It is given polynomial \(P(x)=(x\sin a+\cos a)^n\) for natural number \(n\). Find the remainder when \(P(x)\) is divided by \(x^2+1\).

Let \(R(x)\) be the remainder when \(P(x)\) is divided by \(x^2+1\). If \(n=1\), then \(P(x)=x\sin a+\cos a\), so \(R(x)=x\sin a+\cos a\). For \(n\ge2\), let \(P(x)=(x^2+1)Q(x)+Ax+B\). Taking \(x=i\), we get \((i\sin a+\cos a)^n=(\cos a+i\sin a)^n=\cos(na)+i\sin(na)=B+iA\), hence \(A=\sin(na)\) and \(B=\cos(na)\). Therefore, the remainder is \(R(x)=x\sin(na)+\cos(na)\).