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Confused about double angle identities

  1. Dec 4, 2007 #1
    1. The problem statement, all variables and given/known data
    sin4x-sin2x/sin2x=cos3x/cosx


    2. Relevant equations
    sin2x = 2sinxcosx


    3. The attempt at a solution

    LS = sin(2x + 2x) - sin2x/sin2x
    = sin2xcos2x + cos2xsin2x - sin2x/sin2x
    = 2sin2xcos2x - sin2x/sin2x
    This is where i get stuck...
    I don't know what happens if you try to: 2sinx2x - sin2x or 2sin2x/sin2x, is that possible or not? Can you guys help me solve this question? Thanks.
     
  2. jcsd
  3. Dec 4, 2007 #2

    rock.freak667

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    Quick question you are to prove that
    [tex]\frac{sin4x-sin2x}{sin2x}=\frac{cos3x}{cosx}[/tex]

    ?
     
  4. Dec 4, 2007 #3
    yup.
     
  5. Dec 4, 2007 #4

    dlgoff

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    I believe there are two more Relevant equations you need.
     
  6. Dec 4, 2007 #5
    dosen't matter i was able to solve it, but here's another problem which I'm now stuck with. [tex]cosx+cos2x+cos3x=cos2x(1+2cosx)[/tex].
    I've tried many things with this problem, but always seem to get lost.
    The only given identites which I'm given to use:
    [tex]cos2x=cos^2x-sin^2x[/tex]
    [tex]cos2x=2cos^2x-1[/tex]
    [tex]cos2x=1-2sin^2x[/tex]
    [tex]sin2x=2sinxcosx[/tex]
     
    Last edited: Dec 4, 2007
  7. Dec 4, 2007 #6

    rock.freak667

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    OK, since you have to SOLVE and not prove....
    if you expand the RHS you would see that the cos2x cancels out and you are left with

    cos(x)+cos(3x)=2cos[itex]^2[/itex](x)

    then expand out cos(3x) and see if anything gets simpler
     
  8. Dec 4, 2007 #7
    lol, sorry for not stating this, but i have to prove not solve.
     
  9. Dec 4, 2007 #8

    rock.freak667

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    well then expand out the LHS
    Recall that cos(3x)=cos(2x+x)
     
  10. Dec 4, 2007 #9
    yea then i get: cos2xcosx + sin2xsinx + cos2x + cosx
     
  11. Dec 4, 2007 #10

    rock.freak667

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    cos2x=2cos^2(x)-1
    and sin2x=2sinxcosx
    cos^2(x)+sin^2(x)=1

    expand out and find all in terms of cosx and hopefully it will work
     
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