Question

The terminal side of $\mathrm{\theta }$ in standard position contains $\left(-6,8\right)$, how do you find the exact values of the six trigonometric functions of $\mathrm{\theta }$?

Answer 1

Step 1: Find the distance between the origin and the given point

Given that terminal side of $\mathrm{\theta }$ in standard position contains $\left(-6,8\right)$

If the point in the angle’s terminal side is $\mathrm{P}=\left(\mathrm{x},\mathrm{y}\right)$ then the trigonometric functions can be calculated as:

$\sin \left(\theta \right)=\frac{\mathrm{y}}{\mathrm{r}}$, $\cos \left(\theta \right)=\frac{\mathrm{x}}{\mathrm{r}}$, $\tan \left(\theta \right)=\frac{\mathrm{y}}{\mathrm{x}}$, $\cot \left(\theta \right)=\frac{\mathrm{x}}{\mathrm{y}}$, $\sec \left(\theta \right)=\frac{\mathrm{r}}{\mathrm{x}}$, $\mathrm{cosec}\left(\theta \right)=\frac{\mathrm{r}}{\mathrm{y}}$, where $\mathrm{r}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}$

Here, $\mathrm{x}=-6,\mathrm{y}=8$

$\Rightarrow$$\mathrm{r}=\sqrt{{\left(-6\right)}^2+{\left(8\right)}^2}$

$\Rightarrow$$\mathrm{r}=\sqrt{36+64}$

$\Rightarrow$$\mathrm{r}=\sqrt{100}$

$\Rightarrow$$\mathrm{r}=10$

Step 2: Find the values of the six trigonometric functions

Now substitute the values of $\mathrm{x},\mathrm{y}$ and $\mathrm{r}$ to find the value of all the trigonometric functions.

So, $\sin \left(\theta \right)=\frac{8}{10}$ $\left[\because \sin \left(\theta \right)=\frac{\mathrm{y}}{\mathrm{r}}\right]$

$\Rightarrow$ $\sin \left(\theta \right)=\frac{4}{5}$

$\cos \left(\theta \right)=\frac{-6}{10}$ $\left[\because \cos \left(\theta \right)=\frac{\mathrm{x}}{\mathrm{r}}\right]$

$\Rightarrow$ $\cos \left(\theta \right)=\frac{-3}{5}$

$\tan \left(\theta \right)=\frac{8}{-6}$ $\left[\because \tan \left(\theta \right)=\frac{\mathrm{y}}{\mathrm{x}}\right]$

$\Rightarrow$ $\tan \left(\theta \right)=-\frac{4}{3}$

$\cot \left(\theta \right)=\frac{-6}{8}$ $\left[\because \cot \left(\theta \right)=\frac{\mathrm{x}}{\mathrm{y}}\right]$

$\Rightarrow$ $\cot \left(\theta \right)=\frac{-3}{4}$

$\sec \left(\theta \right)=\frac{10}{-6}$ $\left[\because \sec \left(\theta \right)=\frac{\mathrm{r}}{\mathrm{x}}\right]$

$\Rightarrow$ $\sec \left(\theta \right)=\frac{-5}{3}$

$\mathrm{cosec}\left(\theta \right)=\frac{10}{8}$ $\left[\because \mathrm{cosec}\left(\theta \right)=\frac{\mathrm{r}}{\mathrm{y}}\right]$

$\Rightarrow$ $\mathrm{cosec}\left(\theta \right)=\frac{5}{4}$

Hence, all six trigonometric functions are calculated above.