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Double Angle Formula Problem

  1. May 11, 2014 #1
    1. The problem statement, all variables and given/known data

    Given that cos([itex]\pi[/itex]/6) =[itex]\sqrt{}3[/itex]/2, use the double angle formula for the cosine function to find cos([itex]\pi[/itex]/12) and sin([itex]\pi[/itex]/12) explicitly.

    2. Relevant equations

    cos(2x)=cos2x - sin2x
    cos2x + sin2x = 1

    3. The attempt at a solution

    So it wants me to find cos([itex]\pi[/itex]/12) which is half the angle of cos([itex]\pi[/itex]/6). So I called these cosx and cos 2x.

    I then said [itex]\sqrt{}3[/itex]/2 = cos2x - sin2x

    I used cos2x + sin2x = 1 and got sin2x on its own and subbed into the first formula and then got cosx on its own.

    For sin([itex]\pi[/itex]/12) I subbed in sin2x = 1- cos2x and got sinx on its own.

    Is this the correct method for finding the answers?

    The inverse of cosx and sinx were [itex]\pi[/itex]/12 so I assume I am but not sure. I'd be thankful to anyone who could clear this up.
  2. jcsd
  3. May 11, 2014 #2


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    Yes, [itex]cos(2x)= cos^2(x)- sin^2(x)[/itex] so that [itex]cos(2x)= cos^2(x)- (1- cos^2(x))= 2cos^2(x)- 1[/itex] and [itex]cos(2x)= (1- sin^2(x))- sin^2(x)= 1- 2sin^2(x)[/itex]. Set cos(2x) equal to [itex]\sqrt{3}/2[/itex] and solve for sin(x) and cos(x). Of course, you take the positive root.

    I am not clear why you would question your reasoning.
    Last edited by a moderator: May 11, 2014
  4. May 11, 2014 #3


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    That looks correct. Did you get exact radical values? You have$$
    \cos(2x) = \cos^2x -\sin^2 x = 2\cos^2x - 1 = 1-2\sin^2 x$$You are just using the last two equations to solve for ##\cos x## and ##\sin x## in terms of ##\cos(2x)##.
  5. May 11, 2014 #4
    Thanks guys! HallsofIvy its just sometimes when I do these questions I think I'm right and then I only get it partly correct or not correct at all. I was just wanting to be sure as these type of questions may come up in my finals.
  6. May 11, 2014 #5
    @teme92. You have to get into the habit of not doubting your reasoning. Yes, self-criticism will help you improve your abilities at problem solving, but when it gets in the way of you being confident in your answers it can be a problem.
  7. May 12, 2014 #6
    Hey xiavatar, when it comes down to exams I will go with my instincts unquestionably but I just want to be safe in the run up to them. As you said, self-criticism has improved my understanding of a lot of topics in mathematics so I'd prefer to be safe than sorry in this instant :)
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