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If Sin(x)=1/3 in te 2nd quadrant find sin(2x) and tan(2x)?

3 Answers

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  • Como
    Lv 7
    5 years ago

    sin x = 1/3

    cos x = √8 / 3 = 2√2 / 3

    tan x = 1 / [ 2√2 ] = √2 / 4

    sin 2x = 2 sin x cos x

    sin 2x = 2 [1/3] [ 2√2 / 3 ] = 4√2 / 9

    cos 2x = 1 - 2sin²x = 1 - 2/9 = 7/9

    tan 2x = 4√2 / 7

  • Karl
    Lv 6
    5 years ago

    Draw an unit circle.

    sin(x) = 1/3 , Q2 => x ≈ 160° --> 2x ≈ 320° , Q4

    cosx = +/- √(1 - sin²x)

    cosx = +/- √(1 - 1/9)

    cosx =(+)/- √(8/9) < 0 , Q2

    cos(x) = -√8/3

    sin(2x) = 2*sin(x) * cos(x)

    sin(2x) = 2* (1/3)*(-√8/3)

    sin(2x) = - 2√8 /9

    cos(2x) = 2cos²(x) - 1

    cos(2x) = 2*(-√8/3)² - 1

    cos(2x) = 2*(8/9) - 1

    cos(2x) = 7/9

    tan(2x) = sin(2x)/cos(2x)

    tan(2x) = - (2√8 /9) / (7/9)

    tan(2x) = - 2√8 /7

    tan(2x) = - (4/7)*√2

    ===============

  • sin(x) = (1/3)

    x is in Q2, so cos(x) < 0

    sin(2x) =>

    2 * sin(x) * cos(x) =>

    2 * (1/3) * sqrt(1 - sin(x)^2) =>

    2 * (1/3) * (-sqrt(1 - (1/3)^2)) =>

    (-2/3) * sqrt(1 - 1/9) =>

    (-2/3) * sqrt(8/9) =>

    (-2/3) * (2/3) * sqrt(2) =>

    -4 * sqrt(2) / 9

    tan(2x) =>

    sin(2x) / cos(2x) =>

    (-4 * sqrt(2) / 9) / (1 - 2 * sin(x)^2) =>

    (-4 * sqrt(2) / 9) / (1 - 2 * (1/9)) =>

    (-4 * sqrt(2) / 9) / (7/9) =>

    -4 * sqrt(2) / 7

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