use given info and double angle identity to find sinx=1/3 if 0<x<pi/2, find sin2x?

Answer 1

If sin(x) = 1/3 and 0 < x < pi/2, it follows that x is in the first quadrant.

By SOHCAHTOA,

sin(x) = 1/3 = opp/hyp
opp = 1
hyp = 3, so by Pythagoras,
adj = sqrt(3^2 - 1^2) = sqrt(9 - 1) = sqrt(8) = 2sqrt(2)

cos(x) = adj/hyp = 2sqrt(2)/3, which is positive because it lies in the first quadrant.

By an identity,
sin(2x) = 2sin(x)cos(x), so if sin(x) = 1/3 and cos(x) = 2sqrt(2)/3, then

sin(2x) = 2[1/3] [2sqrt(2)/3]
= (2/3) [2sqrt(2)/3]
= 4sqrt(2)/9

Answer 2

sin x =1/3
cos x =sqrt(1-(sin x)^2) =sqrt(1-1/9) =sqrt(8/9)
cos x =2/3*(2)^0.5

sin 2x =2 sinx cosx
sin 2x =2*1/3*2/3*(2)^0.5
sin 2x =4/9*(2)^0.5