Understanding Trig Problems: When Can You Divide by Cosα?

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Homework Help Overview

The discussion revolves around the conditions under which division by cosine is permissible in trigonometric equations, specifically in the context of the equation sin2α = cos2α. Participants are exploring the implications of dividing by cos2α and the potential for it to equal zero.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the validity of dividing by cos2α in the equation sin2α = cos2α, considering the case where cos(2α) might equal zero. They discuss the implications of this condition and whether it affects the equality of sin(2α) and cos(2α).

Discussion Status

The discussion is active, with participants examining the reasoning behind the division by cos2α and its consequences. Some have provided insights into the relationship between sine and cosine when cos(2α) equals zero, suggesting that this condition prevents the equality from holding.

Contextual Notes

Participants are operating under the assumption that dividing by zero is undefined, leading to a careful examination of when cos(2α) could be zero in the context of the problem.

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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?
 
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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
 

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