\sin (P\pi)=\sum_{i=0}^{\infty}{(-1)^n\over {(2n+1)!}}(P\pi)^{2n+1}=P(\sum_{i=0}^{\infty}{(-1)^n\over {(2n+1)!}}\pi^{2n+1})=P\sin\pi=P\cdot0=0