# Math Help - Can I get help with this fraction!

1. ## Can I get help with this fraction!

I'm trying to solve this cotangent $\frac{1} {2}\cdot\frac{2} {\sqrt3}$When I take the tangent of this I get the right answer of $\frac{\sqrt3}{3}$Can someone work out $\frac{1} {2}\cdot\frac{2} {\sqrt3}$for me? I get $\frac{1} {\sqrt3}$ as an answer.

Thanks
Jason

2. This is very, very difficult to understand.

$\frac{1}{2}*\frac{2}{\sqrt{3}}\;=\;\frac{1}{\sqrt{ 3}}\;=\;\frac{\sqrt{3}}{3}$

I'm not seeing any tangents or cotangents in there. This leads me to believe that you really are not understanding the problem or that you have reported the problem inaccurately.

Just EXACTLY what is the problem statement and what EXACTLY were your efforts to solve it?

Note: Try to remember that honesty does not imply insult.

3. Originally Posted by TKHunny
This is very, very difficult to understand.

$\frac{1}{2}*\frac{2}{\sqrt{3}}\;=\;\frac{1}{\sqrt{ 3}}\;=\;\frac{\sqrt{3}}{3}$

I'm not seeing any tangents or cotangents in there. This leads me to believe that you really are not understanding the problem or that you have reported the problem inaccurately.

Just EXACTLY what is the problem statement and what EXACTLY were your efforts to solve it?

Note: Try to remember that honesty does not imply insult.
I was working on a trig problem and the fraction was end of the problem. I didn't know $\frac{1}{\sqrt{3}}\;=;\frac{\sqrt{3}}{3}$
The original problem was finding the cotangent of $\frac{7pie} {3}$
Thanks

Jason.

4. Therein lies my earlier conclusion. Some time ago, you should have encountered the manipulation of simple radicals. In particular, "Rationalizing the Denominator" is a common idea. I find the process generally silly, but it shows up everywhere.

$\frac{1}{\sqrt{3}}\;=\;\frac{1}{\sqrt{3}}*\frac{\s qrt{3}}{\sqrt{3}}\;=\;\frac{\sqrt{3}}{3}$

Find an algebra book. Find this section. Study it. You will need it in order to move on.

Note: "pie" is for eating.
Note: "pi" is for math and Greeks.

Exact values often require the concept of the "Reference Angle". $\frac{7\pi}{3}\;=\;\frac{\pi}{3}\;+\;2\pi$. Thus, you do not need to find the cotangent of the original angle, you can find the cotangent of this angle that might be more familiar. In fact, if you suffer an exam on this material, you probably will be expected to know at least the sine and cosine of a few very convenient reference angles. pi/3 is one of them that you will need. Memorize the values of sine and cosine for 0, pi/6, pi/4, pi/3, and pi/2.

Philosophy: Most things are not easy when they are new.