# Homework Help: Simple trig problem

1. Nov 1, 2012

### phospho

In class my teacher said in general if you have a equation such as sinαcosα = cosα you shouldn't divide through by cosα as cosα can be 0 and dividing by 0 Is undefined, instead we should factorise, which makes sense.

However I was going through a question which gave sin2α = cos2α and in the solutions they divided by cos2α to get tan2α = 1 and solved, why is it allowed to divide through by cos2α here?

2. Nov 1, 2012

### micromass

Since if $\cos(2\alpha)=0$, then $\sin^2(2\alpha)=1-\cos^2 (2\alpha)=1$. So if $\cos(2\alpha)=0$, then $\sin(2\alpha)=\pm 1$. So we can never have $\sin(2\alpha)=\cos(2\alpha)$.

3. Nov 1, 2012

### phospho

so cos2α is not equal to 0?

4. Nov 1, 2012

### micromass

If $\cos(2\alpha)=0$, then $\cos(2\alpha)=\sin(2\alpha)$ could not hold.

5. Nov 1, 2012

thanks