# How do you find sin, cos, tan, sec, csc, and cot given (-4,-4)?

Mar 2, 2018

See below.

#### Explanation:

If we are given coordinates of the form $\left(x , y\right)$, where $x$ and $y$ are negative then we are in the III quadrant.

Since $\left(- 4 , - 4\right)$ are the sides of a right triangle, then the length of the terminal side ( the hypotenuse ) is given by Pythagoras' theorem:

Let the terminal side be $\boldsymbol{r}$

${r}^{2} = {\left(- 4\right)}^{2} + {\left(- 4\right)}^{2}$

$\implies r = \sqrt{{\left(- 4\right)}^{2} + {\left(- 4\right)}^{2}} = 4 \sqrt{2}$

So for the right triangle $\boldsymbol{A B C}$, we have:

$c = 4 \sqrt{2}$

$a = - 4$

$b = - 4$

$\sin \left(\theta\right) = \text{opposite"/"hypotenuse} = \frac{a}{c} = - \frac{4}{4 \sqrt{2}} = \textcolor{b l u e}{- \frac{\sqrt{2}}{2}}$

$\cos \left(\theta\right) = \text{adjacent"/"hypotenuse} = \frac{b}{c} = - \frac{4}{4 \sqrt{2}} = \textcolor{b l u e}{- \frac{\sqrt{2}}{2}}$

$\tan \left(\theta\right) = \text{opposite"/"adjacent} = \frac{a}{b} = \frac{- 4}{-} 4 = \textcolor{b l u e}{1}$

Since:

$\textcolor{red}{\boldsymbol{\csc \left(\theta\right) = \frac{1}{\sin} \left(\theta\right)}}$

$\textcolor{red}{\boldsymbol{\sec \left(\theta\right) = \frac{1}{\cos} \left(\theta\right)}}$

$\textcolor{red}{\boldsymbol{\cot \left(\theta\right) = \frac{1}{\tan} \left(\theta\right)}}$

We have:

$\csc \left(\theta\right) = \frac{1}{- \frac{\sqrt{2}}{2}} = \textcolor{b l u e}{- \sqrt{2}}$

$\sec \left(\theta\right) = \frac{1}{- \frac{\sqrt{2}}{2}} = \textcolor{b l u e}{- \sqrt{2}}$

$\cot \left(\theta\right) = \frac{1}{1} = \textcolor{b l u e}{1}$