Question

1. if tanθ = -2 in quadrant iv, find secθ. 6. use the unit circle and the period of the function to evaluate the expression. simplify your answer if possible. cos(29π/6)

Explanation:

Step1: Recall trigonometric identity

We know that $\tan\theta=\frac{\sin\theta}{\cos\theta}$ and $\sec\theta = \frac{1}{\cos\theta}$, also $\sin^{2}\theta+\cos^{2}\theta = 1$. Since $\tan\theta=- 2=\frac{\sin\theta}{\cos\theta}$, then $\sin\theta=-2\cos\theta$.

Step2: Substitute into Pythagorean identity

Substitute $\sin\theta=-2\cos\theta$ into $\sin^{2}\theta+\cos^{2}\theta = 1$. We get $(-2\cos\theta)^{2}+\cos^{2}\theta=1$, which simplifies to $4\cos^{2}\theta+\cos^{2}\theta = 1$, so $5\cos^{2}\theta=1$, and $\cos^{2}\theta=\frac{1}{5}$, then $\cos\theta=\pm\frac{1}{\sqrt{5}}$.

Step3: Determine sign of cosine

In Quadrant IV, $\cos\theta>0$. So $\cos\theta=\frac{1}{\sqrt{5}}$. Then $\sec\theta=\sqrt{5}$.

Step4: Evaluate $\cos(\frac{29\pi}{6})$

First, rewrite $\frac{29\pi}{6}$ as $\frac{24\pi + 5\pi}{6}=4\pi+\frac{5\pi}{6}$.
Since the period of the cosine - function is $2\pi$, $\cos(4\pi+\frac{5\pi}{6})=\cos(\frac{5\pi}{6})$.
On the unit - circle, $\cos(\frac{5\pi}{6})=-\frac{\sqrt{3}}{2}$.

Answer:

1. $\sec\theta=\sqrt{5}$
2. $\cos(\frac{29\pi}{6})=-\frac{\sqrt{3}}{2}$