Question

If $\sin A+\sin B+\sin C=0$ and $\cos A+\cos B+\cos C=0,$ and A+B+C+180${}^o$ then $\cos \left(3A\right)+\cos \left(3B\right)+\cos \left(3C\right)=$

• A. $-3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is A $-3$

$(\cos A+\cos B+\cos C)+i(\sin A+\sin B+\sin C)=0$

$\Rightarrow \cos A+i\sin A+\cos B+i\sin B+\cos C+i\sin C=0$

By Demoivre's theorem,

$\cos \theta +i\sin \theta =e^{i\theta }$

$\Rightarrow e^{iA}+e^{iB}+e^{iC}=0$

We know that $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$\Rightarrow {\left(e^{iA}\right)}^3+{\left(e^{iB}\right)}^3+{\left(e^{iC}\right)}^3=0$

$\Rightarrow {\left(e^{iA}\right)}^3+{\left(e^{iB}\right)}^3+{\left(e^{iC}\right)}^3=(e^{iA}+e^{iB}+e^{iC})(e^{2iA}+e^{2iB}+e^{2iC}-e^{iA+iB}-e^{iB+iC}-e^{iC+iA})+3e^{i(A+B+C)}$

$\Rightarrow {\left(e^{iA}\right)}^3+{\left(e^{iB}\right)}^3+{\left(e^{iC}\right)}^3=3e^{i(A+B+C)}$

$\Rightarrow (\cos 3A+\cos 3B+\cos 3C)+i(\sin 3A+\sin 3B+\sin 3C)=3[\cos (A+B+C)+i\sin (A+B+C)]$

$\Rightarrow (\cos 3A+\cos 3B+\cos 3C)+i(\sin 3A+\sin 3B+\sin 3C)=3[\cos \pi +i\sin \pi ]$ since $A+B+C=\pi$

$\Rightarrow (\cos 3A+\cos 3B+\cos 3C)+i(\sin 3A+\sin 3B+\sin 3C)=3[-1+0]$

$\Rightarrow (\cos 3A+\cos 3B+\cos 3C)+i(\sin 3A+\sin 3B+\sin 3C)=-3+0$

$\therefore \cos 3A+\cos 3B+\cos 3C=-3$