Let z = cosθ + isinθ. Then the value of ∑15m=1 Im(z2m-1) at θ = 2° is

z=cosθ+isinθ⇒z2m𕒵=cos(2m𕒵)θ+isin(2m𕒵)θLetX=15∑m=1Im(z2m𕒵)∴X=sinθ+sin3θ+…+sin29θ2(sinθ)X=2sinθsinθ+2sinθsin3θ...+2sinθsin29θ⇒2(sinθ)X=1−cos2θ+cos2θ−cos4θ+…+cos28θ−cos30θ∴X=1−cos30θ2sinθ=14sin2o