1) Let sin A = − 5/13 with A in QIII and find sin 2A. 2)Let cos x = 5/6 with x in QIV and find cos 2x. 3) Let tan θ = 2 with θ in QI and find sin 2θ.

Answer 1

sinA=-5/13

cosA=-12/13, as this is a 5-12-13 triangle

sines and cosines are all negative in Quadrant III

sin2A = 2sinAcosA = 2 x -5/13 x -12/13 = 120/169

cosA = 5/6

co2A = 2cos²A - 1 = 2(5/6)²-1 = 50/36 - 36/36 = 14/36 = 7/18

tanA=2 = sinA/cosA

sinA = 2cosA = 2(sqr5/5)

sin2A = 2sinAcosA = 2(2)cosAcosA =2(2)(1/5) = 4/5

cosA = 1/sqr5 = sqr5/5

tanA = 2 = 2/1 the hypotenuse is square root of sum of squares of 1 and 2

cosA = 1 over that hypotenuse = 1/sqr5