How to solve (1+cosA)^2 - (1-cosA)^2 - sin^2A = ctgA*sinA*cosA

[makarov1901]

Homework Statement

(1+cosA)^2-(1-cosA)^2-sin^2A=ctgA*sinA*cosA

Homework Equations

The Attempt at a Solution

I moved sin^2 to the right side, then expanded the left side and got to:
1+2cosA+cos^2A-1+2cosA-cos^2A=1
When I cancel the left side I get:
4cosA=1

I'd be very grateful if anyone help me.

 

[vela]

Looks good so far. What's the problem?

(I assume when you wrote sin^2, you meant sin² a.)

 

[makarov1901]

You assumed right.
The problem is that both parts of the identity should be equal.
As far as I know, 4cosA is not equal to 1.

 

[vela]

It's not an identity. For example, when a=pi/2, the lefthand side is -1 while the righthand side equals 0.

Perhaps there's a typo in the problem.

 

[makarov1901]

My task is to transform both parts so that they're equal. I don't know why you say it's not an identity.

 

[vela]

If it were an identity, it would hold for all values of a. It clearly doesn't; therefore, it's not an identity.