# Question #990d6

Mar 31, 2017

We know that $\cos \left(x\right) = - \frac{1}{2}$ and that $\sin \left(x\right) > 0$, which means that $\sin \left(x\right)$ is a negative number.

Based on this information, let's make a coordinate grid for this problem. That will help us draw an accurate picture.
$\textcolor{w h i t e}{- - - -} \textcolor{w h i t e}{S \in} \textcolor{w h i t e}{- - - - -} \textcolor{b l a c k}{|} \textcolor{w h i t e}{- - - -}$$\textcolor{w h i t e}{A l l}$$\textcolor{w h i t e}{- - - - - - - - -}$
$\textcolor{w h i t e}{- - - -}$Sin$\textcolor{w h i t e}{- - - - -} \textcolor{b l a c k}{|} \textcolor{w h i t e}{- - - -}$All$\textcolor{w h i t e}{- - - - - - - - -}$
$\textcolor{w h i t e}{- - - -} \textcolor{w h i t e}{S \in} \textcolor{w h i t e}{- - - - -} \textcolor{b l a c k}{|} \textcolor{w h i t e}{- - - -}$$\textcolor{w h i t e}{A l l}$$\textcolor{w h i t e}{- - - - - - - - -}$
$\textcolor{b l a c k}{- - - -} \textcolor{w h i t e}{S \in} \textcolor{b l a c k}{- - - - -} \textcolor{b l a c k}{|} \textcolor{b l a c k}{- - - - - - - - -}$
$\textcolor{w h i t e}{- - - -} \textcolor{w h i t e}{T a n} \textcolor{w h i t e}{- - - - -} \textcolor{b l a c k}{|} \textcolor{w h i t e}{- - - -}$$\textcolor{w h i t e}{C o s}$$\textcolor{w h i t e}{- - - - - - - - -}$
$\textcolor{w h i t e}{- - - -}$Tan$\textcolor{w h i t e}{- - - - -} \textcolor{b l a c k}{|} \textcolor{w h i t e}{- - - -}$Cos$\textcolor{w h i t e}{- - - - - - - - -}$
$\textcolor{w h i t e}{- - - -} \textcolor{w h i t e}{T a n} \textcolor{w h i t e}{- - - - -} \textcolor{b l a c k}{|} \textcolor{w h i t e}{- - - -}$$\textcolor{w h i t e}{C o s}$$\textcolor{w h i t e}{- - - - - - - - -}$

This shows the quadrant in which the answer is positive. I like to remember it by using the mnemonic device All Students Take Calculus.
In our problem, $S \in$ in negative and so is $C o s$, so the picture has to lie in the ${3}^{r d}$ quadrant, where $T a n$ is. That means all our ratios will be negative, except for those concerning tangent

Now, I like to draw a picture, but I can't do that on this program, so I'll just write it out.

$\cos = \left(\text{adjacent")/("hypotenuse}\right)$ or $- \frac{1}{2}$

If the hypotenuse is $2$ and the leg is $1$, we can find the remaining side using Pythagorean's theorem $\left({c}^{2} - {b}^{2} = {a}^{2}\right)$. That gives us $\sqrt{3}$ as the other side.

So, just to summirize:

• adjacent $= 1$
• hypotenuse $= 2$
• opposite $= \sqrt{3}$

Now we can solve everything:

$\sin = \left(\text{opposite")/("hypotenuse}\right)$ or $- \frac{\sqrt{3}}{2}$

$\cos = \left(\text{adjacent")/("hypotenuse}\right)$ or $- \frac{1}{2}$

$\tan = \left(\text{opposite")/("adjacent}\right)$ or $\frac{\sqrt{3}}{1}$ or just $\sqrt{3}$

Now we find the inverses:

$\csc = \frac{1}{\sin} = - \frac{2}{\sqrt{3}}$ or $\frac{- 2 \sqrt{3}}{3}$

$\sec = \frac{1}{\cos} = - \frac{2}{1}$ or jsut $- 2$

$\cot = \frac{1}{\tan} = \frac{1}{\sqrt{3}}$ or $\frac{\sqrt{3}}{3}$

Mar 31, 2017

Since $\cos x = - \frac{1}{2}$ and sin x > 0, then, x is in Quadrant 2.
${\sin}^{2} x = 1 - {\cos}^{2} x = 1 - \frac{1}{4} = \frac{3}{4}$
$\sin x = \pm \frac{\sqrt{3}}{2}$
Take the positive answer because sin x > 0
$\sin x = \frac{\sqrt{3}}{2}$
$\tan x = \frac{\sin}{\cos} = \left(\frac{\sqrt{3}}{2}\right) \left(- \frac{2}{1}\right) = - \sqrt{3}$
$\cot x = \frac{1}{\tan} = - \frac{1}{\sqrt{3}} = - \frac{\sqrt{3}}{3}$
$\sec x = \frac{1}{\cos} = - 2$
$\csc x = \frac{1}{\sin} = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}$