The correct options are
A 3∑i=1sin2αi=0, 3∑i=1cos2αi=0
B 3∑i=1sin3αi=3sin(3∑i=1αi)
C 3∑i=1sin4αi=23∑i=1,j=1 i≠jsin2(αi+αj)
D 3∑i=1,j=1 i≠jsin(αi+αj)=0
Let
a=cosα1+isinα1=cis α1b=cosα2+isinα2=cis α2c=cosα3+isinα3=cis α3
⇒a+b+c=(cosα1+cosα2+cosα3) +i(sinα1+sinα2+sinα3)⇒a+b+c=0 ...(1)
1a+1b+1c=(cosα1+isinα1)−1+(cosα2+isinα2)−1+(cosα2+isinα2)−1∵(cosθ+isinθ)n=cosnθ+isinnθab+bc+caabc=(cosα1+cosα2+cosα3)−i(sinα1+sinα2+sinα3)⇒ab+bc+ca=0 ...(2)
a2+b2+c2=(a+b+c)2−2(ab+bc+ca)⇒(cis α1)2+(cis α2)2+(cis α3)2=0⇒cis 2α1+cis 2α2+cis 2α3=0
Comparing real and imaginary part, we get
3∑i=1sin2αi=0, 3∑i=1cos2αi=0
⇒a3+b3+c3=3abc [∵a+b+c=0](cis α1)3+(cis α2)3+(cis α3)3=3cis α1 cis α2 cis α3⇒cis 3α1+cis 3α2+cis 3α3=3 cis (α1+α2+α3)
Comparing real and imaginary part, we get
3∑i=1sin3αi=3sin(3∑i=1αi)
From eqn(1),
a+b=−c⇒(a+b)2=c2⇒a2+b2−c2=−2ab
Squaring both the sides,
a4+b4+c4=2(a2b2+b2c2+c2a2)⇒(cis α1)4+(cis α2)4+(cis α3)4=2∑(cis αi)2(cis αj)2⇒cis 4α1+cis 4α2+cis 4α3=2∑(cis 2αi)(cis 2αj)⇒cis 4α1+cis 4α2+cis 4α3=2∑cis 2(αi+αj)
Comparing real and imaginary part, we get
3∑i=1sin4αi=23∑i=1,j=1 i≠jsin2(αi+αj)
From eqn(2),
ab+bc+ca=0⇒(cis α1)(cis α2)+(cis α2)(cis α3)+(cis α3)(cis α1)=0⇒cis (α1+α2)+cis (α2+α3)+cis (α3+α1)=0
Comparing real and imaginary part, we get
3∑i=1,j=1 i≠jsin(αi+αj)=0