Assume that $\{\alpha, \beta, \gamma\} \subset \left[0,\frac{\pi}{2}\right]$, $\sin\alpha+\sin\gamma=\sin\beta$ and $\cos\beta+\cos\gamma=\cos\alpha$.

Try to find a value of $\alpha-\beta$.

Actually I have gotten that $\alpha+2\gamma+\beta=\pi$ and $\sin2\alpha+\sin2\beta=\sin(\alpha+\beta)$, $\cos2\alpha+\cos2\beta=\cos(\alpha+\beta)$

But I can't get further more.

  • $\begingroup$ I fixed the MathJax for the functions. $\endgroup$ – Claude Leibovici Sep 9 '17 at 8:57

We have $$\sin 2\alpha+\sin2\beta=\sin(\alpha+\beta)$$ and $$\cos2\alpha+\cos2\beta=\cos(\alpha+\beta)$$ So by squaring and then adding the above equations, we get $$(\sin2\alpha+\sin2\beta)^2+(\cos2\alpha+\cos2\beta)^2=\sin^2(\alpha+\beta)+\cos^2(\alpha+\beta)$$ $$\Rightarrow \sin^22\alpha+\sin^22\beta+2\sin2\alpha\sin2\beta+\cos^22\alpha+\cos^22\beta+2\cos2\alpha\cos2\beta=1$$ $$\Rightarrow2\sin2\alpha\sin2\beta+2\cos2\alpha\cos2\beta+2=1$$ $$\Rightarrow\sin2\alpha\sin2\beta+\cos2\alpha\cos2\beta=-\frac12$$ Now we know that $\cos x\cos y+\sin x\sin y=\cos(x-y)$. So $$\cos(2\alpha-2\beta)=-\frac12$$ $$\Rightarrow2\alpha-2\beta=\frac{2\pi}{3}$$ $$\alpha -\beta=\frac\pi3\ rad=60^\circ$$


we have $$\sin(\gamma)=\sin(\beta)-\sin(\alpha)$$ and $$\cos(\gamma)=\cos(\alpha)-\cos(\beta)$$ squaring and adding both we get $$-1=-2(\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta))$$ therefore we obtain $$\cos(\alpha-\beta)=\frac{1}{2}$$

  • $\begingroup$ So it looks like that the describe of the problem is not right. $\endgroup$ – Intoks Liobein Sep 9 '17 at 9:08
  • $\begingroup$ why that? you can compute $$\alpha-\beta$$ now from the last line $\endgroup$ – Dr. Sonnhard Graubner Sep 9 '17 at 9:10
  • $\begingroup$ But 2\pi/3 is larger than \pi/2 $\endgroup$ – Intoks Liobein Sep 9 '17 at 9:11
  • $\begingroup$ on moment i will look at my calculations $\endgroup$ – Dr. Sonnhard Graubner Sep 9 '17 at 9:19
  • $\begingroup$ sorry i forgot the minus sign $\endgroup$ – Dr. Sonnhard Graubner Sep 9 '17 at 9:23

$$1=\sin^2\gamma+\cos^2\gamma=(\sin\beta-\sin\alpha)^2+(\cos\alpha-\cos\beta)^2=2-2\cos(\alpha-\beta).$$ Thus, $$\cos(\alpha-\beta)=\frac{1}{2}.$$ In another hand, $$\sin\gamma=\sin\beta-\sin\alpha\geq0,$$ which says $\beta\geq\alpha.$

Id est, $\alpha-\beta=-60^{\circ}$ and we are done!

  • $\begingroup$ @Intoks Liobein Why did you accepted a wrong solution? Maybe have you a question about my solution? I am ready to explain. $\endgroup$ – Michael Rozenberg Oct 10 '17 at 7:31

If you have $$\sin2\alpha+\sin2\beta=\sin(\alpha+\beta) \to 2\sin(\alpha+\beta)\sin(\alpha-\beta)=\sin(\alpha+\beta)$$and $$\cos2\alpha+\cos2\beta=\cos(\alpha+\beta)\to \cos(\alpha+\beta)\cos(\alpha-\beta)=\cos(\alpha+\beta) $$divide them $$\frac{2\sin(\alpha+\beta)\sin(\alpha-\beta)}{2\cos(\alpha+\beta)\cos(\alpha-\beta)}=\frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}\\\to \tan(\alpha-\beta)=1 \to \alpha-\beta=\frac{\pi}{4}$$


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