Given that sin(θ)=1013, and θ is in Quadrant II, what is cos(2θ)? Give an exact answer in the form of a fraction.

Answer 1

sinθ = 10/13 and θ lies in quadrant 2, so cosθ < 0 and cosθ = -√(1 - sin²θ) = -√69 / 13

cos(2θ) = cos²θ - sin²θ = 69/169 - 100/169 = -31/169

Answer 2

1013=293 (subtracted 720) asin=-6.3733.............

This put's θ in the 4th quad. To get to quad2 add 180;=172.39..............

Is a decimal supposed to be somewhere in1013????????????????????