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Homework Help: Simple trig problem

  1. Nov 1, 2012 #1
    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. jcsd
  3. Nov 1, 2012 #2
    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].
  4. Nov 1, 2012 #3
    so cos2α is not equal to 0?
  5. Nov 1, 2012 #4
    If [itex]\cos(2\alpha)=0[/itex], then [itex]\cos(2\alpha)=\sin(2\alpha)[/itex] could not hold.
  6. Nov 1, 2012 #5
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