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Math Help - Which trigonometric identity do I use?

  1. #1
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    Which trigonometric identity do I use?

    cosx - cosy = -2sin(x+y / 2)sin(x-y / 2)

    I can't seem to find any that would work for it.
    Thanks in advance!
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  2. #2
    Super Member
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    I realised that what you wanted was to prove it.

    \displaystyle -2\sin\left(\frac{x+y}{ 2}\right)\sin\left(\frac{x-y}{ 2}\right) = -2\sin\left(\frac{x}{2}+\frac{y}{2}\right)\sin\left  (\frac{x}{2}-\frac{y}{2}\right)

    Using the identity \sin\left(\alpha \pm \beta\right) = \sin{\alpha}\cos{\beta}\pm \cos{\alpha}\sin{\beta} , we have:

    \displaystyle -2\sin\left(\frac{x}{2}+\frac{y}{2}\right)\sin\left  (\frac{x}{2}-\frac{y}{2}\right)= -2\left(\sin{\frac{x}{2}}\cos{\frac{y}{2}}+\cos{\fr  ac{x}{2}}\sin{\frac{y}{2}}\right)\left(\sin{\frac{  x}{2}}\cos{\frac{y}{2}}-\cos{\frac{x}{2}}\sin{\frac{y}{2}}\right)

    \displaystyle =  -2\sin^2{\frac{x}{2}}\cos^2{\frac{y}{2}}\right)+2\c  os^2{\frac{x}{2}}\sin^2{\frac{y}{2}} = -2\left(1-\cos^2{\frac{x}{2}}\right)\cos^2{\frac{y}{2}}\righ  t)+2\cos^2{\frac{x}{2}}\left(1-\cos^2{\frac{y}{2}}\right)

    \displaystyle =  -2\cos^2{\frac{y}{2}}+2\cos^2{\frac{x}{2}}\cos^2{\f  rac{y}{2}}+2\cos^2{\frac{x}{2}}-2\cos^2{\frac{x}{2}}\cos^2{\frac{y}{2}} = -2\cos^2{\frac{y}{2}}+2\cos^2{\frac{x}{2}}

     \displaystyle = -2\left(\frac{\cos{y}+1}{2}\right)+2\left(\frac{\co  s{x}+1}{2}\right) = -\left(\cos{y}+1}\right)+\left(\cos{x}+1\right) = \cos{x}-\cos{y}. \ \ \ \square
    Last edited by TheCoffeeMachine; November 20th 2010 at 11:30 AM.
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