Question

Find the values of the six trigonometric functions for the angle in standard position having the point $\left(3,-5\right)$on its terminal side.

Answer 1

Find the values of the six trigonometric functions.

The angle in the standard position has the point $\left(3,-5\right)$on its terminal side.

Step 1: Calculate all three sides of the triangle.

Given, point $\left(3,-5\right)$

i.e. $x=3,y=-5$

$\begin{array}{rcl}r& =& \sqrt{x^2+y^2}\\ & =& \sqrt{3^2+{\left(-5\right)}^2}\\ & =& \sqrt{34}\end{array}$

Step 2: Calculate the trigonometric ratio values.

Using trigonometric ratios,

$$\begin{gathered} \Rightarrow \sin \mathrm{\theta }=\frac{\mathrm{y}}{\mathrm{r}}=\frac{-5}{\sqrt{34}} \\ \Rightarrow \cos \mathrm{\theta }=\frac{\mathrm{x}}{\mathrm{r}}=\frac{3}{\sqrt{34}} \\ \Rightarrow \tan \mathrm{\theta }=\frac{\mathrm{y}}{\mathrm{x}}=\frac{-5}{3} \\ \Rightarrow \mathrm{cosec}\mathrm{\theta }=\frac{\mathrm{r}}{\mathrm{y}}=\frac{\sqrt{34}}{-5}=-\frac{\sqrt{34}}{5} \\ \Rightarrow \cos \mathrm{\theta }=\frac{\mathrm{r}}{\mathrm{x}}=\frac{\sqrt{34}}{3} \\ \Rightarrow \cot \mathrm{\theta }=\frac{\mathrm{x}}{\mathrm{y}}=\frac{3}{-5}=-\frac{3}{5} \end{gathered}$$

Hence, all the trigonometric ratio values have been found.