# How do you find the exact value of the five remaining trigonometric function given cottheta=1/2 and the angle is in standard position in quadrant III?

Feb 9, 2018

$\sin \theta = - \frac{2 \sqrt{5}}{5} \text{ "" "" } \cos \theta = - \frac{\sqrt{5}}{5}$

$\csc \theta = - \frac{\sqrt{5}}{2} \text{ "" "" "color(white)"x} \sec \theta = - \sqrt{5}$

$\tan \theta = 2 \text{ "" "" "" "" "color(white)".-} \cot \theta = \frac{1}{2}$

#### Explanation:

First, let's find $\tan \theta$. We know that it will be $\frac{1}{\cot} \theta$, so:

$\textcolor{red}{\tan \theta} = \frac{1}{\cot} \theta = \frac{1}{\frac{1}{2}} = \textcolor{red}{2}$

Now, using the identities $\tan \theta = \sin \frac{\theta}{\cos} \theta$ and ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$, we can find the cosine:

$\tan \theta = 2$

$\sin \frac{\theta}{\cos} \theta = 2$

$\frac{\sqrt{{\sin}^{2} \theta}}{\cos} \theta = 2$

$\frac{\sqrt{{\sin}^{2} \theta + {\cos}^{2} \theta - {\cos}^{2} \theta}}{\cos} \theta = 2$

$\sqrt{1 - {\cos}^{2} \theta} = 2 \cos \theta$

$1 - {\cos}^{2} \theta = 4 {\cos}^{2} \theta$

$1 = 5 {\cos}^{2} \theta$

$\frac{1}{5} = {\cos}^{2} \theta$

$\pm \frac{\sqrt{5}}{5} = \cos \theta$

Now, since the angle is in Q3 where both coordinates are negative, we can say that $\cos \theta$ is negative, so:

color(red)(costheta = -sqrt5/5

To find the sine, remember the identity ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$.

${\sin}^{2} \theta + {\left(- \frac{\sqrt{5}}{5}\right)}^{2} = 1$

${\sin}^{2} \theta + \frac{1}{5} = 1$

${\sin}^{2} \theta = \frac{4}{5}$

$\sin \theta = \pm \frac{2 \sqrt{5}}{5}$

Again, both coordinates are negative, so sine must be negative.

color(red)(sintheta = -(2sqrt5)/5

Remember that $\csc \theta = \frac{1}{\sin} \theta$, and $\sec \theta = \frac{1}{\cos} \theta$

color(red)(csctheta) = 1/sintheta = 1/(-(2sqrt5)/5) = color(red)(-sqrt5/2

color(red)(sectheta) = 1/costheta = 1/(-sqrt5/5) = color(red)(-sqrt5

The values of all five remaining trig function have been found at this point, and their solutions are highlighted above in red.