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given sinθ=-5/13 and π<θ<3π/2 find

sin2θ
cos( θ-4π/3)
sin(θ/2)

can some1 explain to me how to do these?

  • math -

    sinA = -5/13 = Y/r,

    X^2 + Y^2 = r^2,
    X^2 + (-5)^2 = (13)^2,
    X^2 + 25 = 169,
    X^2 = 169 - 25 = 144,
    X = +- 12.
    X = -12 Because it places our resultant
    vector in the required quadrant(Q3).
    (X,Y) = (-12,-5).
    tanA = Y/X = --5/-12 = 0.416666,
    A = 22.62 Deg. ,
    A = 22.62 + 180 = 202.62(Q3).

    180 < 202.62 < 270. Q3.

    sin(2A) = sin(405.24) = 0.710.

    cos(A-4pi/3) = cos(202.62-240) = 0.7946

    sin(A/2) = +- sqrt((1-cosA)/2) =
    sqrt((1-cos202.62)/2) = 0.9806

  • math -

    Correction:

    sinA = -5/13 = Y/r,
    A = 157.38 Deg.

    sin(2A) = sin(314.76) = -0.710.

    cos(A-4pi/3)=cos(157.38-240)=0.1284

    sin(A/2) = +- sqrt((1-cosA)/2) =
    sqrt((1-cos157.38)/2 = 0.9806.

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