Question

math 110 exam 1 dr. eggenberger (show all your work to receive credit) the terminal side of angle θ in standard position passes through (-2,5). find and simplify the values: 1. cos θ tan θ reference angles to find the exact values and simplify: sin(7π/3) sin(11π/6)

Explanation:

Step1: Find the radius $r$

Given the point $(-2,5)$ on the terminal - side of the angle $\theta$, use the formula $r=\sqrt{x^{2}+y^{2}}$. Here, $x = - 2$ and $y = 5$, so $r=\sqrt{(-2)^{2}+5^{2}}=\sqrt{4 + 25}=\sqrt{29}$.

Step2: Calculate $\cos\theta$

The formula for $\cos\theta$ is $\cos\theta=\frac{x}{r}$. Substituting $x=-2$ and $r = \sqrt{29}$, we get $\cos\theta=-\frac{2}{\sqrt{29}}=-\frac{2\sqrt{29}}{29}$.

Step3: Calculate $\tan\theta$

The formula for $\tan\theta$ is $\tan\theta=\frac{y}{x}$. Substituting $x=-2$ and $y = 5$, we get $\tan\theta=-\frac{5}{2}$.

Step4: Find the reference - angle for $\frac{7\pi}{3}$

First, find an equivalent angle between $0$ and $2\pi$. $\frac{7\pi}{3}=2\pi+\frac{\pi}{3}$. The reference - angle for $\frac{7\pi}{3}$ is $\frac{\pi}{3}$. And $\sin(\frac{7\pi}{3})=\sin(2\pi+\frac{\pi}{3})=\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$.

Step5: Find the reference - angle for $\frac{11\pi}{6}$

Since $\frac{11\pi}{6}=2\pi-\frac{\pi}{6}$, the reference - angle is $\frac{\pi}{6}$. And $\sin(\frac{11\pi}{6})=-\sin\frac{\pi}{6}=-\frac{1}{2}$.

Answer:

$\cos\theta=-\frac{2\sqrt{29}}{29}$
$\tan\theta=-\frac{5}{2}$
$\sin(\frac{7\pi}{3})=\frac{\sqrt{3}}{2}$
$\sin(\frac{11\pi}{6})=-\frac{1}{2}$