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Proving trigonometric equation?

  1. Sep 20, 2007 #1
    for x between 0 and 2pie, solve cos(4x)=sin(2x)....

    is this a proving trigonometric equation? i dont think it is
     
  2. jcsd
  3. Sep 20, 2007 #2

    D H

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    This is not a tautology, which you should be able to see if you expand cos(4x) in terms of sin(2x). The problem is find the value(s) of x between 0 and 2 pi for which the given expression is true.
     
  4. Sep 20, 2007 #3
    so how do u do that algebraically
     
  5. Sep 20, 2007 #4

    D H

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    I gave you a big hint in my first post. I'll repeat to make it blatantly obvious: Expand cos(4x) in terms of sin(2x).
     
  6. Sep 20, 2007 #5
    i understand that,i'm not an idiot i'm just missing something very crucial to be able to determine the answer.

    i dont know how to expand it into terms of sin, i have checked through my textbook i have looked everywhere. I know this is probably really easy but i'm missing that key concept which i can't figure out to be able to put cos in terms of sin
     
  7. Sep 20, 2007 #6

    D H

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    Start with cos(4x) = cos(2x+2x). Can you proceed from here?
     
  8. Sep 20, 2007 #7
    ok i think that helped me realise the identity..

    Cos(A+B)=CosACosB-sinAsinB
    cos(2x+2x)=cos2xcos2x-sin2xsin2x
    =2cos2x-2sin2x
    2cos2x-2sin2x=sin2x
    2cos2x=3sin2x
    2/3=tan2x?

    right?
     
  9. Sep 20, 2007 #8

    D H

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    Good so far.
    You are saying the equivalent of [itex]u*u = 2*u[/tex], which is obviously incorrect.
     
  10. Sep 20, 2007 #9
    Ok so this is what i have so far
    cos4x=sin2x
    cos(2x+2x)=sin2x
    cos^22x-sin^22x=sin2x

    as far as i know, cos^2x+sin^2x=1, so when i have -sin^22x, i can't complete that property correct? or is it just -1
     
  11. Sep 20, 2007 #10

    D H

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    Use the identity [itex]\cos^2(2x)+\sin^2(2x) = 1[/itex] to eliminate the [itex]\cos^2(2x)[/itex] term: [itex]\cos^2(2x)-\sin^2(2x) = 1 - 2\sin^2(2x)[/itex]. Applying this to the original problem yields

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

    which is a quadratic equation in [itex]\sin(2x)[/itex].
     
  12. Sep 22, 2007 #11

    Curious3141

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    A more elementary solution would be to utilise the identity

    [tex]\sin(\frac{\pi}{2} \pm y) = \cos y[/tex]

    Since multiple angles are in play here, add 2n*pi to the argument, for integral n.

    [tex]\sin(2n\pi + \frac{\pi}{2} \pm y) = \cos y[/tex]

    giving [tex]\sin(\frac{1}{2}(4n + 1)\pi \pm y) = \cos y[/tex]

    Now substitute that into the cosine expression in the LHS of the orig. equation (y = 4x), remove the sines on both sides, and you have a linear equation to solve. Simply list the multiple solutions in the required range by varying n (n can be zero, positive or negative). Don't have to worry about the plus/minus part too much, since all the solutions with one sign are included when solving for the other, but you need to establish this.
     
  13. Sep 23, 2007 #12
    Furthering D H's help,

    After subbing cos (4x) for 1- sin^2 (2x), make a substitution to form a simple quadratic equation which you can solve (ie. let sin 2x = u)
     
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