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How to find sin(ax)

  1. Nov 10, 2004 #1
    What is the standard procedure if one wants to find an expression for for example sin(4x)? Is this procedure the same also for cos and tan? :smile:
  2. jcsd
  3. Nov 10, 2004 #2


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    I'm not entirely sure if there is a general formula for all [tex] \sin {(xa)} [/tex]
    or not, but I do know there are formulas for all the double angle varieties:

    \sin{(2a)} = 2\sin{(a)}\cos{(a)}

    \cos{(2a)} = 1 - 2\sin^2{(a)}

  4. Nov 10, 2004 #3


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    Starting with trigonomtric identities:

    sin(x+y) = sin x cos y + sin y cos x

    cos(x+y) = cos x cos y - sin x sin y

    tan(x+y) = ( tan x + tan y )/( 1 - tan x tan y )

    One can generalize by letting x = mx and y = ny or nx.

    Then one can find recursion relationship.

    Hint: nx = (n-1)x + x
    Last edited: Nov 10, 2004
  5. Nov 10, 2004 #4


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    if you know de Moivre's rule:

    [tex]e^{i \theta} = cos(\theta) + i sin(\theta) [/tex]

    Then you get the identities you want like this:

    [tex] (e^{i \theta})^4 = e^{4 i \theta} [/tex]

    [tex] [cos(\theta) + i sin(\theta)]^4 = cos(4 \theta) + i sin(4 \theta) [/tex]

    If you multiply out the left side, then you can equate the real and imaginary terms, since sin and cos are pure real here.

    youll get something like:

    [tex] cos(4 \theta) = cos^4(\theta) - 6 sin^2(\theta)cos^2(\theta) + sin^4(\theta) [/tex]

    and a similar expression for sin. tan is just sin/cos. i just did this in my head, so you should probably check to make sure its right.
  6. Nov 11, 2004 #5


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    Extending from what StatusX just said:

    Taking the imaginary parts of both sides:

    [tex](\cos \theta + i \sin \theta)^a = \cos a \theta + i \sin a \theta [/tex]

    We get:

    [tex]\sin a \theta = \text{Im} \left( \sum_{k=0}^a \left( \begin{array}{c}a \\k \end{array} \right) \left( \cos \theta \right)^k \left( i \sin \theta \right)^{a-k} \right)[/tex]

    Therefore when a is odd:

    [tex]\sin a \theta = \text{Im} \left( (i\sin \theta)^a + \frac{a}{(a-2)!2!} (i\sin \theta)^{a-2} \cos^2 \theta + \frac{a}{(a-4)!4!} (i\sin \theta)^{a-4} \cos^4 \theta + \ldots \right)[/tex]

    When a is even:

    [tex]\sin a \theta = \text{Im} \left( \frac{a!}{(a-1)!1!} (i\sin \theta)^{a-1} \cos \theta + \frac{a!}{(a-3)!3!} (i\sin \theta)^{a-3} \cos^3 \theta + \ldots \right)[/tex]

    If you have time and patience you can rearrange the equation for the cosine(ax) function and express purely in terms of cosine(x) by using the simple identity:

    [tex]\cos^2 \theta + \sin^2 \theta \equiv 1[/tex]

    I always find formulas like this give you some appreciation for the very simple and powerful fact that if:

    [tex]x + iy = u + iv[/tex]




    Last edited: Nov 12, 2004
  7. Nov 12, 2004 #6


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    Sorry made a big mistake in the post above, edited it out now. Also note you can remove the i's from the above equations by looking at a=4n, a=4n+1, a=4n+2 and a=4n+3.
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