Question

use the unit circle to find the value of tan $\frac{7pi}{4}$ and periodic properties of trigonometric functions to find the value of tan $\frac{19pi}{4}$.

Explanation:

Step1: Recall tangent formula

$\tan\theta=\frac{\sin\theta}{\cos\theta}$.

Step2: Find $\tan\frac{7\pi}{4}$

For $\theta = \frac{7\pi}{4}$, on the unit - circle, $\sin\frac{7\pi}{4}=-\frac{\sqrt{2}}{2}$ and $\cos\frac{7\pi}{4}=\frac{\sqrt{2}}{2}$. Then $\tan\frac{7\pi}{4}=\frac{\sin\frac{7\pi}{4}}{\cos\frac{7\pi}{4}}=\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=- 1$.

Step3: Use periodic property of tangent

The period of the tangent function $y = \tan x$ is $\pi$, i.e., $\tan(x + n\pi)=\tan x$, where $n\in\mathbb{Z}$.
We can rewrite $\frac{19\pi}{4}$ as $\frac{19\pi}{4}=4\pi+\frac{3\pi}{4}$. Since $\tan(x + 4\pi)=\tan x$ (because $4\pi$ is a multiple of the period $\pi$ of the tangent function), $\tan\frac{19\pi}{4}=\tan(4\pi+\frac{3\pi}{4})=\tan\frac{3\pi}{4}$.
For $\theta=\frac{3\pi}{4}$, $\sin\frac{3\pi}{4}=\frac{\sqrt{2}}{2}$ and $\cos\frac{3\pi}{4}=-\frac{\sqrt{2}}{2}$. Then $\tan\frac{3\pi}{4}=\frac{\sin\frac{3\pi}{4}}{\cos\frac{3\pi}{4}}=\frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}}=-1$.

Answer:

$\tan\frac{7\pi}{4}=-1$, $\tan\frac{19\pi}{4}=-1$