Question

If there exist three values of ${\alpha }_i,$ $-\frac{\pi }{2}\le {\alpha }_i\le \pi$, such that $\sum _{i=1}^3\sin {\alpha }_i=\sum _{i=1}^3\cos {\alpha }_i=0,$ then which of the following is/are correct?

• A. $\sum _{i=1}^3\sin 2{\alpha }_i=0,\sum _{i=1}^3\cos 2{\alpha }_i=0$
• B. $\sum _{i=1}^3\sin 3{\alpha }_i=3\sin \left(\sum _{i=1}^3{\alpha }_i\right)$
• C. $\sum _{i=1}^3\sin 4{\alpha }_i=2\sum _{{}_{i\ne j}^{i=1,j=1}}^3\sin 2({\alpha }_i+{\alpha }_j)$
• D. $\sum _{{}_{i\ne j}^{i=1,j=1}}^3\sin ({\alpha }_i+{\alpha }_j)=0$

Answer 1

Answer: (A), (B), (C), (D)

$\sum _{{}_{i\ne j}^{i=1,j=1}}^3\sin ({\alpha }_i+{\alpha }_j)=0$

Let
$a=\cos {\alpha }_1+i\sin {\alpha }_1=cis{\alpha }_{1}b=\cos {\alpha }_2+i\sin {\alpha }_2=cis{\alpha }_{2}c=\cos {\alpha }_3+i\sin {\alpha }_3=cis{\alpha }_3$

$\Rightarrow a+b+c=(\cos {\alpha }_1+\cos {\alpha }_2+\cos {\alpha }_3)+i(\sin {\alpha }_1+\sin {\alpha }_2+\sin {\alpha }_3)\Rightarrow a+b+c=0...\left(1\right)$

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=(\cos {\alpha }_1+i\sin {\alpha }_1{)}^{-1}+(\cos {\alpha }_2+i\sin {\alpha }_2{)}^{-1}+(\cos {\alpha }_2+i\sin {\alpha }_2{)}^{-1}\because (\cos \theta +i\sin \theta {)}^n=\cos n\theta +i\sin n\theta \frac{ab+bc+ca}{abc}=(\cos {\alpha }_1+\cos {\alpha }_2+\cos {\alpha }_3)-i(\sin {\alpha }_1+\sin {\alpha }_2+\sin {\alpha }_3)\Rightarrow ab+bc+ca=0...\left(2\right)$

$a^2+b^2+c^2=(a+b+c{)}^2-2(ab+bc+ca)\Rightarrow (cis{\alpha }_1{)}^2+(cis{\alpha }_2{)}^2+(cis{\alpha }_3{)}^2=0\Rightarrow cis2{\alpha }_1+cis2{\alpha }_2+cis2{\alpha }_3=0$

Comparing real and imaginary part, we get
$\sum _{i=1}^3\sin 2{\alpha }_i=0,\sum _{i=1}^3\cos 2{\alpha }_i=0$

$\Rightarrow a^3+b^3+c^3=3abc[\because a+b+c=0](cis{\alpha }_1{)}^3+(cis{\alpha }_2{)}^3+(cis{\alpha }_3{)}^3=3cis{\alpha }_{1}cis{\alpha }_{2}cis{\alpha }_3\Rightarrow cis3{\alpha }_1+cis3{\alpha }_2+cis3{\alpha }_3=3cis({\alpha }_1+{\alpha }_2+{\alpha }_3)$

Comparing real and imaginary part, we get
$\sum _{i=1}^3\sin 3{\alpha }_i=3\sin \left(\sum _{i=1}^3{\alpha }_i\right)$

From eqn$\left(1\right)$,
$a+b=-c\Rightarrow (a+b{)}^2=c^2\Rightarrow a^2+b^2-c^2=-2ab$

Squaring both the sides,
$a^4+b^4+c^4=2(a^2{b}^2+b^2{c}^2+c^2{a}^2)\Rightarrow (cis{\alpha }_1{)}^4+(cis{\alpha }_2{)}^4+(cis{\alpha }_3{)}^4=2\sum (cis{\alpha }_i{)}^2(cis{\alpha }_j{)}^2\Rightarrow cis4{\alpha }_1+cis4{\alpha }_2+cis4{\alpha }_3=2\sum (cis2{\alpha }_i\left)\right(cis2{\alpha }_j)\Rightarrow cis4{\alpha }_1+cis4{\alpha }_2+cis4{\alpha }_3=2\sum cis2({\alpha }_i+{\alpha }_j)$

Comparing real and imaginary part, we get
$\sum _{i=1}^3\sin 4{\alpha }_i=2\sum _{{}_{i\ne j}^{i=1,j=1}}^3\sin 2({\alpha }_i+{\alpha }_j)$

From eqn$\left(2\right)$,
$ab+bc+ca=0\Rightarrow \left(cis{\alpha }_1\right)\left(cis{\alpha }_2\right)+\left(cis{\alpha }_2\right)\left(cis{\alpha }_3\right)+\left(cis{\alpha }_3\right)\left(cis{\alpha }_1\right)=0\Rightarrow cis({\alpha }_1+{\alpha }_2)+cis({\alpha }_2+{\alpha }_3)+cis({\alpha }_3+{\alpha }_1)=0$
Comparing real and imaginary part, we get
$\sum _{{}_{i\ne j}^{i=1,j=1}}^3\sin ({\alpha }_i+{\alpha }_j)=0$