Answer to Question #155022 in Trigonometry for Bonevie tutor

Question

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

Answer 1

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

1. $\cos\theta = \frac{3}{\sqrt{3^2+4^2}}=\frac{3}{5}=0.6$ , $\sin\theta=\frac{4}{5}=0.8$
2. $\cos\theta = \frac{-2}{\sqrt{2^2+2^2}} = \frac{-2}{2\sqrt{2}} = -\frac{1}{\sqrt{2}}$, $\sin\theta = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$

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

1. $\cos^2\theta +\frac{1}{4}=1, \cos^2\theta=\frac{3}{4}$, in Q1 $\cos \theta\geq 0$, so $\cos\theta = \frac{\sqrt{3}}{2}$
2. $\frac{25}{36}+\sin^2\theta=1, \sin^2\theta = \frac{11}{36}$, Q11 (which is the same as Q3, as every 4 quadrants we do a full turn) $\sin\theta\leq 0$, so $\sin\theta = -\frac{\sqrt{11}}{6}$