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

Jul 15, 2016

$\sin \theta = - 3 \frac{\sqrt{13}}{13}$

$\cos \theta = - 2 \frac{\sqrt{13}}{13}$

$\tan \theta = \frac{3}{2}$

$\sec \theta = - \frac{\sqrt{13}}{2}$

$\csc \theta = - \frac{\sqrt{13}}{3}$

$c t n \theta = \frac{2}{3}$

#### Explanation:

For the point $\left(- 4 , - 6\right)$ we can create a right triangle with legs of $x = - 4$ and $y = - 6$

Using the pythagorean theorem we get the hypotenuse.

${a}^{2} + {b}^{2} = {c}^{2}$

${\left(- 4\right)}^{2} + {\left(- 6\right)}^{2} = h y {p}^{2}$

$16 + 36 = h y {p}^{2}$

$52 = h y {p}^{2}$

$\sqrt{52} = h y p$

$2 \sqrt{13} = h y p$

The six trig functions are

$\sin \theta = \frac{o p p}{h y p}$

$\cos \theta = \frac{a \mathrm{dj}}{h y p}$

$\tan \theta = \frac{o p p}{a \mathrm{dj}}$

$\sec \theta = \frac{h y p}{a \mathrm{dj}}$

$\csc \theta = \frac{h y p}{o p p}$

$c t n \theta = \frac{a \mathrm{dj}}{o p p}$

$\sin \theta = \frac{- 6}{2 \sqrt{13}} = - \frac{3}{\sqrt{13}} \cdot \frac{\sqrt{13}}{\sqrt{13}} = - 3 \frac{\sqrt{13}}{13}$

$\cos \theta = \frac{- 4}{2 \sqrt{13}} = - \frac{2}{\sqrt{13}} \cdot \frac{\sqrt{13}}{\sqrt{13}} = - 2 \frac{\sqrt{13}}{13}$

$\tan \theta = \frac{- 6}{-} 4 = \frac{3}{2}$

$\sec \theta = \frac{2 \sqrt{13}}{-} 4 = - \frac{\sqrt{13}}{2}$

$\csc \theta = \frac{2 \sqrt{13}}{-} 6 = - \frac{\sqrt{13}}{3}$

$c t n \theta = \frac{- 4}{-} 6 = \frac{2}{3}$