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

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

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

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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