(*) $P(\sin x)+P(\cos x)=1$ First $P(-y)=P(\sin (-x))=1-P(\cos (-x))=1-P(\cos x)=P(y)$ for all $y=\sin x\in [-1,1]$ using (*). Then $P(t)-P(-t)$ has infinitely many roots, and so $P(t)=P(-t)$ is even. Therefore $P(t)=Q(t^{2})$ for some polynomial $Q$. Then $1=P(\sin x)+P(\cos x)=Q(\sin^{2}x)+Q(\cos^{2}x)=Q({1\over 2}+y)+Q({1\over 2}-y)$ for all $y=\sin^{2}x-{1\over 2}\in [-{1\over 2},{1\over 2}]$. So the polynomial $Q({1\over 2}+t)+Q({1\over 2}-t)-1$ has infinitely many roots, and therefore the polynomial $Q({1\over 2}+t)-{1\over 2}=-(Q({1\over 2}-t)-{1\over 2})$ is odd. So $Q({1\over 2}+t)-{1\over 2}=tR(t^{2})$ for some polynomial $R$. But now $Q(t)=(t-{1\over 2})R((t-{1\over 2})^{2})+{1\over 2}$ and $P(t)=Q(t^{2})=(t^{2}-{1\over 2})R((t^{2}-{1\over 2})^{2})+{1\over 2}$. Conversely every so defined polynomial satisfies (*), where $\deg P=2+4\deg R$ runs through all natural numbers $\equiv 2 \pmod{4}$ for $R\not \equiv 0$, and $\deg P=0$ for $R\equiv 0$.