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Math Help - trig identities

  1. #16
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    In post#11 Archie Meade reached the step

    sinA + sinB = 2sin(A+B)/2*cos(A-B)/2

    sinA - sinB = 2cos(A+B)/2*sin(A-B)/2.

    From that step

    (sinA + sinB)(sinA - sinB) = [2sin(A+B)/2*cos(A+B)/2][2sin(A-B)/2*cos(A-B)/2]

    sin^2A - sin^2B = sin(A+B)sin(A-B)

    When you try to solve the equation, it is better to take A>B. So I sugested to write the equation as

    sin^(2x) - sin^2(x-30) = 0
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  2. #17
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    (sinA + sinB)(sinA - sinB) = [2sin(A+B)/2*cos(A+B)/2][2sin(A-B)/2*cos(A-B)/2]
    I believe (sinA + sinB)(sinA - sinB) = [2sin(A+B)/2*cos(A-B)/2][2sin(A-B)/2*cos(A+B)/2]

    and further... how do we get from [2sin(A+B)/2*cos(A+B)/2][2sin(A-B)/2*cos(A-B)/2] to sin(A+B)sin(A-B)
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  3. #18
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    What I'm trying to say is that it's easy to go from sin(A+B)sin(A-B) to sin^2A - sin^2B
    but I can't figure out how to go from sin^2A - sin^2B to sin(A+B)sin(A-B). Can someone please show me the proof??
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  4. #19
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    Quote Originally Posted by jayshizwiz View Post
    I believe (sinA + sinB)(sinA - sinB) = [2sin(A+B)/2*cos(A-B)/2][2sin(A-B)/2*cos(A+B)/2]

    and further... how do we get from [2sin(A+B)/2*cos(A+B)/2][2sin(A-B)/2*cos(A-B)/2] to sin(A+B)sin(A-B)
    I am using sin(2θ) = 2sinθcos(θ). So [2sin(A+B)/2*cos(A+B)/2] = sin(A+B)
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