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Given tan(theta)=-2 and (pi/2) < theta < pie; find cos 2(theta)
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This problem is nonsensical. If π < θ < 3π/2, then cosθ is negative, not positive. I am assuming that cosθ = -4/5. Since π < θ < 3π/2, then θ is in Quadrant III, which implies that sinθ < 0 and tanθ > 0 (because sinθ < 0 in Quadrants III and IV and tanθ > 0 in Quadrants I and III). By drawing a 3-4-5 triangle in Quadrant III such that cosθ = -adj/hyp = -4/5, we see that: adj = 4, opp = 3, and hyp = 5 ==> sinθ = -opp/hyp = -3/5 and tanθ = opp/adj = 3/4. (You could also find tanθ using sinθ/cosθ = (-3/5)/(-4/5) = 3/4.) I hope this helps!
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