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Math Help - Identity question

  1. #1
    Member wiseguy's Avatar
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    Identity question

    "Prove (algebraically) that: cos6x+cos4x=2cos5xcosx"

    I'm not sure if I should handle this like an identity or not... I'm well out of the identity chapter, so I don't know why I'd be asked this now.

    I'm thinking the composite argument property might be of use?
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  2. #2
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    Hello, wiseguy!

    Prove (algebraically) that: . \cos6x+\cos4x\:=\:2\cos5x\cos x

    I'm not sure what "algebraically" means . . .
    . . some Trigonometry must be used.


    We have the identity: . \cos(A \pm B) \;=\;\cos A\cos B \mp \sin A\sin B


    Then: . \begin{array}{cccccccc}<br />
\cos(6x) &=& \cos (5x + x) &=& \cos5x\cos x - \sin5x\sin x & [1] \\<br />
\cos(4x) &=& \cos(5x - x) &=& \cos5x\cos x + \sin5x\sin x & [2] \end{array}


    Add [1] and [2]: . \cos6x + \cos 4x \;=\;2\cos5x \cos x

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  3. #3
    Member wiseguy's Avatar
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    Wow my brain must of been asleep. Thank you!!!
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  4. #4
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    Or...

    2cosAcosB=cos(A+B)+cos(A-B)

    Alternatively..

    cosA+cosB=2cos\left(\frac{A+B}{2}\right)cos\left(\  frac{A-B}{2}\right)
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