# How to find sin(ax)

1. Nov 10, 2004

### 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?

2. Nov 10, 2004

### 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:

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

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

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

3. Nov 10, 2004

### Astronuc

Staff Emeritus
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
4. Nov 10, 2004

### 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.

5. Nov 11, 2004

### 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:

$$y=v$$

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

### 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.