Understanding Trig Problems: When Can You Divide by Cosα?

  • Thread starter Thread starter phospho
  • Start date Start date
  • Tags Tags
    Trig
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 2K views
phospho
Messages
250
Reaction score
0
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?
 
Physics news on Phys.org
phospho said:
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 [itex]\cos(2\alpha)=0[/itex], then [itex]\sin^2(2\alpha)=1-\cos^2 (2\alpha)=1[/itex]. So if [itex]\cos(2\alpha)=0[/itex], then [itex]\sin(2\alpha)=\pm 1[/itex]. So we can never have [itex]\sin(2\alpha)=\cos(2\alpha)[/itex].
 
micromass said:
Since if [itex]\cos(2\alpha)=0[/itex], then [itex]\sin^2(2\alpha)=1-\cos^2 (2\alpha)=1[/itex]. So if [itex]\cos(2\alpha)=0[/itex], then [itex]\sin(2\alpha)=\pm 1[/itex]. So we can never have [itex]\sin(2\alpha)=\cos(2\alpha)[/itex].

so cos2α is not equal to 0?
 
phospho said:
so cos2α is not equal to 0?

If [itex]\cos(2\alpha)=0[/itex], then [itex]\cos(2\alpha)=\sin(2\alpha)[/itex] could not hold.
 
micromass said:
If [itex]\cos(2\alpha)=0[/itex], then [itex]\cos(2\alpha)=\sin(2\alpha)[/itex] could not hold.

thanks