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If $x,y,z \in\mathbb{R}$ and $$\frac{\cos x+\cos y+\cos z}{\cos \left(x+y+z\right)}=\frac{\sin x+\sin y+\sin z}{\sin \left(x+y+z\right)}=k$$ Find the value of $\cos(x+y)+\cos(y+z)+\cos(z+x)$.

I was working through this problem and found an interesting solution to it. How would you solve it?

Edit: I found the same question here and here. So, this should be marked as duplicate.

V.G
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Rewrite the equations as $$\cos x +\cos y+\cos z =k\cos(x+y+z) \\ \sin x+\sin y+\sin z=k\sin(x+y+z)$$

Multiply the second equation by $i$ (where $i=\sqrt{-1}$) , and add both the equations. Using Euler's formula, we are left with a single equation, that is,$$e^{ix}+e^{iy}+e^{iz}=ke^{i(x+y+z)}$$ Now, we have to find $\cos(x+y)+\cos(y+z)+\cos(z+x)=\Re\left(e^{i(x+y)}+e^{i(y+z)}+e^{i(z+x)}\right)$.

Now, we let $\alpha=e^{ix},\beta=e^{iy},\gamma=e^{iz}$, so that we have $\alpha+\beta+\gamma=k\alpha \beta\gamma $ and we need to find $\alpha\beta+\beta\gamma+\gamma\alpha$.

Dividing by $\alpha\beta\gamma$ in the last equation results in $(\alpha\beta)^{-1}+(\beta\gamma)^{-1}+(\gamma\alpha)^{-1}=k$.

But notice that $\Re(\alpha\beta)=\Re\left((\alpha\beta)^{-1}\right)$ since cosine is an even function and hence $\Re\left((\alpha\beta)^{-1}+(\beta\gamma)^{-1}+(\gamma\alpha)^{-1}\right)=\Re(\alpha\beta+\beta\gamma+\gamma\alpha)=k$.

If instead of real part, we take the imaginary parts, we get $\Im\left((\alpha\beta)^{-1}+(\beta\gamma)^{-1}+(\gamma\alpha)^{-1}\right)=-\Im\left(\alpha\beta+\beta\gamma+\gamma\alpha\right) \implies \sin(x+y)+\sin(y+z)+\sin(z+x)=0$.

V.G
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    Usually you should wait for some time before even posting the solution give time for someone to comeup with an answer if not post the answer – Albus Dumbledore Jan 10 '21 at 14:57
  • I warned in the question itself to give it a try yourself.. I can hide the solution if you want.. – V.G Jan 10 '21 at 14:59
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    @AlbusDumbledore: Self-answered questions are welcome and encouraged: https://math.stackexchange.com/help/self-answer – Martin R Jan 10 '21 at 15:00
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    @MartinR please dont mistake me that i am discouraging this ,My only point was the OP might want to give more time for users to comeup with solutions before self answering,it is hard to just concentrate in the question without your eyes drifting to the solution especially when the third line (using complex number) is a give away – Albus Dumbledore Jan 10 '21 at 15:03
  • @AlbusDumbledore, see if this helps, although it is spoiled for you already :(. – V.G Jan 10 '21 at 15:05
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    @ Albus Dombledore true. He's hidden it now. Also, there are likely other solutions... – Adam Rubinson Jan 10 '21 at 15:05
  • @MartinR OK ,I didnt think of that, all the same now that the OP has put spoilers I guess it wont spoil the fun, I apologise both to the OP and you if my first comment was discouraging in tone,but i swear that was not my intention – Albus Dumbledore Jan 10 '21 at 15:12
  • @LightYagami It actually didnt spoil Why? i have seen a very similar problem in some book and it used complex numbers :) Any way thank you for replying and adding the spoilers.Bye – Albus Dumbledore Jan 10 '21 at 15:15
  • @AlbusDumbledore: Oh, I thought you read the solution and wanted to think yourself first, so you were angry. – V.G Jan 10 '21 at 15:38
  • @LightYagami angry? not at all, it is hard to decipher the tone in comments so i can understand your misunderstanding ,i only said that because it is difficult to avert your eyes from a solution just below a question.I hope you are not angry with me :) :) – Albus Dumbledore Jan 10 '21 at 15:42
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    @AlbusDumbledore: Of course not :). Anyways, do you have any other ideas for the problem? – V.G Jan 10 '21 at 15:43
  • @LightYagami spent half an hour but it is getting very complicated ?I guess complex numbers is the way to go, will let you know if there is something different, – Albus Dumbledore Jan 10 '21 at 15:45
  • @LightYagami I did componendo diivdendo and used the results $$\sin x+\sin y+\sin z+\sin(x+y+z)=4\prod \sin((x+y)/2)$$ and $$\cos x+\cos y+\cos z+\cos (x+y+z)=4\prod cos((x+y)/2)$$ Do you see something nice ? not me! – Albus Dumbledore Jan 10 '21 at 15:53
  • @AlbusDumbledore: As of now, I too cannot see anything useful from this, but I will keep trying, maybe something emerges. – V.G Jan 10 '21 at 15:57
  • @AlbusDumbledore: Check out the links that I added. – V.G Jan 10 '21 at 16:41
  • @LightYagami wow! nice find Math 110 's solutions is just awesome!thanks for notifying me, hope to meet you again! – Albus Dumbledore Jan 10 '21 at 16:47