# When does cos(A+B)=cosA +cosB

1. Nov 23, 2009

### tlub77987

I recently read an article in a math magazine where some guy found anlges (A and B) when sin(A+B)=sinA + sinB. I have been trying to work with cos(A+B) and see if there are instances when cos(A+B) would equal cosA+cosB. Any ideas?

2. Nov 23, 2009

### tauon

assuming he didn't do it by brute force I suppose he may have used the formulae

$$\sin(A+B)=\sin A \cos B + \cos A \sin B$$

$$\sin A + \sin B =2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$$

so he possibly assumed

$$\sin A \cos B + \cos A \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$$

and played around with it or other formulae?

but honestly I have no idea how he did it, I didn't really "think"/"work" this over.
btw, can you post a link or something to that article?

Last edited: Nov 23, 2009
3. Nov 23, 2009

### Staff: Mentor

If you set sinAcosB + sinBcosA = sinA + sinB, this will be true if cosB = 1 and cosA = 1, which means that A and B can be 0, pi, 2pi, etc. Any integer multiple of pi works.

It's much harder to find a solution for cos(A + B) = cosA + cosB, since cos(A + B) = cosAcosB - sinAsinB.

4. Nov 23, 2009

### tauon

I think he means angles besides those (which are trivial cases... I don't think such a "discovery" would be that interesting to be published in a mathematics magazine like the OP said)... 0=0 isn't that fascinating...

(btw, $$\cos \pi = -1$$ so only angles of the form $$2k\pi , \ k \in \mathbb{Z}$$ work for this solution)

5. Nov 23, 2009

### Staff: Mentor

$k\pi$, $2k\pi$, what's the difference?

6. Nov 23, 2009

### Staff: Mentor

π/4 & 3π/2

7. Nov 23, 2009

### tauon

you said we must have cosA=cosB=1 , I was replying to that...

but anyway in this case yeah, it doesn't matter since it's 0=0 anyway...

anyway @OP:

if this is not a farce/joke (which I suspect to be the case) I'd like to know about that "result", thanks.

lol :rofl:
nice one.

edit: I just noticed you have a "best humour" badge. indeed, you're good. :surprised

8. Nov 23, 2009

### Gerenuk

Giving one particular solution is useless from a mathematicians point of view. It's important to find all solutions.

I'm not sure what that guy did, but one easily gets
cos(a+b)=cos(a)+cos(b)
$$a=\arccos t+\arccos\left(t-\frac{1}{2t}\right)$$
$$b=\arccos t-\arccos\left(t-\frac{1}{2t}\right)$$
$$\frac{\sqrt{3}-1}{2}<|t|<1$$
which is a continous set of solutions.

And btw, sin(a+b)=sin(a)+sin(b) is solved with
$$a=\pi+t, b=\pi-t$$
or
$$a=t, b=0$$
or
$$a=0, b=t$$

Last edited: Nov 23, 2009