Question #155022

Find the exact value of the sum and cosine functions if angle θ is in standard position with the given coordinates of the terminal side.

1.) (3,4)


2.) (-2,2)


Angle θ is in standard position and lies in the given quadrant. When sin and cos θ is given, find the exact value of sin θ and cos θ.

1.) sin θ =½ ; Q1


2.) cos θ= -⅚; Q11





1
Expert's answer
2021-01-12T15:01:50-0500

For the first part we will use that cosθ=xx2+y2,sinθ=yx2+y2\cos\theta = \frac{x}{\sqrt{x^2+y^2}}, \sin\theta=\frac{y}{\sqrt{x^2+y^2}} :

  1. cosθ=332+42=35=0.6\cos\theta = \frac{3}{\sqrt{3^2+4^2}}=\frac{3}{5}=0.6 , sinθ=45=0.8\sin\theta=\frac{4}{5}=0.8
  2. cosθ=222+22=222=12\cos\theta = \frac{-2}{\sqrt{2^2+2^2}} = \frac{-2}{2\sqrt{2}} = -\frac{1}{\sqrt{2}}, sinθ=222=12\sin\theta = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}

For the second part we will use that cos2θ+sin2θ=1\cos^2\theta+\sin^2\theta=1 and quadrant to determine the sign :

  1. cos2θ+14=1,cos2θ=34\cos^2\theta +\frac{1}{4}=1, \cos^2\theta=\frac{3}{4}, in Q1 cosθ0\cos \theta\geq 0, so cosθ=32\cos\theta = \frac{\sqrt{3}}{2}
  2. 2536+sin2θ=1,sin2θ=1136\frac{25}{36}+\sin^2\theta=1, \sin^2\theta = \frac{11}{36}, Q11 (which is the same as Q3, as every 4 quadrants we do a full turn) sinθ0\sin\theta\leq 0, so sinθ=116\sin\theta = -\frac{\sqrt{11}}{6}

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