Understanding Trig Problems: When Can You Divide by Cosα?

[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?

 

[micromass]

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?

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).

 

[phospho]

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).

so cos2α is not equal to 0?

 

[micromass]

so cos2α is not equal to 0?

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

 

[phospho]

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

thanks