Question 590596
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You want to convert from *[tex \LARGE (x,\,y)] to *[tex \LARGE (r,\,\theta)]


Use the following relationships:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r^2\ =\ x^2\ +\ y^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \theta\ =\ \tan^{-1}\left(\Large{\frac{y}{x}}\LARGE\right)]


For


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (-\sqrt{3},\,-1)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r^2\ =\ \left(-\sqrt{3}\right)^2\ +\ \left(-1\right)^2\ =\ 3\ +\ 1\ =\ 4\ \Rightarrow\ r\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \theta\ =\ \tan^{-1}\left(\Large{\frac{-1}{-\sqrt{3}}}\LARGE\right)]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\theta\ =\ \Large{\frac{-1}{-\sqrt{3}}]


but tangent is the ratio of sine to cosine, so:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin\theta}{\cos\theta}\ =\ \Large{\frac{-1}{-\sqrt{3}}]


by the way, before we get too far along, notice that both *[tex \LARGE x] and *[tex \LARGE y] are negative which tells us that our point is in the third quadrant meaning that, whatever *[tex \LARGE \theta] turns out to be, we are certain that it is in the interval *[tex \LARGE \pi\ <\ \theta\ <\ \Large\frac{3\pi}{2}]


Square both sides of:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin\theta}{\cos\theta}\ =\ \Large{\frac{-1}{-\sqrt{3}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin^2\theta}{\cos^2\theta}\ =\ \Large{\frac{1}{3}]


Then cross multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\theta\ =\ 3\left(\sin^2\theta\right)]


A little algebra music, Sammy:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\theta\ -\ 3\left(\sin^2\theta\right)\ =\ 0]


Apply the Pythagorean Identity:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\theta\ -\ 3\left(1\ -\ \cos^2\theta\right)\ =\ 0]


Another chorus, Sammy:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\cos^2\theta\ -\ 3\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\theta\ =\ \frac{3}{4}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\theta\ =\ \pm\frac{\sqrt{3}}{2}]



Use the unit circle:


<img src="http://www.math.ucsd.edu/~jarmel/math4c/Unit_Circle_Angles.png">


To see that the 3rd quadrant angle where *[tex \LARGE \cos\theta\ =\ -\frac{\sqrt{3}}{2}]


is *[tex \LARGE \frac{4\pi}{3}]


So the polar coordinate point is *[tex \LARGE \left(2, \frac{4\pi}{3}\right)]


Do the other ones the same way, including the two that you did that are, in sum, 75% wrong.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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