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If $$cos(A-B)+cos(B-C)+cos(C-A)=\frac {-3}{2}$$ prove that $$cosA+cosB+cosC=sinA+sinB+sinC=0$$

My Attempt:

$$cos(A-B)+cos(B-C)+cos(C-A)=\frac {-3}{2}$$ $$cosA \cdot cosB+sinA \cdot sinB+cosB \cdot cosC+sinB \cdot sinC+cosC \cdot cosA+sinC \cdot sinA=\frac {-3}{2}$$.

How should I go further? Please help me with a simple method.

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    First thing I thought of: with the sum-to-product formula $\cos(a) + \cos(b) = 2 \cos(\frac{a+b}{2})\cos(\frac{a-b}{2})$, we have $$ 2 \left[ \cos(A - B) + \cos(B - C) + \cos(C - A) \right] = \\ [\cos(A - B) + \cos(B - C)]+ [\cos(B - C) + \cos(C - A)] + [\cos(C - A) + \cos(A - B)] =\\ 2\left[\cos[(C - A)/2]\cos(B) + \cos[(B - A)/2]\cos(C) + \cos[(C-B)/2]\cos(A) \right] $$ I don't know if that helps, though – 2017-02-06
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    $A, B, C$ are the angles of a triangle? – 2017-02-06
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    @ Omnomnomnom, I could not understand. Please elaborate a bit more. – 2017-02-06
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    @Eugen Covaci, Not specified in the question! – 2017-02-06
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    What is up with the title. what goes after that first equals sign? – 2017-02-06
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    @The Count, Please check the edited version. – 2017-02-06

1 Answers 1

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Continuing from where we left off, we have that

$$ 2(\cos a \cos b + \sin a \sin b + \cos b \cos c + \sin b \sin c + \cos c \cos a + \sin c \sin a) + 3 = 0. $$

Now we write $3$ as $(\cos^2 a + \sin^2 a) + (\cos^2 b + \sin^2 b) + (\cos^2 c + \sin^2 c) $ and substitute this in. We then make use of the identity

$$ (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) $$

to obtain that

$$ (\cos a + \cos b + \cos c)^2 + (\sin a + \sin b + \sin c)^2 = 0 $$

from which the claim follows easily.

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    I used a similar technique as Dylan's above but he got in first :-) I also looked at this geometrically. There is a similar claim for the complex third roots of unity and the complex third roots of z. The result will be true if A = x , B = x + $2 \pi /3$, and C = x - $2 \pi /3$ or (coterminal) C = x + $4 \pi/3$, so that A - B = $-2 \pi /3$, B - C = $4 \pi/3$ – 2017-02-07
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    Re-try I used a similar technique as Dylan's above but he got in first :-) I also looked at this geometrically. There is a similar claim for the complex third roots of unity and the complex third roots of z. The above claimswill be true if A = x + $2 \pi /3$ , B = $x $, and C = x - $2 \pi /3$ or (coterminal) C = x + $4 \pi/3$, so that A - B = $2 \pi /3$, B - C = $2 \pi/3$, and C - A = $-4\pi /3$ or $2 \pi /3$ (Think I got the signs right this time.) Question -- is there any other solution? – 2017-02-07
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    @victoria Those are indeed the only solutions. (Up to adding multiples of $2 \pi$.) – 2017-02-12
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    Thank you. Is there any not-too-advanced way to prove uniqueness? – 2017-02-13
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    Yes. Let $x = e^{i A}, y = e^{iB}$, and $z = e^{iC}$. Then the answer shows that $x + y + z = 0$, and so $x + y$ lies on the unit circle. (Since it is equal to $-z$.) Let $\theta$ be the angle between the segment from $0$ to $x$ and the segment from $0$ to $y$. Then $\theta$ is also the angle formed by the segment from $x$ to $x + y$, and the segment from $y$ to $x +y$, and is twice the angle formed by the segments from $x$ to $z$ and from $y$ to $z$. We get that $\theta + \theta/2 = \pi$, and so $\theta = \pi/3$. (It would be a lot clearer with a picture.) – 2017-02-13