Question

I'm looking for the intersection points of a pair of polar curves r=cos(3*theta) and r= sin(3*theta) on the interval of 0 <= theta <= pi.

I know that to find the points I set the equations equal to one another. So I did that and got tan(3*theta) = 1 and I got that theta = pi/12.

There are 2 other intersection points that I need to find, but I forgot the method for solving for multiple values. Can somebody give me a refresher?

Answer 1

when is cos(u)=sin(u)?

Answer 2

think about the unit circle

Answer 3

When tan(u) = 1.... So at 45 degree angles?

Answer 4

well there are other numbers where cos and sin values are the same

Answer 5

so if we replace 3 theta with u
then theta=u/3
so that means 0<=u/3<=pi
this means you need to solve cos(u)=sin(u) on the interval u in [0,3pi]

Answer 6

so you should get 3 answers

Answer 7

you already have one which is at pi/4

Answer 8

u=pi/4
you could go ahead and say u=pi/4+2pi
but there is one more look in the quadrant where both sin and cos are negative

Answer 9

Ahh I see what you're saying. Ok, so there's pi/4, 5pi/4, and 9pi/4?

Answer 10

\[u=\frac{\pi}{4} \\ u=\frac{5\pi}{4} \\ u=\frac{9 \pi}{4} \\ \text{ so } u=3 \theta \\ 3 \theta= \frac{\pi}{4} \\ 3 \theta=\frac{5 \pi}{4} \\ 3 \theta=\frac{ 9 \pi}{4}\]

Answer 11

solve those three equations

Answer 12

pi/12, 5pi/12, and 9pi/12.

So the method for this or any problem would be to set your theta with coefficient value equal to u, solve for theta, and then substitute that into the boundary inequality?

Answer 13

yeah when i have something like sin(6 x)=1/2
and i want to solve it on the interval 0<x<9 pi
I replace 6x with u
so so x=u/6
replace x in the inequality above to find out what u needs to be between and then solve
sin(u)=1/2 on 0<x<54 pi
which this question would be ridiculous
because that domain is huge

Answer 14

\[u=\frac{\pi}{6}+2 \pi n \\ u=\frac{5 \pi}{6}+ 2 \pi n \]
where n=0,1,...,27

Answer 15

then replace u with 6x and then solve for x

Answer 16

Ok great, thanks a lot Freckles

Answer 17

actually I think that should be n=0,..,26

Answer 18

but yeah I think you get the idea