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Math Help - please help me to prove this...

  1. #1
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    please help me to prove this...

    Prove that (cos 2A - cos 2B)^2 + (sin 2A - sin 2B)^2=4sin^2(A-B)....please help me to solve this trigo question...thank u
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: please help me to prove this...

    Hint: within the parentheses, use sum to product identities, and you will have it in a few steps.
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  3. #3
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    Re: please help me to prove this...

    can anyone please show me the first step...please
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: please help me to prove this...

    (\cos 2A - \cos 2B)^2 + (\sin 2A - \sin 2B)^2 = 4 \sin^2 (A-B)

    Use
    \cos(2A) - \cos(2B) = -2 \sin \left(\frac{2A+2B}{2}\right)\sin\left(\frac{2A-2B}{2}\right)
    = - 2 \sin(A+B)\sin(A-B)
    \sin(2A) - \sin(2B) = 2 \cos \left(\frac{2A+2B}{2}\right)\sin\left(\frac{2A-2B}{2}\right)
     = 2 \cos(A+B)\sin(A-B)

    Can you proceed now?
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  5. #5
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    Re: please help me to prove this...

    im still having problem with this question....
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: please help me to prove this...

    Quote Originally Posted by sharmala View Post
    im still having problem with this question....
    Show your problem.
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  7. #7
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    Re: please help me to prove this...

    Hello, sharmala!

    We will need these two identities:

    . . \cos x\cos y + \sin x\sin y \:=\:\cos(x-y)

    . . \frac{1-\cos2\theta}{2} \:=\:\sin^2\theta



    \text{Prove: }\:(\cos2A - \cos2B)^2 + (\sin2A - \sin2B)^2\;=\;4\sin^2(A\!-\!B)

    Expand the left side:

    \cos^22A - 2\cos2A\cos2B + \cos^22B + \sin^22A - 2\sin2A\sin2B + \sin^22B

    . . . =\;\underbrace{\sin^22A + \cos^22A}_1 \:+\: \underbrace{\sin^22B + \cos^22B}_1 \:-\: 2\big(\cos2A\cos2B + \sin2A\sin2B\big)

    . . . =\;2 - 2\cos(2A\!-\!2B)

    . . . =\;2\big[1 - \cos(2[A\!-\!B])\big]

    . . . =\;4\left[\frac{1-\cos(2[A\!-\!B])}{2}\right]

    . . . =\;4\sin^2(A\!-\!B)

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  8. #8
    MHF Contributor Siron's Avatar
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    Re: please help me to prove this...

    In case you wanted to continue with the formulas of Simpson.
    [- 2 \sin(A+B)\sin(A-B)]^2+[2 \cos(A+B)\sin(A-B)]^2
    = 4\sin^2(A+B)\sin^2(A-B)+4\cos^2(A+B)\sin^2(A-B)
    = 4 \sin^2(A-B)[\sin^2(A+B)+\cos^2(A+B)]=4\sin^2(A-B)
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