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Math Help - compound-angle formulae with trig identity question

  1. #1
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    compound-angle formulae with trig identity question

    sin(A+B)sin(A-B) = sin^2A-sin^2B is what I have to prove

    So
    LHS= (sinAcosB+cosAsinB)(sinAcosB-cosAsinB)

    This is obviously a quadratic. I just don't know how it would multiply

    e.g sinAcosB x sinAcosB = ???
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  2. #2
    Junior Member bebrave's Avatar
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    ==>sin^2A.cos^2B-sin^2Bcos^2A
    ===> sin^2A(1-sin^2B)-sin^2B(1-sin^2A)
    ===> sin^2A-sin^2Asin^2 B-sin^2B+sin^2Asin^2B
    ===> sin^2A-sin^2B.
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  3. #3
    Junior Member bebrave's Avatar
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    sin(A+B)=sinAcosB+cosAsinB
    sin(x) = cos(90- x) and cos(x) = sin(90 - x)
    let x = A + B,
    we have,
    sin(A+B)=cos (90-(A+B))
    then i can write this
    cos90-A-B)as cos((90-A)-B)
    cos(A-B) =cos(A)cos(B) + sin(A)sin(B)
    sin(A+B)=cos(90-A)cosB +sin(90-A)sin(B)
    sin(A+B)=sinAcosB+cosAsinB

    i think these prooves can help you...
    Last edited by bebrave; May 8th 2009 at 10:01 PM.
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  4. #4
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    Thankyou
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  5. #5
    Junior Member bebrave's Avatar
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    you are welcome
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