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Trig identity help

  1. May 27, 2013 #1
    Hello there,

    I have a problem i'm hoping someone can help me with. I'm writing a bit of code for computing the value of pi that converges faster than a previous piece that relies on the leibniz series.

    Anyway, i'm struggling with showing how this identity arises. tan(2t) = 2 * tan(t) / 1 - tan2(t)

    So far i've got to this point;

    sin(2θ) = 2sin(θ)cos(θ) and,
    cos(2θ) = cos2(θ) - sin2(θ)

    tan(2θ) = sin(2θ) / cos(2θ)

    = 2sin(θ) cos(θ) / cos2(θ) - sin2(θ)

    From there, I know that this is the step i'm supposed to take but i'm struggling to make sense of it. =[ 2sin(θ)cos(θ) / cos2(θ) ] * 1 / 1 - tan2

    Doing that step brings me back to the identity, but why it works is what I don't understand. I have been told that it is dividing by cos2θ but I must be doing something wrong because it doesn't seem to work out for me. when dividing through by cos2θ I get 2sin(θ)cos(θ) / 1 - tan2θ

    I would really appreciate someone explaining this without making any large jumps in the reasoning. (I shouldn't be doing maths at this time, but it's really bugging me)

    Thanks!
     
  2. jcsd
  3. May 27, 2013 #2

    SteamKing

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    Your algebra is OK to this step:

    tan(2θ) = sin(2θ) / cos(2θ)

    = 2sin(θ) cos(θ) / (cos^2(θ) - sin^2(θ))

    Now, what you need to do is divide both the numerator and the denominator by cos^2(theta) and then simplify.
     
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