If f(θ)=sin2θ+sin2(θ+2π3)+sin2(θ+4π3),thenf(π15)
23
32
12
13
f(θ)=sin2θ+sin2(θ+2π3)+sin2(θ+4π3)=1−cos2θ2+1−cos(2θ+4π3)2+1−cos(2θ+8π3)2=12(3−cos2θ−cos(4π3+2θ)−cos(8π3+2θ))=12(3−cos2θ−2cos(2π+2θ)cos2π3)=12(3−cos2θ+cos2θ)f(θ)=32∴f(π15)=32
f(θ)=sin2θ+sin2(θ+2π3)+sin2(θ+4π3) then f(π15) is equal to