# Deriving equation for circle using sin and cos identities

spanglophile

## Homework Statement

I have been given these two equations:
x=2acos^2(x) , y = 2a(cos(x))(sin(x)) where a ranges from 0 to 5 and -2π < x < 2π
I need to prove that these equations (when you plug in values for x) create points that when plotted, give you a circle with center (x-a) and radius a.

## Homework Equations

I have gotten the correct answer, however I am missing a step and I'm not sure what I did to get the correct answer.

## The Attempt at a Solution

I wanted to get an equation for a circle out of the two equations I have been given. So using the sin and cos identities, I can get rid of the x's. I know 2cos^2x = cos2x + 1 and that 2(cos(x))(sin(x)) = sin2x. I then added the a values back in so I have these two equations:
x= a+a(cos(2x)) , y= a(sin(2x))
From there, I fixed the x value to look like this:
(x-a) = a(cos(2x)) , y = a(sin(2x))
Then I remembered the circle equation of x^2 + y^2 = 1
So I did this:
(x-a)^2 + y^2 = a^2(cos^2(2x)) + a^2(sin^2(2x))
I know I'm supposed to get (x-a)^2 + y^2 = a^2 , but I'm not sure what to do to the above equation to get this result. Any help would be greatly appreciated!

Try factoring out a^2 from the right hand side.

Staff Emeritus
Gold Member

$$\sin^2 \theta + \cos^2 \theta$$ ?

spanglophile
Oh, I get it. So I could write it like this:

(x-a)^2 + y^2 = a^2 (cos^2(2x) + sin^2(2x))

Then the identity takes care of the rest so I'm left with

(x-a)^2 + y^2 = ^2 , right?

spanglophile
I meant to write the above post as:

(x-a)^2 + y^2 = a^2