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

**tlub77987** · Nov23-09, 11:20 AM

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?

**tauon** · Nov23-09, 02:51 PM

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

$LaTeX Code: \\sin(A+B)=\\sin A \\cos B + \\cos A \\sin B$

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

so he possibly assumed

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

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?

**Mark44** · Nov23-09, 03:53 PM

Originally Posted by 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?

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.

**tauon** · Nov23-09, 04:13 PM

Originally Posted by Mark44

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.

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

**Mark44** · Nov23-09, 04:50 PM

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

**Borek** · Nov23-09, 06:05 PM

π/4 & 3π/2

**tauon** · Nov23-09, 06:35 PM

Originally Posted by Mark44

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

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.

Originally Posted by Borek

π/4 & 3π/2

lol

nice one.

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

**Gerenuk** · Nov24-09, 12:12 AM

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)
$LaTeX Code: a=\\arccos t+\\arccos\\left(t-\\frac{1}{2t}\\right)$
$LaTeX Code: b=\\arccos t-\\arccos\\left(t-\\frac{1}{2t}\\right)$
$LaTeX Code: \\frac{\\sqrt{3}-1}{2}<|t|<1$
which is a continous set of solutions.

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

or