How to find sin(ax)

[TSN79]

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:

[kreil]

I'm not entirely sure if there is a general formula for all \sin {(xa)}
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)}



\tan{(2a)}=\frac{2\tan{(a)}}{1-\tan^2{(a)}}

[Astronuc]

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

[StatusX]

if you know de Moivre's rule:

e^{i \theta} = cos(\theta) + i sin(\theta)

Then you get the identities you want like this:

(e^{i \theta})^4 = e^{4 i \theta}

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

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:

cos(4 \theta) = cos^4(\theta) - 6 sin^2(\theta)cos^2(\theta) + sin^4(\theta)

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.

[Zurtex]

Extending from what StatusX just said:

Taking the imaginary parts of both sides:

(\cos \theta + i \sin \theta)^a = \cos a \theta + i \sin a \theta

We get:

\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)

Therefore when a is odd:

\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)

When a is even:

\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)


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:

\cos^2 \theta + \sin^2 \theta \equiv 1

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

x + iy = u + iv

then:

x=u

and:

y=v

[Zurtex]

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.