My Math Forum I can't find the third answer in the polar equations.

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 March 6th, 2017, 06:59 PM #1 Senior Member   Joined: Apr 2008 Posts: 194 Thanks: 3 I can't find the third answer in the polar equations. I am going to write down a problem below. (ex) Find the points of intersection of r=1-sin(u) and r=1+sin(u) without drawing their graphs. This is how I do it. I equate the two equations and get 1-sin(u)=1+sin(u). Simplifying it gives sin(u)=0. Solving for u gives 0 and pi. Then, the two points are (1 , pi) and (1 , 0). They are correct. The answer key gives one more answer (0 , pi/2). How do you find the third answer without graphing? Thanks for your help.
 March 6th, 2017, 07:47 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs You get: $\displaystyle \sin(u)=-\sin(u)$ which is indeed true for: $\displaystyle \sin(u)=0$ or $\displaystyle u=k\pi$ where $\displaystyle k\in\mathbb{Z}$ But we also need to consider points of intersection that occur for different parameters for the two functions...so we could consider the identity: $\displaystyle \sin(u+(2k-1)\pi)=-\sin(u)=\sin(-u)$ Hence: $\displaystyle u+(2k-1)\pi=-u$ $\displaystyle u=\frac{\pi}{2}(1-2k)$ Since the period of the two functions is $2\pi$, we are interesting in: $\displaystyle 0\le u<2\pi$ $\displaystyle 0\le \frac{\pi}{2}(1-2k)<2\pi$ $\displaystyle 0\le 1-2k<4$ $\displaystyle -1\le -2k<3$ $\displaystyle 1\ge 2k>-3$ $\displaystyle \frac{1}{2}\ge k>-\frac{3}{2}$ Thus: $\displaystyle k\in\{-1,0\}$ And so: $\displaystyle u=\frac{\pi}{2},\,\frac{3\pi}{2}$ We find: $\displaystyle r_1\left(\frac{\pi}{2}\right)=1-\sin\left(\frac{\pi}{2}\right)=0$ $\displaystyle r_2\left(\frac{3\pi}{2}\right)=1+\sin\left(\frac{3 \pi}{2}\right)=0$ And so the two functions intersect at the origin, but for different values of $u$. We also find that: $\displaystyle r_1\left(\frac{3\pi}{2}\right)=1-\sin\left(\frac{3\pi}{2}\right)=2$ $\displaystyle r_2\left(\frac{\pi}{2}\right)= 1+\sin\left(\frac{\pi}{2}\right)=2$ But, we observe that the first function corresponds with the Cartesian point (0,-2) while the second corresponds with (0,2). Thanks from davedave and romsek Last edited by MarkFL; March 6th, 2017 at 07:50 PM.
 March 7th, 2017, 09:51 PM #3 Senior Member   Joined: Apr 2008 Posts: 194 Thanks: 3 Thanks a lot, MarkFL. Thanks from MarkFL

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