Question

If $\sin A+\sin B+\sin C=0$ and $\cos A+\cos B+\cos C=0$, then $\cos (A+B)+\cos (B+C)+\cos (C+A)=$

• A. $\cos (A+B+C)$
• B. $2$
• C. $1$
• D. $0$

Answer 1

Answer: (D)

$0$

Given

$\sin A+\sin B+\sin C=0$

$\cos A+\cos B+\cos C=0$

Let $Z_1=e^{iA},Z_2=e^{iB},Z_3=e^{iC}$

Consider

$Z_1+Z_2+Z_3$

$⟹e^{iA}+e^{iB}+e^{iC}$

$⟹\cos A+i\sin A+\cos B+i\sin B+\cos C+i\sin C$

$⟹(\cos A+\cos B+\cos C)+i(\sin A+\sin B+\sin C)$

$⟹0$

As $Z_1+Z_2+Z_3=0$

$⟹Z_1^3+Z_2^3+Z_3^3-3Z_1{Z}_2{Z}_3=0$

$⟹(Z_1+Z_2+Z_3)(Z_1^2+Z_2^2+Z_3^2-Z_1{Z}_2-Z_2{Z}_3-Z_1{Z}_3)=0$

$⟹(Z_1^2+Z_2^2+Z_3^2-Z_1{Z}_2-Z_2{Z}_3-Z_1{Z}_3)=0$

$⟹Z_1^2+Z_2^2+Z_3^2=Z_1{Z}_2+Z_2{Z}_3+Z_1{Z}_3$

$⟹(Z_1+Z_2+Z_3{)}^2-2(Z_1{Z}_2+Z_2{Z}_3+Z_1{Z}_3)=Z_1{Z}_2+Z_2{Z}_3+Z_1{Z}_3$

$⟹3(Z_1{Z}_2+Z_2{Z}_3+Z_1{Z}_3)=0$

$⟹3(e^{iA}{e}^{iB}+e^{iB}{e}^{iC}+e^{iA}{e}^{iB})=0$

$⟹e^{i(A+B)}+e^{i(B+C)}+e^{i(A+C)}=0$

$⟹\cos (A+B)+\cos (B+C)+\cos (A+C)=0$