Question

How do you find the exact value of the six trigonometric functions of $\theta$ when your given a point $(-4,-6)$?

Answer 1

Find the exact value of the six trigonometric functions of $\theta$ :

Given points are $(-4,-6)$

Use the Pythagorean formula $c=\sqrt{a^2+b^2}$.

Where $c$ is the hypotenuse, $a$ be the adjacent side and $b$ be the opposite side.

Substitute $\mathrm{Adjacent}=-4$ and $\mathrm{Opposite}=-6$ in the above formula:

$\begin{array}{rcl}\mathrm{Hypotenuse}& =& \sqrt{{(-4)}^2+{(-6)}^2}\\ & =& \sqrt{16+36}\\ & =& \sqrt{52}\\ & =& 2\sqrt{13}\end{array}$

So, the values are:

$\mathrm{Adjacent}=-4$

$\mathrm{Opposite}=-6$

$\mathrm{Hypotenuse}=2\sqrt{13}$

Step-1: Find the sin functions of $\theta$.

Substitute the known values in sin formula $\sin \left(\mathrm{\theta }\right)=\frac{\mathrm{Opposite}}{\mathrm{Hypotenuse}}$:

$\begin{array}{rcl}& \Rightarrow & \sin \left(\mathrm{\theta }\right)=\frac{-6}{2\sqrt{13}}\\ & =& \frac{-3}{\sqrt{13}}\left[\mathrm{Divide}-6\mathrm{by}2\right]\\ & =& \frac{-3}{\sqrt{13}}\times \frac{\sqrt{13}}{\sqrt{13}}\\ & =& -3\frac{\sqrt{13}}{13}\end{array}$

Step-2: Find the cos functions of $\theta$.

Substitute the known values in cos formula $\cos \left(\mathrm{\theta }\right)=\frac{\mathrm{Adjacent}}{\mathrm{Hypotenuse}}$:

$\begin{array}{rcl}& \Rightarrow & \cos \left(\mathrm{\theta }\right)=\frac{-4}{2\sqrt{13}}\\ & =& \frac{-2}{\sqrt{13}}\left[\mathrm{Divide}-4\mathrm{by}2\right]\\ & =& \frac{-2}{\sqrt{13}}\times \frac{\sqrt{13}}{\sqrt{13}}\\ & =& -2\frac{\sqrt{13}}{13}\end{array}$

Step-3: Find the tan functions of $\theta$.

Substitute the known values in tan formula $\tan \left(\mathrm{\theta }\right)=\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$:

$\begin{array}{rcl}& \Rightarrow & \tan \left(\mathrm{\theta }\right)=\frac{-6}{-4}\\ & =& \frac{3}{2}\end{array}$

Step 4. Find the sec functions of $\theta$.

Substitute the known values in sec formula $\sec \left(\mathrm{\theta }\right)=\frac{\mathrm{Hypotenuse}}{\mathrm{Adjacent}}$:

$\begin{array}{rcl}& \Rightarrow & \sec \left(\mathrm{\theta }\right)=\frac{2\sqrt{13}}{-4}\\ & =& -\frac{\sqrt{13}}{2}\end{array}$

Step-5: Find the cosec functions of $\theta$.

Substitute the known values in cosec formula $\mathrm{cosec}\left(\mathrm{\theta }\right)=\frac{\mathrm{Hypotenuse}}{\mathrm{Opposite}}$:

$\begin{array}{rcl}& \Rightarrow & \mathrm{cosec}\left(\mathrm{\theta }\right)=\frac{2\sqrt{13}}{-6}\\ & =& -\frac{\sqrt{13}}{3}\end{array}$

Step-6: Find the cot functions of $\theta$.

Substitute the known values in cot formula $\cot \left(\mathrm{\theta }\right)=\frac{\mathrm{Adjacent}}{\mathrm{Opposite}}$:

$\begin{array}{rcl}& \Rightarrow & \cot \left(\mathrm{\theta }\right)=\frac{-4}{-6}\\ & =& \frac{2}{3}\end{array}$

Hence, the exact value of the six trigonometric functions of $\theta$ are:

$\begin{array}{rcl}\mathbf{sin}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{}& \mathbf{=}& \mathbf{-}\mathbf{3}\frac{\sqrt{\mathbf{13}}}{\mathbf{13}}\\ \mathbf{cos}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{}& \mathbf{=}& \mathbf{-}\mathbf{2}\frac{\sqrt{\mathbf{13}}}{\mathbf{13}}\\ \mathbf{}\mathbf{tan}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{}& \mathbf{=}& \frac{\mathbf{}\mathbf{3}}{\mathbf{}\mathbf{2}\mathbf{}}\\ \mathbf{sec}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{}& \mathbf{=}& \mathbf{-}\frac{\sqrt{\mathbf{13}}}{\mathbf{2}\mathbf{}}\\ \mathbf{cosec}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{}& \mathbf{=}& \mathbf{-}\frac{\sqrt{\mathbf{13}}}{\mathbf{3}\mathbf{}}\\ \mathbf{cot}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{}& \mathbf{=}& \frac{\mathbf{2}}{\mathbf{3}\mathbf{}}\end{array}$