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March 26, 2017

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If A is between 270degrees and 360degrees, and cos A=12/13, find the exact value of:

sin2a, cos2a, sin1/2a, cos1/2a

  • trig/math - ,

    since acos(12/13) = .39479, a = 2*pi - .39479 = 5.88839

    2a = 11.77678 is in the 4th quadrant, so

    sin 2a = 2sin(a)cos(a) = 2(-5/13)(12/13) = -120/169
    cos(2a) = cos^2(a) - sin^2(a) = 144/169 - 25/169 = 119/169

    a/2 is in the 2nd quadrant

    sin(a/2) = √(1/26)
    cos(a/2) = -√(25/26)

  • trig/math - ,

    if cosA = 12/13 in quadr IV, then
    sinA = -5/13

    sin 2A = 2sinAcosA = 2(12/13)(-5/13) = -120/169 =

    cos 2A = cos^2 A - sin^2 A
    = 144/169 - 25/169 = 119/169

    we know cos2A = 1 - 2sin^2 A
    or
    cos A = 1 - 2sin^2 (A/2)
    12/13 = 1 - 2sin^2 (A/2)
    2sin^2 (A/2) = 1 - 12/13 = 1/13
    sin^2 (A/2) = 1/26
    sin (A/2) = ±1/√26
    but if A is in IV, then A/2 is in II
    so sin(A/2) = 1/√26

    using
    cos A = 2cos^2 (A/2) -1
    try the last one.

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