For the first part we will use that cos θ = x x 2 + y 2 , sin θ = y x 2 + y 2 \cos\theta = \frac{x}{\sqrt{x^2+y^2}}, \sin\theta=\frac{y}{\sqrt{x^2+y^2}} cos θ = x 2 + y 2 x , sin θ = x 2 + y 2 y :
cos θ = 3 3 2 + 4 2 = 3 5 = 0.6 \cos\theta = \frac{3}{\sqrt{3^2+4^2}}=\frac{3}{5}=0.6 cos θ = 3 2 + 4 2 3 = 5 3 = 0.6 , sin θ = 4 5 = 0.8 \sin\theta=\frac{4}{5}=0.8 sin θ = 5 4 = 0.8 cos θ = − 2 2 2 + 2 2 = − 2 2 2 = − 1 2 \cos\theta = \frac{-2}{\sqrt{2^2+2^2}} = \frac{-2}{2\sqrt{2}} = -\frac{1}{\sqrt{2}} cos θ = 2 2 + 2 2 − 2 = 2 2 − 2 = − 2 1 , sin θ = 2 2 2 = 1 2 \sin\theta = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} sin θ = 2 2 2 = 2 1 For the second part we will use that cos 2 θ + sin 2 θ = 1 \cos^2\theta+\sin^2\theta=1 cos 2 θ + sin 2 θ = 1 and quadrant to determine the sign :
cos 2 θ + 1 4 = 1 , cos 2 θ = 3 4 \cos^2\theta +\frac{1}{4}=1, \cos^2\theta=\frac{3}{4} cos 2 θ + 4 1 = 1 , cos 2 θ = 4 3 , in Q1 cos θ ≥ 0 \cos \theta\geq 0 cos θ ≥ 0 , so cos θ = 3 2 \cos\theta = \frac{\sqrt{3}}{2} cos θ = 2 3 25 36 + sin 2 θ = 1 , sin 2 θ = 11 36 \frac{25}{36}+\sin^2\theta=1, \sin^2\theta = \frac{11}{36} 36 25 + sin 2 θ = 1 , sin 2 θ = 36 11 , Q11 (which is the same as Q3, as every 4 quadrants we do a full turn) sin θ ≤ 0 \sin\theta\leq 0 sin θ ≤ 0 , so sin θ = − 11 6 \sin\theta = -\frac{\sqrt{11}}{6} sin θ = − 6 11
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