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Given that tan¤=3/4 and ¤ is in the second quadrant find sin2¤ (¤=theta)

  • TRIG -

    see other post.

  • TRIG -

    cos theta = + OR - 1 / sqrt ( 1 + tan ^ 2 theta )

    In Quadrant II, cosine are negative so :

    cos theta = - 1 / sqrt ( 1 + tan ^ 2 theta )

    cos theta = - 1 / sqrt [ 1 + ( 3 / 4 ) ^ 2 theta ]

    cos theta = - 1 / sqrt ( 1 + 9 / 16 )

    cos theta = - 1 / sqrt ( 16 / 16 + 9 / 16 )

    cos theta = - 1 / sqrt ( 25 / 16 )

    cos theta = - 1 / ( 5 / 4 )

    cos theta = - 4 / 5


    sin ^ 2 theta + cos ^ 2 theta = 1

    sin ^ 2 theta = 1 - cos ^ 2 theta

    sin theta = + OR - sqrt ( 1 - cos ^ 2 theta )

    In Quadrant II, sine are positive so :

    sin theta = sqrt ( 1 - cos ^ 2 theta )

    sin theta = sqrt [ 1 - ( - 4 / 5 ) ^ 2 ]

    sin theta = sqrt ( 1 - 16 / 25 )

    sin theta = sqrt ( 25 / 25 - 16 / 25 )

    sin theta = sqrt ( 9 / 25 )

    sin theta = 3 / 5


    sin 2 theta = 2 sin theta cos theta

    sin 2 theta = 2 * 3 / 5 * - 4 / 5

    sin 2 theta = - 24 / 25

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