How to find when two curves intersect? I need help :0

[Ineedhelp777]

Hi I need help with this problem: Find all points at which the two curves intersect. r=1-2sintheta and r=2costheta.
I tried setting the two equations equal to each other but then I get stuck.

 

[Steven G]

Please post your work so we can see where you got stuck and then we can direct you in the correct direction. Thanks.

 

[Ineedhelp777]

I attached my work. I tried setting them equal to eachother. Then I tried solving sin and cos at one but my answer I got was different fro the one on the key. I will also attach the way it was solved on the key.Its problem #11.

 

[HallsofIvy]

I presume you quickly got to $\displaystyle sin(\theta)+ cos(\theta)= \frac{1}{2}$. To go beyond that you will need a trig identity such as $\displaystyle sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b)$ taking $\displaystyle b= \theta$. Of course, "sin(a)" and "cos(a)" can't both be one. Sin(a) and cos(a) are the same for $\displaystyle a= \pi/4$ and then are both equal to $\displaystyle \frac{\sqrt{2}}{2}$. Multiplying both sides by that, and taking $\displaystyle a= \pi/4$, $\displaystyle \frac{\sqrt{2}}{2}sin(b+ \pi/4)= \frac{\sqrt{2}}{2}sin(b)+ \frac{\sqrt{2}}{2}cos(b)$.

So, going back to the original equation, $\displaystyle \frac{\sqrt{2}}{2}sin(\theta)+ \frac{\sqrt{2}}{2}cos(\theta)= sin(\theta+ \pi/4)= \frac{\sqrt{2}}{4}$..edited. Can you solve that equation?

 

[Steven G]

Are you sure that you don't have an extra sqrt(2)/2? I checked it three times.

 

[Ineedhelp777]

Thank you for helping. I tried graping each equation and finding the intersection. I then got two values. I plugged those two values into r. I then multiplied the two values of r by y=rsintheta and x=costheta. I then found the two points. Thank you for your help.

 

[Deleted member 4993]

Are you sure that you don't have an extra sqrt(2)/2? I checked it three times.

Jomo,

It will be very helpful if you use the "reply" button in the "response to be corrected".

 

[Steven G]

Jomo,

It will be very helpful if you use the "reply" button in the "response to be corrected".

...It was immediately after Halls post. But you are a Super Moderator and I should listen to you. Thanks for the comment/instructions