# Converting Polar Coordinates to Regular Coordinates

If you guys didn't know, I have my quiz tomorrow and I have one last thing to ask to this Community! I am completely confused on how to convert polar coordinates to regular coordinates. The teacher gave us this example:

Convert to Polar Coordinates: $(3 , -45^\circ)$

$x = r\cos(\theta) = 2\cos(-45^\circ) = 3(\frac{\sqrt{2}}{2})$

$y = r\cos(\theta) = 3\sin(-45 ^\circ) = 3(\frac{-\sqrt{2}}{2})$

$(\frac{3\sqrt{2}}{2} , \frac{-3\sqrt{2}}{2})$

Ok, she did that and gave us this (one of of the two) for a review:

$(6 , \frac{-2\pi}{3})$

Well then I had completely no idea..(I know the equation though)

I did this:

$x = r\cos(\theta)$ $x = r\sin(\theta)$

I basically didn't know what to put at the coefficient of $\cos$ and $\sin$. please help. Thanks a lot for reading!

• You've copied the example down incorrectly (either here or in your notes). You should have: $$x = r\cos(\theta) = \color{red}3\cos(-45^\circ) = 3(\frac{\sqrt{2}}{2})$$ – Ian Miller May 13 '16 at 0:21
• Guess i wrote it incorrectly in my notes... wow @IanMiller – amanuel2 May 13 '16 at 0:22
• The example you gave is a conversion from polar coordinates to rectangular coordinates. The polar coordinates are $(r, \theta) = (3, -45^\circ)$. The rectangular coordinates are $(x, y) = (\frac{3\sqrt{2}}{2}, -\frac{3\sqrt{2}}{2})$. Your formula for converting from polar coordinates to rectangular coordinates should read $x = r\cos\theta$, $\color{red}{y} = r\sin\theta$. The coefficient for $\cos\theta$ and $\sin\theta$ is the value of $r$. Does this help? – N. F. Taussig May 13 '16 at 0:23
• Yes @N.F.Taussig It does thanks a lot, for you guys helping me through this! I got this concept! – amanuel2 May 13 '16 at 0:26

When writing coordinates in polar notation you've written them as $(r,\theta)$. So these are the values you should stick into your formulae for $x$ and $y$.

The transformation from polar to rectangular can be seen as:

$$(r,\theta)\rightarrow(r\cos\theta,r\sin\theta)$$

So for your example you get:

$$\left(6,-\frac{2\pi}{3}\right)\rightarrow\left(6\cos\left(-\frac{2\pi}{3}\right),6\sin\left(-\frac{2\pi}{3}\right)\right)$$

You can then use your knowledge of exact values to get:

$$(-3,-3\sqrt{3})$$

• Wow that way is much easier than my teacher's way... Ian thanks a lot sir! Thanks a lot for helping me through guys @N.F.Taussig , Ian – amanuel2 May 13 '16 at 0:25