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Math Help - Trigonomic Identity Problem

  1. #1
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    Nov 2008
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    Exclamation Trigonomic Identity Problem

    Hello

    I am new at this forum, but I was wondering if I could get some help.

    The identity problem I am trying to solve is:

    sinx-sin3x
    ---------- = -cot2x
    cosx-cos3x

    (note, the sin3x, cos3x, and cot2x do not mean to the square root of)

    Thanks for any help with this
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  2. #2
    Super Member

    Joined
    May 2006
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    Lexington, MA (USA)
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    Hello, themartinacster!

    Welcome aboard!

    There are two ways (at least) to prove the identity,
    . . neither is pleasant.


    Prove: . \frac{\sin x - \sin3x}{\cos x - \cos 3x} \:=\:-\cot2x

    [1] Use the Sum-to-Product identities:

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


    We have: . \begin{array}{ccccc}\sin x - \sin3x &=& 2\cos2x\sin(-x) &=&\text{-}2\cos2x\sin x \\<br /> <br />
\cos3x - \cos x &=& \text{-}2\sin2x\sin(-x) &=& 2\sin2x\sin x \end{array}

    The problem becomes: . \frac{\sin x-\sin3x}{\cos x - \cos3x} \;=\;\frac{\text{-}2\cos2x\sin x}{2\sin2x\sin x} \;=\;-\frac{\cos2x}{\sin2x} \;=\;-\cot2x


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    [2] Use the Compound Angle identities:

    . . . \begin{array}{ccc}\sin(A + B) &=& \sin A\cos B + \sin B\cos A \\ \cos(A + B) &=& \cos A\cos B - \sin A\sin B \end{array}


    Use them repeatedly to derive these statements:

    . . \begin{array}{ccc}\sin3x &=& 3\sin x - 4\sin^3\!x \\ \cos3x &=& 4\cos^3\!x - 3\cos x \end{array}


    We have: . \frac{\sin x - \sin3x}{\cos x - \cos3x} \;=\;\frac{\sin x - (3\sin x - 4\sin^3\!x)}{\cos x - (4\cos^3\!x - 3\cos x)} \;=\;\frac{4\sin^3\!x - 2\sin x}{4\cos x-4\cos^3\!x}

    . . . = \;\frac{2\sin x(2\sin^2\!x - 1)}{4\cos x\underbrace{(1 - \cos^2x)}_{\text{This is }\sin^2\!x}} \;=\;-\frac{\sin x\overbrace{(1 - 2\sin^2\!x)}^{\text{This is }\cos2x}}{2\cos x\sin^2\!x}

    . . . = \;-\frac{\cos2x}{\underbrace{2\sin x\cos x}_{\text{This is }\sin2x}} \;=\;-\frac{\cos2x}{\sin2x} \;=\;-\cot2x


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