Question

use the unit circle to find the value of cos $\frac{3pi}{4}$ and periodic properties of trigonometric functions to find the value of cos $\frac{19pi}{4}$.

Explanation:

Step1: Find $\cos\frac{3\pi}{4}$ using unit - circle

On the unit - circle, for an angle $\theta$, the $x$ - coordinate of the point on the unit - circle corresponding to the angle $\theta$ is $\cos\theta$. For $\theta=\frac{3\pi}{4}$, the corresponding point on the unit - circle is $(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$. So, $\cos\frac{3\pi}{4}=-\frac{\sqrt{2}}{2}$.

Step2: Rewrite $\frac{19\pi}{4}$ using periodicity

The cosine function has a period of $2\pi$, i.e., $\cos(x + 2k\pi)=\cos x$ for any real number $x$ and integer $k$. We can rewrite $\frac{19\pi}{4}$ as $\frac{19\pi}{4}=4\pi+\frac{3\pi}{4}$. Here, $x = \frac{3\pi}{4}$ and $k = 2$.

Step3: Find $\cos\frac{19\pi}{4}$

Since $\cos\frac{19\pi}{4}=\cos(4\pi+\frac{3\pi}{4})$ and using the periodicity property $\cos(x + 2k\pi)=\cos x$, we have $\cos(4\pi+\frac{3\pi}{4})=\cos\frac{3\pi}{4}$. So, $\cos\frac{19\pi}{4}=-\frac{\sqrt{2}}{2}$.

Answer:

$\cos\frac{3\pi}{4}=-\frac{\sqrt{2}}{2}$, $\cos\frac{19\pi}{4}=-\frac{\sqrt{2}}{2}$