Question

Show that $\cos (B-C)+\cos (C-A)+\cos (A-B)=-\frac{3}{2}$ if and only if $\cos A+\cos B+\cos C=0$ and $\sin A+\sin B+\sin C=0$

Answer 1

Given :

$\Rightarrow$$\cos (B-C)+\cos (C-A)+\cos (A-B)=-\frac{3}{2}$

$\Rightarrow$$(\cos B\cos C+\sin B\sin C)+(\cos C\cos A+\sin C\sin A)+(\cos A\cos B+\sin A\sin B)=-\frac{3}{2}$

$\Rightarrow$$2(\cos B\cos C+\sin B\sin C)+2(\cos C\cos A+\sin C\sin A)+2(\cos A\cos B+\sin A\sin B)+3=0\to 1$

We know that ${\sin }^{2}A+{\cos }^{2}A=1$

$\Rightarrow$${\sin }^{2}B+{\cos }^{2}B=1$ & ${\sin }^{2}C+{\cos }^{2}C=1$

$\therefore {\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C+{\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C=3$

$1\Rightarrow \therefore {\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C+2\sin A\sin B+2\sin B\sin C+2\sin C\sin A+{\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C+$
$2\cos A\cos B+2\cos B\cos C+2\cos C\cos A=0$

$\Rightarrow$${(\sin A+\sin B+\sin C)}^2+{(\cos A+\cos B+\cos C)}^2=0$

If sum of the positive quantities$=0$. Then both must be zero $\left(0\right)$

$\therefore \sin A+\sin B+\sin C=0$ & $\cos A+\cos B+\cos C=0$

$\Rightarrow$If we square and add $(\sin A+\sin B+\sin C=0)$ & $(\cos A+\cos B+\cos C=0)$

$\Rightarrow$Then we get $\cos (B-C)+\cos (C-A)+\cos (A-B)=-\frac{3}{2}$