

    cos2A/1 + sin2A = cotA - 1 / cotA + 1



    2. Relevant equations



    3. The attempt at a solution

    I proved the right side, which eventually lead up to cosA - sinA / cosA + sinA

    I have NO idea how to do the left side. I have wasted roughly 10-15 sheets of paper now trying to prove it, but it all leads up to nothing. I tried MANY different solutions...
     
  2. jcsd
  3. Nov 18, 2008 #2
    did u use the double angle formulas of sin2u=2sinucosu and cos2u=1-2sin^2u?
     
  4. Nov 18, 2008 #3
    [​IMG]
     
  5. Nov 18, 2008 #4
    Vitaly, thanks for trying BUT I have a question about where sin^2a(cot^2 - 1) came from.. could you add me to MSN please? supaflygt@hotmail.com
     
  6. Nov 18, 2008 #5
    [​IMG]
    no probs :wink:
    Look carefully, it's sine squared, not sin(2A). My 2A are big
     
  7. Nov 18, 2008 #6
    Yeah, I know. I'm curious how you got sin^2a/sin^2a + sin^2a in the equation (2nd step) :P
     
  8. Nov 18, 2008 #7
    Please use brackets where necessary. (sinA)^2/(sinA)^2=1 and as you must know if you multiply something by 1, it doesn't change anything. It's a very common trick in math, e.g. integrating logs, taking inverse of a matrix, etc..
    So, I muultipled (cosA)^2 by 1. And (sinA)^2 is therefore a common factor.
    It's like (5+7)=(5*7/7+7)=7(5/7+1)
     
  9. Nov 18, 2008 #8
    Well, I tried it doing my own thing, and here is the result:
    EDIT: it's not -1 or 0, but it must be something.
    [​IMG]
     
  10. Nov 18, 2008 #9
    Solved it now, thanks for your help.
     
  11. Nov 19, 2008 #10
    my friend they are not equal to negative one. however you can prove both sides are
    (cos A-sin A)/(cos A+sin A)
     
  12. Nov 19, 2008 #11
    My God, seeing those divisions just scares me.
    Revisit fractions please and formulas of (a+b)2, a2-b2

    And consider this: (52-32)/(5+3)2 is NOT equal -1.
    (52-32)/(5+3)2=(5-3)(5+3)/((5+3)(5+3)). Divide by a common factor of 5+3, leaving you with (5-3)/(5+3)

    Learn (a+b)2=(a+b)(a+b)=a2+2ab+b2, a2-b2=(a-b)(a+b) NOW
     
  13. Nov 19, 2008 #12
    I know they aren't equal to -1. I proved both sides were (cosa-sina)/(cosa+sina), though, which was why I said it was solved
     
  14. Nov 20, 2008 #13
    He says so because of your amazing cancellation.
     
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