Question

math 110 exam 1 dr. eggenberger name: john taylor (show all your work to receive credit) the terminal side of angle $\theta$ in standard - position passes through (-2,5). find and simplify the values: 1. $cos\theta$ 2. $\tan\theta$ use reference angles to find the exact values and simplify: 3. $sin(\frac{7pi}{3})$ 4. $\tan(\frac{11pi}{6})$

Explanation:

Step1: Recall trigonometric definitions

For a point \((x,y)\) on the terminal - side of an angle \(\theta\) in standard position, \(r=\sqrt{x^{2}+y^{2}}\), \(\cos\theta=\frac{x}{r}\), and \(\tan\theta = \frac{y}{x}\). Given \(x=-2\) and \(y = 5\), then \(r=\sqrt{(-2)^{2}+5^{2}}=\sqrt{4 + 25}=\sqrt{29}\).

Step2: Calculate \(\cos\theta\)

\(\cos\theta=\frac{x}{r}=\frac{-2}{\sqrt{29}}=-\frac{2\sqrt{29}}{29}\)

Step3: Calculate \(\tan\theta\)

\(\tan\theta=\frac{y}{x}=\frac{5}{-2}=-\frac{5}{2}\)

Step4: Use periodicity of sine function

The sine function has a period of \(2\pi\), i.e., \(\sin(\alpha + 2k\pi)=\sin\alpha\), \(k\in\mathbb{Z}\). For \(\alpha=\frac{\pi}{3}\) and \(k = 1\), \(\frac{7\pi}{3}=2\pi+\frac{\pi}{3}\), so \(\sin(\frac{7\pi}{3})=\sin(2\pi+\frac{\pi}{3})=\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}\)

Step5: Use periodicity of tangent function

The tangent function has a period of \(\pi\), i.e., \(\tan(\alpha + k\pi)=\tan\alpha\), \(k\in\mathbb{Z}\). \(\frac{11\pi}{6}=2\pi-\frac{\pi}{6}\), and \(\tan(\frac{11\pi}{6})=\tan(2\pi - \frac{\pi}{6})=-\tan\frac{\pi}{6}=-\frac{\sqrt{3}}{3}\)

Answer:

1. \(-\frac{2\sqrt{29}}{29}\)
2. \(-\frac{5}{2}\)
3. \(\frac{\sqrt{3}}{2}\)
4. \(-\frac{\sqrt{3}}{3}\)