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Question

If there exist three values of αi , π2αiπ, such that 3i=1sinαi=3i=1cosαi=0, then which of the following is/are correct?

A
3i=1sin2αi=0, 3i=1cos2αi=0
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B
3i=1sin3αi=3sin(3i=1αi)
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C
3i=1sin4αi=23i=1,j=1 ijsin2(αi+αj)
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D
3i=1,j=1 ijsin(αi+αj)=0
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Solution

The correct options are
A 3i=1sin2αi=0, 3i=1cos2αi=0
B 3i=1sin3αi=3sin(3i=1αi)
C 3i=1sin4αi=23i=1,j=1 ijsin2(αi+αj)
D 3i=1,j=1 ijsin(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)22(ab+bc+ca)(cis α1)2+(cis α2)2+(cis α3)2=0cis 2α1+cis 2α2+cis 2α3=0

Comparing real and imaginary part, we get
3i=1sin2αi=0, 3i=1cos2αi=0

a3+b3+c3=3abc [a+b+c=0](cis α1)3+(cis α2)3+(cis α3)3=3cis α1 cis α2 cis α3cis 3α1+cis 3α2+cis 3α3=3 cis (α1+α2+α3)

Comparing real and imaginary part, we get
3i=1sin3αi=3sin(3i=1αi)

From eqn(1),
a+b=c(a+b)2=c2a2+b2c2=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)2cis 4α1+cis 4α2+cis 4α3=2(cis 2αi)(cis 2αj)cis 4α1+cis 4α2+cis 4α3=2cis 2(αi+αj)

Comparing real and imaginary part, we get
3i=1sin4αi=23i=1,j=1 ijsin2(αi+αj)

From eqn(2),
ab+bc+ca=0(cis α1)(cis α2)+(cis α2)(cis α3)+(cis α3)(cis α1)=0cis (α1+α2)+cis (α2+α3)+cis (α3+α1)=0
Comparing real and imaginary part, we get
3i=1,j=1 ijsin(αi+αj)=0

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