Find the exact values of sin 2θ, cos 2θ, and tan 2θ given the following: cos θ =-5/8 where θ is in quadrant III.

Answer 1

sinx^2 + cosx^2 = 1

sinx^2 + (-5/8)^2 = 1

sinx^2 +25/64 = 1

sinx^2 = 39/64

sinx = +/- 0.78

since in QIII, sinx = -0.78

cosx = -5/8 = -0.625

tanx = sinx/cosx

Use above values to compute:

sin2x = 2sinxcosx

cos2x = 1 - 2sinx^2

tan2x = 2tanx/(1-tanx^2)