Given that sinθ=0.3431 and θ is in quadrant​ I, find sin2θ.

Answer 1

sin2θ=2sinθcosθ

cosθ = √(1-sin²θ) = √(1-0.3431²) = 0.93929 (Positive since QI)

sin2θ = 2(0.3431)(0.93929) = 0.6445

Answer 2

sinx = 0.3432, x = about 20.06 degrees

2x = about 40.12 degrees

sin2x = 0.644

or another way to do this problem is use the trig identity for double sine

sin2x = 2sinxcosx = 2(.3431(.939) = 0.644

if sinx = .3431 use a calculator with an inverse trig function to find x, and then find cosx

cos20.06 = about .939