A1A2=modofA1A2=|A1A2|
|A1Ar|2=|z1|2[2−2cos2(r−1)πn]
=a2⋅4sin22(r−1)πnby(A)
∴|A1Ar|=2asin(r−1)πn
∵1−cosθ=2sin2(θ/2)
Putting r=2,3,4 we have from the given relation,
1sinπn=1sin2πn+1sin3πn
2sinπncosπnsin3πn=sinπn[sin3πn+sin2πn]
Cancelsinπnassinπn≠0asn≠1
∴sin4πn+sin2πn=sin3πn+sin2πn
orsin4πn=sin3πn∴4πn=π−3πn
7πn=πorn=7