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Math Help - Difficult Trig Proof

  1. #1
    C^2
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    Difficult Trig Proof

    Hey guys, I was given a hard extra credit assignment and was wondering if anyone could do this proof:
    cos2x-cos10x=(tan4x)(sin2x+sin10x)

    Yeah, this seems like a really tricky problem.
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  2. #2
    MHF Contributor
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    using a substitution, like u = 2x, may cut down the work some.
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  3. #3
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    Dear C^2,

    my first tipp would be that you express the equation only with the term sin(2x) and cos(2x).
    e. g.
    cos(10x) = cos(8x + 2x) = cos(8x)sin(2x) - sin(8x)cos(2x)
    and again...and again

    and finally you can substitution cos(2x) = y and sin(2x) = sqrt(1-y^2)

    So you get an algebraic equation. The problem is now the too high degree of the polinom but maybe you get a special polinom which is solvable, i dunno.
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  4. #4
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    Hello, Cē!

    Normally, we are not allowed to work on both sides of the identity.
    But I cannot find a way to transform one side into the other.



    Prove: . \cos2x-\cos10x\:=\:\tan4x(\sin2x+\sin10x)

    We can use two Sum-to-Product Identities:

    . . \begin{array}{ccc}\cos A - \cos B &=& \text{-}2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \\ \\[-3mm]<br />
\sin A + \sin B &=& 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \end{array}


    The left side is: . -2\sin(6x)\sin(-4x) \;=\;2\sin(6x)\sin(4x)


    The right side is: . \frac{\sin(4x)}{\cos(4x)}\cdot2\sin(6x)\cos(4x) \;=\;2\sin(6x)\sin(4x)


    The two sides are equal, but it is not a satisfactory proof of an identity.

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