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Math Help - Simplifying Trig Equation

  1. #1
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    Question Simplifying Trig Equation

    Hi can anyone show me step by step simplification of this equation

    ((A+C)/2)^2 - (((A-C)/2)cos2x + Bsin2x)^2 -(Bcos2x-((A-C)/2)sin2x)^2


    I know its suppose to come out to equal AC-B^2

    Thank you so much in advance... I can't seem to get the write answer.
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  2. #2
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    Hello, mikec5!

    Simplify: . \left(\frac{A+C}{2}\right)^2 - \bigg[\left(\frac{A-C}{2}\right)\cos2x + B\sin2x\bigg]^2 -\bigg[B\cos2x-\left(\frac{A-C}{2}\right)\sin2x\bigg]^2

    Answer: . AC-B^2
    We have:


    \left(\frac{A+C}{2}\right)^2 -\bigg[\left(\frac{A-C}{2}\right)^2\cos^2\!2x + B(A-C)\sin2x\cos2x + B^2\sin^2\!2x\bigg]

    . . . . -\bigg[B^2\cos^2\!2x - B(A-C)\sin2x\cos2x + \left(\frac{A-C}{2}\right)^2\sin^2\!2x\bigg]



    =\;\left(\frac{A+C}{2}\right)^2 - \left(\frac{A-C}{2}\right)^2\cos^2\!2x - {\color{red}\rlap{||||||||||}}B(A-C)\sin2x\cos2x - B^2\sin^2\!2x

    . . . . -B^2\cos^2\!2x + {\color{red}\rlap{||||||||||}}B(A-C)\sin2x\cos2x - \left(\frac{A-C}{2}\right)^2\sin^2\!2x<br />



    Re-arrange the terms:

    . . \left(\frac{A+C}{2}\right)^2 -\;\left(\frac{A-C}{2}\right)^2\!\cos^2\!2x \;- \;\left(\frac{A-C}{2}\right)^2\!\sin^2\!2x \;-\; B^2\sin^2\!2x \;-\; B^2\cos^2\!2x


    Factor:

    . . \left(\frac{A+C}{2}\right)^2 - \left(\frac{A-C}{2}\right)^2\!\underbrace{\left(\cos^2\!2x + \sin^2\!2x\right)}_{\text{This is 1}} \;-\; B^2\underbrace{\left(\sin^2\!2x + \cos^2\!2x\right)}_{\text{This is 1}}

    . . = \;\left(\frac{A+C}{2}\right)^2 - \left(\frac{A-C}{2}\right)^2 - B^2

    . . =\;\frac{A^2 + 2AC + C^2}{4} - \frac{A^2 - 2AC + C^2}{4} - B^2

    . . = \;\frac{4AC}{4} - B^2

    . . =\;AC - B^2

    Last edited by Soroban; October 26th 2008 at 08:33 PM.
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