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Math Help - Trig Equations

  1. #1
    Member >_<SHY_GUY>_<'s Avatar
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    Question Trig Equations

    Cosx*Cos3x - Sinx*Sin3x = 0

    I have tried sing the addition formula, and I end up with a lot of work. I'm trying it for the third time, and nothing seems to cancel.

    Is there another way to do this?
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  2. #2
    Master Of Puppets
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    Maybe use some triple angle formula? Look here.

    Double & Triple Angle formulas
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  3. #3
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    I would suggest try playing around with the Sum and Difference and Product-to-Sum Formulas.

    \displaystyle sin(u\pm v)=sin(u)cos(v)\pm cos(u)sin(v), \ cos(u\pm v)=cos(u)cos(v)\mp sin(u)sin(v),  \ sin(u)sin(v)=\frac{1}{2}[cos(u-v)-cos(u+v), \ cos(u)cos(v)]=\frac{1}{2}[cos(u-v)+cos(u+v)]
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  4. #4
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    Quote Originally Posted by dwsmith View Post
    I would suggest try playing around with the Sum and Difference and Product-to-Sum Formulas.

    \displaystyle sin(u\pm v)=sin(u)cos(v)\pm cos(u)sin(v), \ cos(u\pm v)=cos(u)cos(v)\mp sin(u)sin(v),  \ sin(u)sin(v)=\frac{1}{2}[cos(u-v)-cos(u+v), \ cos(u)cos(v)]=\frac{1}{2}[cos(u-v)+cos(u+v)]
    I've tried using the Product-to-sum formula, but it only makes things more confusing (to me). My teacher said that we should be able to solve this only using the addition/subtraction formulas.

    and this is what I have gotten so far

    [Cosx (cos^2x - sin^2x) cosx - (2sinxcosx) sinx] - [sinx (2sinxcosx) cosx + (cos^2x - sin^2x) sinx]

    I don't know if that is right, I'm still playing around with what what I have.
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  5. #5
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    \cos(x)\cos(3x)-\sin(x)\sin(3x)=cos(4x)=0

    x=\dfrac{\pi}{8}
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  6. #6
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    Quote Originally Posted by Plato View Post
    \cos(x)\cos(3x)-\sin(x)\sin(3x)=cos(4x)=0

    x=\dfrac{\pi}{8}
    I should have seen that.

    Thank you so much, you have saved me so much work and time
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