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