# 2 Proofs

1. Jan 25, 2005

### Ryoukomaru

How can you prove:
$$cos(A+B)=cosAcosB-sinAsinB$$
and similarly
$$sin(A+B)=sinAcosB+cosAsinB$$
If the proofs arent very complicated, I would appreciate you giving me hints first so maybe i can work them out on my own.

Btw i am looking for an algebraic proof, not one with graphs & triangles.

2. Jan 25, 2005

### arildno

You'd better stick with a perfectly good graphical proof.
The simplest "algebraical" proof involves the complex exponential.

3. Jan 25, 2005

### dextercioby

I think the graphical proof is much more elegant than the one involving Euler's formula...

Geometry is elegant...

Daniel.

4. Jan 25, 2005

### arildno

I agree; I should have written "perfectly better" rather than "perfectly good"..

5. Jan 25, 2005

### Ryoukomaru

woohh there is a "hint" now. :tongue:

6. Jan 25, 2005

### Kahsi

http://home.tiscali.se/21355861/bilder/proof.PNG [Broken]

$$(PQ)^2=(cosu-cosv)^2+(sinu-sinv)^2$$

and with cosine
$$(PQ)^2=1^2+1^2-2*1*1*cos(u-v)$$

we get

$$(cosu-cosv)^2+(sinu-sinv)^2=1^2+1^2-2*1*1*cos(u-v)$$

$$cos^2u-2cosu*cosv+cos^2v+sin^2u-2sinu*sinv+sin^2v=2-2cos(u-v)$$

You know that $$1 - cos^2x = sin^2x$$ and that $$1-sin^2x=cos^2x$$

$$1-2cosu*cosv+1-2sinu*sinv=2-2cos(u-v)$$

$$2cos(u-v)=2cosu*cosv + 2sinu*sinv$$

$$cos(u-v)=cosu*cosv + sinu*sinv$$

Last edited by a moderator: May 1, 2017
7. Jan 25, 2005

### dextercioby

Believe me,there's a much more elegant way of doing it geometrically...Anyways,he called for an algebraic proof...Without drawigs and angles...

Daniel.

8. Jan 25, 2005

### HallsofIvy

You CAN define cosine and sine by

1. y(x)= sin(x) is the function satisfying the differential equation d<sup>2</sup>y/dx<sup>2</sup>= -y and y(0)= 0, y'(0)= 1.

2. y(x)= cos(x) is the function satisfying the differential equation d<sup>2</sup>y/dx<sup>2</sup> = -y and y(0)= 1, y'(0)= 0.
It's easy to show that sin(x) and cos(x) are independent solutions so any solution to that equation can be written as C1cos(x)+ C2sin(x). In fact, if y satisfies y"= y, y(0)= A, y'(0)= B, then y(x)= Acos(x)+ B sin(x).

Let y= (cos(x))' (the derivative of cosine). Since cosine satisfies a second order equation, it is twice differentiable and y'= (cos(x))"= -cos(x). That means that y is twice differentiable and y"= -(cos(x))'= -y. y(0)= 0 since the derivative of cosine at 0 is 0) and y'(0)= -cos(0)= -1. Thus, y= 0cos(x)+(-1)sin(x)= -sin(x). Similarly, one can prove that (sin(x))'= cos(x).

Now, let y= sin(x+a). Then y'= cos(x+a) and y"= -sin(x+a)= -y. That is, y also satisfies y"= -y. y(0)= sin(a), y'(0)= cos(a) so y(x)= sin(x+a)= sin(a)cos(x)+ cos(a)sin(x). Let x= b and we have sin(a+b)= sin(a)cos(b)+ cos(a)sin(b).

Let y= cos(x+a). Then y'= -sin(x+a) and y"= -cos(x+a)= -y. This y also satisfies y"= -y. y(0)= cos(a), y"(0)= -sin(a) so y(x)= cos(a)cos(x)- sin(a)sin(x). Let x= b and we have cos(a+b)= cos(a)cos(b)- sin(a)sin(b).

Those are the simplest proofs I know.

9. Jan 26, 2005

### arildno

Halls:
You are, of course, right.
However, I'm not quite sure how we might prove the 2(pi)-periodicity of cos&sine if we define these functions (such a proof ought necessarily exist).
I might be dense, but I would appreciate if you could sketch how to prove the periodicity of the two functions.

Last edited: Jan 26, 2005
10. Jan 26, 2005

### dextercioby

Defining $\sin 2\pi=0$ and $\cos 2\pi=1$,he can use the formula he just proved
$$\sin (a+2\pi)=...=\sin a$$

He woul then go on and associate a geometrical interpretation (using the trigonometric circle) and that would be it...

Daniel.

11. Jan 26, 2005

### arildno

And HOW do you legitimize that move??
Why does the diff. eq. definitions accept that?

You have basically placed more restraints upon a second-order differential equation solution than those restraints needed for a unique solution.

Last edited: Jan 26, 2005
12. Jan 26, 2005

### dextercioby

What does that have to do with the diff.eq???I'm discussing the functions that came out as a basis that span the space of solutions...
I'm not putting those constraints in the beginning,before solving the diff.eq.,but afterwards,that is FOR MY PURPOSE I AM SELECTING THE PERIODIC SOLUTIONS...

I'm not affecting the diff.eq. in no way...

Daniel.

13. Jan 26, 2005

### CrankFan

Apostol proves those identities from axioms that he calls fundamental properties of sine and cosine:

1. sine and cosine are defined everywhere on the real line.
2. cos 0 = sin 1/2pi = 1 and cos pi = -1
3. cos (y - x) = cos y cos x + sin y sin x
4. for 0 < x < pi, 0 < cos x < sin x / x < 1 / cos x

I thought it was worth mention since you seemed interested in a deductive proof; if it isn't what you had in mind, ignore it.

14. Jan 26, 2005

### arildno

Sure you are.
1) HallsofIvy's UNIQUE solutions are found by placing the demands:
sin(0)=1, sin'(0)=1, cos(0)=1, cos'(0)=0
You haven't got any solution space to wiggle in here.
It is these basis solutions you need to show are periodic.

You have a solution space of the homogenous diff.eq; those constraints pick out which unique solution you're after.

15. Jan 26, 2005

### HallsofIvy

Arildno, I wrote a paper on defining sine and cosine in terms of an initial value problem in which I proved the periodicity.

Here's a link to an e-copy:

Last edited by a moderator: May 1, 2017
16. Jan 26, 2005

### dextercioby

Nice work,Halls!And useful,too.

Daniel.

17. Jan 26, 2005

### arildno

Last edited by a moderator: May 1, 2017
18. Jan 28, 2005

### Ryoukomaru

Well today, I learned that you can prove the identities by tranformation Matrices. Its a neat way of proving them without any drawings.

19. Jan 28, 2005

### HallsofIvy

Good.

What tranformation Matrices are you talking about. I learned "rotation matrices" by using the sum formulas but I imagine it could be done the other way around!

20. Jan 28, 2005

### dextercioby

The use of "sine" and "cosine" to parametrize finite angle rotations round an axis is dependent of the way the functions are defined...
And viceversa...You can define the (circular) trigonometrical functions using finite angle rotations. (actually the SO(2) group (it's the group axiom regarding matrix multiplication that proves addition formula)).
So it's an equivalence.It woudln't be fair if u said A->B and forget about B->A...

Daniel.