Let $A,B,C$ satisfy $0<A<B<C<2\pi$. If $$\cos(x+A)+\cos(x+B)+\cos(x+C)=0$$ for all x $\in \mathbb{R}$, then i have to find possible values of $(C-A)$.
i don't know how to begin
Thanks
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See also: http://math.stackexchange.com/questions/1397066/clarification-regarding-a-question and http://math.stackexchange.com/questions/2132148/if-cosa-bcosb-ccosc-a-frac-32-prove-that-cosacosbcosc-sinas – lab bhattacharjee Feb 07 '17 at 08:15
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We have that $$\cos (x+A) + \cos (x+B) + \cos (x + C)=0$$ $$\Rightarrow \cos x (\cos A + \cos B + \cos C) - \sin x (\sin A + \sin B + \sin C)=0$$
We can conclude thus $$\cos A + \cos B + \cos C =0 \tag {1}$$ $$\sin A + \sin B + \sin C =0 \tag {2}$$
Using $(1)$ and $(2)$, we get, $$(\cos A + \cos C)^2 +(\sin A + \sin C)^2 =(-\cos B)^2 +(-\cos C)^2$$ $$\Rightarrow 2 + 2\cos (A-C)=1$$ $$\Rightarrow \cos (C-A) = -\frac {1}{2} $$
Hope you can take it from here.
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Could you please explain, how can one conclude equations $(1)$ and $(2)$? – Student Feb 07 '17 at 07:42
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@ShreyAryan If both $(1) $ and $(2) $ are true then the equation equals zero for all $x \in \mathbb R $. Otherwise $\cdots $ ? – Feb 07 '17 at 07:43
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