math

if A+B+C=180degrees, prove that 2sin(B+C)sinA=1-cos2A

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  1. OK, B+C = 180-A
    so
    LS = 2sin(B+C)sinA
    = 2sin(180 - A)sinA
    = 2(-sinA)sinA
    = -2 sin^2 A

    RS = 1 - cos 2A
    = 1 -(2sin^2 A - 1)
    = -2sin^2 A
    = LS

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  2. in part 1, how does sin(180-A) equal -sinA from line 4-5

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  3. Actually I have 2 errors in my solution

    1.
    sin(180-A) = sinA , not -sinA

    e.g. sin (150) = sin(180-150) = sin 30

    2. cos 2A = 1 - 2sin^2 A, I had it backwards

    so here is the correct solution

    LS = 2sin(B+C)sinA
    = 2sin(180 - A)sinA
    = 2(sinA)sinA
    = 2 sin^2 A

    RS = 1 - cos 2A
    = 1 -(1 - 2sin^2 A)
    = 2sin^2 A
    = LS

    sorry about that

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