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March 27, 2017

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given sin(4x)=5/13 in Q II

find Tan(2x) and cos(2x)

  • Trig - ,

    sin 4 x = 5/13
    then cos 4 x = 12/13 (5,12,13 right triangle)
    and
    tan 4x = sin 4 x /cos 4x = 5/12
    so identities:
    sin 4x = 2 sin 2x cos 2x = 5/13
    cos 4x = cos^2 2x - sin^2 2x = 12/13
    tan 4x = 2 tan 2x / (1-tan^2 2x) = 5/12

    so
    cos^2 2x = 12/13 + sin^2 2x
    = 12/13 + [ 5/(26 cos 2x)]^2
    = 12/13 + 25 /(114244 cos^2 2x)
    114244 cos^4 2x = 105456 cos^2 2x + 25
    cos^4 2x - .9231 cos^2 2x - .00021883=0
    cos^2 2x = (1/2)[.9231 +/- .96124]
    cos^2 2x =.4616 +/- .4806
    use + sign
    cos^2 2x = .94217
    cos 2x = .9707
    check my arithmetic then go on and get the sin 2x and tan 2x

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