Math

If sinx=3/4, with x in quadrant II, then determine the value of sin 2x

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asked by Lana
  1. Help I use

    S|A
    ---
    T|C

    First quad: All trigs. positive
    Sec. quad: Only sin and csc
    Third quad: Only tan and cot
    Fourth quad: Only cos and sec

    So, in the second quadrant makes sin positive.

    Therefor, I think it should be
    sin2x=2*(3/4)

    I may be wrong though.

    I know if I was looking for the sin(theta)=constant
    I would use sin inverse time the constant to find theta the angle.
    EX:
    sin(theta)=3
    theta=sin^(-1)*3

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    posted by John
  2. John: sin(theta)=3 cannot be right... the range of sin(theta) is [-1,1].

    Lana:
    First find the reference angle of x, which is the acute angle formed with the x-axis. In this case, since sin(x)=3/4,
    we have
    reference angle
    =arcsin(3/4)
    =48.6 degrees (approx.)

    Since we know that x is in the second quadrant,
    x=180-reference angle
    =180-48.6

    2x is therefore twice this amount, or
    2(180-48.6)
    =360-2*48.6)
    =-2*48.6
    using the fact that 360-2*48.6 and -2*48.6 are coterminal angles.

    sin(2x)=sin(-2*48.6)
    =sin(-2*48.6)
    =-0.992

    Replace 48.6 degrees in the above expressions with the accurate values of arcsin(3/4).

    Post if you need more information.

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    posted by MathMate

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