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Math Help - Prove(6)

  1. #1
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    Prove(6)

    Prove cos^2A-cos^2B=sin(B-A)sin(B+A)
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  2. #2
    Senior Member Dinkydoe's Avatar
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    hint:

    cos^2(A)-cos^2(B) = [cos(A)+cos(B)]\cdot[cos(A)-cos(B)]

    Now use some identities.
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  3. #3
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    Quote Originally Posted by Punch View Post
    Prove cos^2A-cos^2B=sin(B-A)sin(B+A)
    Use the identity

    \sin{(\alpha \pm \beta)} = \sin{\alpha}\cos{\beta} \pm \cos{\alpha}\sin{\beta}.


    So \sin{(B - A)} = \sin{B}\cos{A} - \cos{B}\sin{A}

    and \sin{(B + A)} = \sin{B}\cos{A} + \cos{B}\sin{A}.


    Therefore \sin{(B - A)}\sin{(B + A)} = (\sin{B}\cos{A} - \cos{B}\sin{A})(\sin{B}\cos{A} + \cos{B}\sin{A})

     = \sin^2{B}\cos^2{A} - \cos^2{B}\sin^2{A}

     = (1 - \cos^2{B})\cos^2{A} - \cos^2{B}(1 - \cos^2{A})

     = \cos^2{A} - \cos^2{A}\cos^2{B} - \cos^2{B} + \cos^2{A}\cos^2{B}

     = \cos^2{A} - \cos^2{B}.
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