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Trigonometric identity from Euler's intro to analysis of infinite

  1. May 2, 2013 #1
    So I'm trying to get through euler's introduction to the analysis of the infinite so I could eventually read his books on calculus but I'm stuck somewhere and can't seem to figure out how he equates this identity euler_trig.png

    so by expanding I get sin(2y) * cos(z) + cos(2y) * sin(z).

    I get that the second term of the equation [cos(2y) * sin(z)] gives cos2y - sin2y which gives - sin z if you take the negative out of the factor but I can't seem to figure out how sin(2y) * cos(z) = 2cosy * sin(y + z). I just get 2siny * cosy * cosz and can't seem to do anything else.

    I'm probably missing something terribly obvious like always happens but I'd really appreciate some help I have nobody I can ask in real life and I don't like to just move on because the book gets more complicated. This is on page 104 if anyone cares to know.
    Last edited: May 2, 2013
  2. jcsd
  3. May 2, 2013 #2


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    Left side:$$
    \sin(2y+z) = \sin(2y)\cos z + \cos(2y)\sin z =
    2\sin y\cos y \cos z + (2\cos^2y - 1)\sin z$$
    Right side:$$
    2\cos y(\sin y \cos z + \cos y \sin z)-\sin z=2\cos y \sin y \cos z
    +2\cos^2 y \sin z -\sin z$$Does that do it for you?
  4. May 2, 2013 #3
    Yes, thank you very much, figures I just had to use basic identify [tex]sin^2x = 1 - cos^2x[/tex] and then simply factor to get to euler's answer.

    I definitely should have seen that.
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