Let f(θ)=sinθsin3θsin5θ , then f(π/14) is equal to
1/8
1/4
1/7
1/14
As π14−π2−3π7,3π14−π2−2π7
and 5π14=π2−π7 , we can write
fπ14=cos3π7cos2π7cosπ7⇒ 2sinπ7fπ14=sin2π7cos2π7cos3π7⇒ 4sinπ7fπ14=sin4π7cos3π7⇒ 8sinπ7fπ14=sin7π7+sinπ7=sinπ7⇒ fπ14=18