# Sine/cosine sum and difference formulas

1. Dec 4, 2005

### Werg22

How to proove those formulas for any angle? So far all the proofs I've found are for angles between 0 and 2pi...

2. Dec 4, 2005

### Cincinnatus

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

3. Dec 4, 2005

### amcavoy

...which ones did you find?

4. Dec 4, 2005

### Werg22

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;

$$\sqrt{(\cos({{\theta} \pm {\sigma}}) - \cos{\theta})^{2} + (\sin({{\theta} \pm {\sigma}}) - \sin{\theta})^{2}} = \sqrt {({cos{\sigma} - 1})^{2} - \sin^{2}{\sigma}}$$

And this later simplifies to

$$cos({{\theta} \pm {\sigma}}) = \cos{\theta}\cos{\sigma} \pm \sin{\theta}\sin{\sigma}$$

But now I can't go further...

Last edited: Dec 4, 2005
5. Dec 4, 2005

### Werg22

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)

$$\sqrt{(\cos({{\theta} \pm {\sigma}}) - \cos{\theta})^{2} + (\sin({{\theta} \pm {\sigma}}) - \sin{\theta})^{2}} = \sqrt {({cos{\sigma} - 1})^{2} - \sin^{2}{\sigma}}$$

We can derive

$$\sin(\theta \pm \sigma) = \cos \sigma \sin \theta \pm \cos \theta \sin \sigma$$

With this identity

$$\sin -a = -\sin a$$

We can compare $$\sin(\theta - \sigma)$$ and $$\sin(\sigma - \theta)$$ As being opposite.

$$(\sin(\theta \pm \sigma) = \cos \sigma \sin \theta \pm \cos \theta \sin \sigma) =-(\sin(\sigma \pm \theta) = \cos \theta \sin \sigma \pm \cos \sigma \sin \theta)$$

The only possible solution is

$$\sin(\theta - \sigma) = \cos \sigma \sin \theta - \cos \theta \sin \sigma$$

and

$$\sin(\sigma - \theta) = \cos \theta \sin \sigma - \cos \sigma \sin \theta$$

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

$$\sin(\theta + \sigma) = \cos \sigma \sin \theta + \cos \theta \sin \sigma$$

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
6. Dec 4, 2005

### matt grime

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