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Sine/cosine sum and difference formulas

  1. Dec 4, 2005 #1
    How to proove those formulas for any angle? So far all the proofs I've found are for angles between 0 and 2pi...
     
  2. jcsd
  3. Dec 4, 2005 #2
    Surely you know that for any angle less than 0 or greater than 2pi there is a corresponsing angle between 0 and 2pi...

    If the angle x is less than 0, Use 2pi-x.
    If the angle x is greater than 2pi, Use 0+x
     
  4. Dec 4, 2005 #3
    ...which ones did you find?
     
  5. Dec 4, 2005 #4
    Sorry, i meant pi/2... (thus maximum of pi for the resultant angle)

    I found one with euler's equation, another with triangles and the other with the trigonometric circle.

    My solution was;

    [tex]\sqrt{(\cos({{\theta} \pm {\sigma}}) - \cos{\theta})^{2} + (\sin({{\theta} \pm {\sigma}}) - \sin{\theta})^{2}} = \sqrt {({cos{\sigma} - 1})^{2} - \sin^{2}{\sigma}} [/tex]

    And this later simplifies to

    [tex]cos({{\theta} \pm {\sigma}}) = \cos{\theta}\cos{\sigma} \pm \sin{\theta}\sin{\sigma}[/tex]

    But now I can't go further...
     
    Last edited: Dec 4, 2005
  6. Dec 4, 2005 #5
    Alright I think I thought of a good proof

    With this identity (this is using the distance formula and the coordinates on the trigonmetric circle)

    [tex]\sqrt{(\cos({{\theta} \pm {\sigma}}) - \cos{\theta})^{2} + (\sin({{\theta} \pm {\sigma}}) - \sin{\theta})^{2}} = \sqrt {({cos{\sigma} - 1})^{2} - \sin^{2}{\sigma}} [/tex]

    We can derive

    [tex]\sin(\theta \pm \sigma) = \cos \sigma \sin \theta \pm \cos \theta \sin \sigma [/tex]

    With this identity

    [tex]\sin -a = -\sin a [/tex]

    We can compare [tex]\sin(\theta - \sigma) [/tex] and [tex]\sin(\sigma - \theta) [/tex] As being opposite.

    [tex](\sin(\theta \pm \sigma) = \cos \sigma \sin \theta \pm \cos \theta \sin \sigma) =-(\sin(\sigma \pm \theta) = \cos \theta \sin \sigma \pm \cos \sigma \sin \theta) [/tex]

    The only possible solution is

    [tex]\sin(\theta - \sigma) = \cos \sigma \sin \theta - \cos \theta \sin \sigma [/tex]

    and

    [tex]\sin(\sigma - \theta) = \cos \theta \sin \sigma - \cos \sigma \sin \theta[/tex]


    And since sin (a+b) is not equal to sin(a-b), exept if one of the angle is pi, thus

    [tex]\sin(\theta + \sigma) = \cos \sigma \sin \theta + \cos \theta \sin \sigma [/tex]

    The result stays the same if one of the angle is pi/2.

    The expansion for cos(a+b) and cos (a-b) is easy to derive once we have established the expansion for sin.

    Q.E.D.?
     
    Last edited: Dec 5, 2005
  7. Dec 4, 2005 #6

    matt grime

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    Draw the sin graph, notice the symmetry, all's well.
     
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