Are there polynomials in $\mathbb{R}[x]$, other than $P(x)=x^2$, such that $P(\sin x)+P(\cos x)=1$ for all real $x$?

Solution 1:

This has been solved on Art of Problem Solving: Polynomial - TUYMAADA-2000.

Assume that $P$ is a polynomial satisfying $$ \tag{*} \forall x \in \Bbb R: P(\sin x) + P(\cos x) = 1 \, . $$

• First show that $P(-y) = P(y)$ for all $y= \sin(x) \in [-1, 1]$ and conclude that $P$ is even, i.e. $P(x) = Q(x^2)$ for some polynomial $Q$.

• Then show that $Q(\frac 12 +y) + Q(\frac 12 - y) = 1$ for all $y = \sin^2(x) - \frac 12 \in [-1/2, 1/2]$, and conclude that $Q(\frac 12 +y) - \frac 12$ is odd, i.e. $Q(\frac 12 +y) - \frac 12 = y R(y^2)$ for some polynomial $R$.

It follows that $$ P(x) = (x^2- \frac 12) R((x^2 - \frac 12)^2) + \frac 12 $$ for a polynomial $R$. Conversely, every such polynomial satisfies $(*)$.

Solution 2:

I'll answer for $P(\sin^2x)+P(\cos^2x)=1.$

With $t:=\sin^2x-\frac12$, we have the functional equation $$P\left(\frac12+t\right)+P\left(\frac12-t\right)=1,$$

so that $Q(t):=P(t+\frac12)+\frac12$ is an odd function of $t$, such as a polynomial with odd terms.

With $Q(t)=t$, $P(t)=t-\frac12+\frac12=t$.

With $Q(t)=t^3$, $P(t)=(t-\frac12)^3+\frac12=t^3-\frac32t^2+\frac34t+\frac38.$

And so on.

Also, with $Q(t)=16t^3-3t$, $P(\sin^2x)=\sin^2x(3-4\sin^2x)^2$.