ive either been studying too long and my brain has fried, or i am an idiot..

im trying to find the points of intersection between the graph of r=sin;r=1+cos

the graphing is easy, but to find the points of intersection i make those statements equal to each other, so...

sin=1+cos

but I need arrange it in such a way where i can find what angles give me the points of intersection

example, on a previous problem i graphed and had sin=cos which i switched to tan=1, and tan=1 at pi/4 and 5pi/4

so how can i arrange sin=1+cos in a way that i can find the correct angles? I mean, i know what the angles are because i can see where they touch on the graph, but i need to prove that

use unit circle
coscos+sinsin=1

square both of your sides

so where does cos^2+sin^2 = 1? without using a unit circle

I don't really understand your question

coscos+sinsin=1 is an identity, it is always true
like aa+bb=cc for right triangles

This post was edited by saber_x3 on Apr 16 2014 06:36pm

If you square both sides, you'll have:

(sin x)^2 = 1 + 2cos x + (cos x)^2

I'm not sure this is the right method to go about this problem.


E: maybe try multiplying both sides by the conjugate? It might lead to an identity or easy simplifying.
btw, the conjugate of (1 + cos x) is (1 - cos x)

This post was edited by TritonV8 on Apr 16 2014 06:41pm

yeah i know, and i thought of doing that but i need to arrange sin=1+cos in a way that i can find what angles make that true.

like if i was to tell you where does sin=cos, you would divide both sides by cos and end up with tan=1, then i ask you where does tan=1? pi/4 and 5pi/4

the thing is that i have to find points of intersection, i need to find the polar coordinates where the two graphs intersect.

sin=1+cos
sinsin=1+2cos+coscos
0=1-sinsin+2cos+coscos
0=coscos+2cos+coscos
2coscos+2cos=0
2cos [cos+1]=0

(cos) (cos+1)=0
cos=0, 90 deg
cos=-1 , 180deg

i dont think thats it but i think youre close, i know they intersect at the origin r=0 and r=1;theta=pi/2 so these have to be the polar coords

how'd you get sin^2= 1+2cos+cos^2?

square both sides

and 1-sinsin= coscos

This post was edited by saber_x3 on Apr 16 2014 06:55pm

yeah but wouldnt that be sin^2=1+cos^2 or sin^2=sin^2

where did you get 2cos?