Trigonometry problem

Question

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 $

Answer 1

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

Answer 2

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

Answer 3

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.