xn+p1xn−1+p2xn−2+.......+pn−1x+pn=0
⇒1+p1x−1+p2x−2+.......+pn−1x−n+1+pnx−n=0
Since, cosθ+isinθ is the root
Therefore, 1+p1(cosθ+isinθ)−1+p2(cosθ+isinθ)−2+.......+pn−1(cosθ+isinθ)−n+1+pn(cosθ+isinθ)−n=0
⇒1+p1(cosθ−isinθ)+p2(cos2θ−isin2θ)+.......+pn−1(cos(n−1)θ−isin(n−1)θ)+pn(cosnθ−isinnθ)=0
On comparing imaginary parts, we get
p1(sinθ)+p2(sin2θ)+.......+pn−1(sin(n−1)θ)+pn(sinnθ)=0
Therefore, 22+7(p1sinθ+p2sin2θ+...+pnsin(nθ))=22
Ans: 22