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Trig Identity problem (double angle formulas incl.)

  1. Nov 10, 2007 #1
    1. The problem statement, all variables and given/known data
    (1) (1+cosx)/(sinx)= cot (x/2)
    (2) 2 csc 2x= sec x csc x
    (3) cos^6 x- sin^6 x= cos 2x(1 - 1/4sin^2 2x) ( I think this has to do something with subtracting -3a^2b^2, since I need to get a-2ab+b to factor it..?)

    2. Relevant equations
    Addition and Subtraction formulas (sin,cos and tan)
    Double angle formulas (Sin 2x, cos 2x (3 formulas), tan 2x)
    Reciprocal id., quotient id., Pythagorean Id.


    3. The attempt at a solution

    Could someone explain the thought process of proving these identities, I have zillion more of these, and I think I am missing something. I think it is the double angle formulas that are confusing me. Thank you.
     
  2. jcsd
  3. Nov 10, 2007 #2

    Dick

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    I don't remember a lot of half/double angle formulas myself. There's too many of them. But lets start with the first one, cos(x)=cos(x/2+x/2). Now use the cos addition formula and get cos(x)=cos(x/2)cos(x/2)-sin(x/2)sin(x/2)=cos^2(x/20-sin^2(x/2). Do the same thing with sin(x)=sin(x/2+x/2). Now put those into the left hand side. Similarly for the second one, csc(2x)=1/sin(2x). sin(2x)=sin(x+x)=2sin(x)cos(x), right? Put that in.
    I'm still thinking about the third...
     
  4. Nov 10, 2007 #3

    Dick

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    For the third one, I think you just have to slog it out. I did the 2x=x+x thing to get the right side completely in terms of sin(x) and cos(x). Then you just have to use sin^2(x)+cos^2(x)=1 enough times in the right places to cancel everything but the sixth powers.
     
  5. Nov 10, 2007 #4
    Thanks, the second one was easy (funny how stupid one can feel for not getting this..=P) . However, I can see where you're going with replacing 2x=x+x for cosx and sinx, but I'm getting stuck at the first step for both 1 and 3.

    I am having trouble with especially the third one, where I'm trying to figure out the last part of the problem (1- 1/4sin^2 2x).. I don't know exactly how to get it in the form of cos2x.


    I appreciate your help =D
     
  6. Nov 10, 2007 #5

    Dick

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    Try 1 again. That's by far the easier of the two. Post what you've done so for. I'm not sure what you are thinking of as 'the first step'.
     
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