(1/(sinα+sin3α))+(1/(sin3α+sin5α))=
(1/2sin2α*cosα)+(1/2sin4α*cosα)=(sin4α+sin2α)/(2cosα*sin2α*sin4α)=
(2sin3α*cosα)/(2cosα*sin2α*sin4α)=sin3α/(sin2α*sin4α)
При α=π/12
получим (sin3π/12)/((sin2π/12)(sin4π/12))=(sinπ/4)/((sinπ/6)(sinπ/3))=
(√2/2)/((1/2)*(√3/2)=2√2/√3
Ответ А) 2√2/√3