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Cos²α+cos²β+cos²γ = 1

  1. Mar 19, 2012 #1

    RK7

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    On pg 6 of http://www.scribd.com/doc/3914281/Crystal-Structure, it quotes this result without proof. My notes from uni also quote this result but I can't see where it comes from. Does anyone know? Thanks
     
    Last edited: Mar 19, 2012
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  3. Mar 19, 2012 #2

    mathman

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    Re: cos²α+cos²β+cos²γ=1

    It looks like the Pythagorean theorem applied in three dimensions.
     
  4. Mar 19, 2012 #3

    HallsofIvy

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    Re: cos²α+cos²β+cos²γ=1

    Well, it isn't true for just any angle [itex]\alpha[/itex], [itex]\beta[/itex], [itex]\gamma[/itex], of course! If, however, you draw a line through the origin of a three dimensional coordinate system, define [itex]\alpha[/itex] to be the angle the line makes with the x-axis, [itex]\beta[/itex] to be the angle the line makes with the y-axis, and [itex]\gamma[/itex] to be the angle the line makes with the z-axis, then this is true.

    To see that, think of a vector of unit length in the direction of that line. If we drop a perpendicular from the tip of the vector to the x-axis, we have a right triangle in which an angle is [itex]\alpha[/itex] and the hypotenuse is 1, the length of the vector. Thus, the projection of the vector on the x-axis, and so the x-component of the vector is [itex]cos(\alpha)[/itex]. Similarly, the y-component of the vector is [itex]cos(\beta)[/itex] and the z-component of the vector is [itex]cos(\gamma)[/itex]. That is, [itex]cos^2(\alpha)+ cos^2(\beta)+ cos^2(\gamma)[/itex] is the square of the length of the vector which is, of course, 1.

    By the way, look what happens if you do this in two dimensions. If you have a line in the plane through the origin making angles [itex]\alpha[/itex] with the x-axis and [itex]\beta[/itex] with the y- axis then [itex]\beta= \pi/2- \alpha[/itex] so [itex]cos^2(\alpha)+ cos^2(\beta)= cos^2(\alpha)+ cos^2(\pi/2- \alpha)= cos^2(\alpha)+ sin^2(\alpha)= 1[/itex].
     
  5. Mar 19, 2012 #4

    RK7

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    Re: cos²α+cos²β+cos²γ=1

    That was embarrassing.. thanks
     
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