If tanθ=−4/3 and θ is in the second quadrant, find the exact value of sin2θ.

Answer 1

sin2θ = 2sinθcosθ; cosθ = - 1/(1 + 16/9)^1/2 = - 3/5; sinθ = tanθcosθ = - 4/3(-3/5) = 4/5

sin2θ = 2(4/5)(-3/5) = - 24/25

Answer 2

sin2θ = 2sinθcosθ; cosθ = - 1/(1 + 16/9)^1/2 = - 3/5; sinθ = tanθcosθ = - 4/3(-3/5) = 4/5

sin2θ = 2(4/5)(-3/5) = - 24/25

Answer 3

tanx = -4/3

sinx =4/5

cosx =-3/5

sin2x = 2sinxcosx = 2(4/5)(-3/5) = -24/25 in quadrant III

2x = about 253.7

x = 126.9 degrees

sin2x=sin253.7 + =-0.9598, -24/25 = -0.96

exact value of sin2x = -24/25 =-0.96

sinx = 4/5, x = about 127 degrees

2x= about 2(127) = 254 degrees

sin2x= about sin254 = -24/25