SOLUTION: Find all points [r, theta] of intersection r=1 and r= 2sin(theta) Multiple choice: a)[1/2, pi/3] & [1/2, 5pi/3] b) [1/2, pi/6] and [1/2, 5pi/6] c) [1, pi/3] & [1,5pi/3] d) [

Algebra ->  Trigonometry-basics -> SOLUTION: Find all points [r, theta] of intersection r=1 and r= 2sin(theta) Multiple choice: a)[1/2, pi/3] & [1/2, 5pi/3] b) [1/2, pi/6] and [1/2, 5pi/6] c) [1, pi/3] & [1,5pi/3] d) [      Log On


   



Question 281690: Find all points [r, theta] of intersection r=1 and r= 2sin(theta)
Multiple choice:
a)[1/2, pi/3] & [1/2, 5pi/3]
b) [1/2, pi/6] and [1/2, 5pi/6]
c) [1, pi/3] & [1,5pi/3]
d) [1, pi/6] and [1,5pi/6]
e) None of these
I want to eliminate A & B because their r-values are 1/2 instead of 1.
If I graph these in Polar mode (in radian mode too), the two graphs form two ellipses. If I trace, I can find intersections at theta = .5236 (pi/6) and 2.618 (5pi/6). So I'm leading toward D as my answer.
First, is it okay to use the graphical approach to solving this problem?
Second, what is the algebraic approach to solving this problem?

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
There's nothing wrong with using a graphical approach. You just need to keep in mind that the solutions will be approximate, which means that you need to figure out a way to get them into exact form. Since you are familiar with 0.5236 and 2.618 as pi%2F6 and 5pi%2F6, the second concern isn't really an issue. However, the approximate solutions may not be so easy to recognize in general. Because of this potential problem, I suggest to use an algebraic approach.


r=2sin%28theta%29 Start with the second equation


1=2sin%28theta%29 Plug in r=1


1%2F2=sin%28theta%29 Divide both sides by 2.


sin%28theta%29=1%2F2 Rearrange the equation.


theta=pi%2F6 or theta=5pi%2F6 Take the arcsine of both sides


Note: Since sin%28pi%2F6%29=sin%285pi%2F6%29=1%2F2, this means that the arcsine of 1%2F2 is either pi%2F6 or 5pi%2F6.


So the exact solutions for theta are theta=pi%2F6 or theta=5pi%2F6


So you are correct.