# Intersection of Polar Curves

• October 27th 2008, 02:57 PM
aeubz
Intersection of Polar Curves
Hello guys!

I'm trying to brush up on my calc. 2.
So I had two questions, find the points of intersections of both;

33) r = 2sin(theta)

r = 2sin(2theta)
I get to sin(theta)/sin(2theta)=1
what do I do with the "sin(2theta)."(Can this 2 be moved out?)
Answer: (0,0), (3^1/2, pi/3), (-(3^1/2), pi/3)

37) r=1
r^2 = 2sin(2theta)
Don't know what to do here...
1 = +/- 2sin(2theta)
...?
Answer: (1, pi/12), (1, 5pi/12), (1, 13pi/12), (1, 17pi/12)
• October 27th 2008, 04:22 PM
mr fantastic
Quote:

Originally Posted by aeubz
Hello guys!

I'm trying to brush up on my calc. 2.
So I had two questions, find the points of intersections of both;

33) r = 2sin(theta)

r = 2sin(2theta)
I get to sin(theta)/sin(2theta)=1
what do I do with the "sin(2theta)."(Can this 2 be moved out?)
Answer: (0,0), (3^1/2, pi/3), (-(3^1/2), pi/3)

37) r=1
r^2 = 2sin(2theta)
Don't know what to do here...
1 = +/- 2sin(2theta)
...?
Answer: (1, pi/12), (1, 5pi/12), (1, 13pi/12), (1, 17pi/12)

General advice not necessarily related to the above questions: It's always wise to draw the curves as well as do the algebra because the algebra doesn't always give all intersection points.

Both problems assume that you have sufficient understanding of trigonometric functions to solve trigonmetric equations. (This would be an assumed prerequisite).

33) $2 \sin (\theta) = 2 \sin (2 \theta)$

$\Rightarrow 2 \sin (\theta) = 4 \sin (\theta) \cos (\theta)$

$\Rightarrow 2 \sin (\theta) [1 - 2 \cos (\theta)] = 0$

$\Rightarrow \sin (\theta) = 0$ or $\cos (\theta) = \frac{1}{2}$.

37) Solve $1 = 2 \sin (2 \theta) \Rightarrow \sin (2 \theta) = \frac{1}{2}$.