Trigonometry problem Okay..this one simple problem but I am really stuck and have no idea how to start..
$\cos(a-b)+\cos(b-c)+\cos(c-a)=-\frac32$ we need to prove $\cos(a)+\cos(b)+\cos(c)=\sin(a)+\sin(b)+\sin(c)=0 $ 
 A: HINT:
$$(\cos A+\cos B+\cos C)^2+(\sin A+\sin B+\sin C)^2$$
$$=3+2\sum\cos(A-B)=0$$
Now use the fact that sum of squares of two real numbers is zero
A: $\cos(a−b)+\cos(b−c)+\cos(c−a)=−3/2$
$\cos a \cos b+\sin a \sin b+\cos b \cos c+\sin b \sin c+\cos c \cos a+\sin c \sin a=-3/2 $
$\frac{1}{2}\cos a(\cos a + \cos b +\cos c)-\frac{1}{2}\cos^2a+\frac{1}{2}\sin a(\sin a+\sin b+\sin c) - \frac{1}{2}\sin^2 a+
\frac{1}{2}\cos b(\cos a + \cos b +\cos c)-\frac{1}{2}\cos^2b+\frac{1}{2}\sin b(\sin a+\sin b+\sin c) - \frac{1}{2}\sin^2 b+
\frac{1}{2}\cos c(\cos a + \cos b +\cos c)-\frac{1}{2}\cos^2c+\frac{1}{2}\sin c(\sin a+\sin b+\sin c) - \frac{1}{2}\sin^2 c=-\frac{3}{2}$
$\cos a(\cos a + \cos b +\cos c)+\sin a(\sin a+\sin b+\sin c) - (sin^2 a+\cos^2a)+
\cos b(\cos a + \cos b +\cos c)+\sin b(\sin a+\sin b+\sin c) - (sin^2 b+\cos^2b)+
\cos c(\cos a + \cos b +\cos c)+\sin a(\sin a+\sin b+\sin c) - (sin^2 c+\cos^2c)=-3$
$(\cos a + \cos b +\cos c)^2+(\sin a+\sin b+\sin c)^2=0$
then $\cos a + \cos b +\cos c=0$ and $\sin a+\sin b+\sin c=0$
A: This is a detour to the solution compared to my original solution
Like Prove that in any triangle $ABC$, $\cos^2A+\cos^2B+\cos^2C\geq\frac{3}{4}$,
or  In a triangle, find the minimum and maximum of $\cos(A-B)\cos(B-C)\cos(C-A)$
let $2x=A-B$ etc.  $\implies x+y+z=0$
Now $$-\dfrac32=\cos2x+\cos2y+\cos2z=2\cos(x-y)\cos(x+y)+2\cos^2z-1$$
As $x+y=-z,$
$$-\dfrac32=2\cos(x-y)\cos z+2\cos^2z-1\iff2\cos^2z+2\cos(x-y)\cos z+\dfrac12=0$$
which is a quadratic equation $\cos z$
As $\cos z$ is real, the discriminant must be $\ge0$
i.e.,  $(2\cos(x-y))^2-4\cdot2\cdot\dfrac12\ge0\iff\sin^2(x-y)\le0\implies\sin(x-y)=0$
Consequently, $\cos z=\dfrac{-2\cos(x-y)}4=\mp\dfrac12\implies\cos2z=2\cos^2z-1=-\dfrac12$
Using Clarification regarding a question,
we can say the angles namely, $A,B,C$ have to differ by $\dfrac{2\pi}3\pmod{2\pi}$
WLOG $A-B=\dfrac{2\pi}3, B-C=\dfrac{2\pi}3, A-C=\dfrac{4\pi}3$
Now $\sin A+\sin B=\sin\left(C+\dfrac{4\pi}3\right)+\sin\left(C+\dfrac{2\pi}3\right)=2\sin(\pi+C)\cos\dfrac\pi3=-\sin C$
$\implies\sin A+\sin B+\sin C=0$
Similarly, for cosines.
