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 $x$? 
Are there any polynomials $P\in\mathbb R[x]$, other than $P(x)=x^2$, such that
    $$ P(\sin x) + P(\cos x) = 1,\quad x\in\mathbb R? $$

Notice that there are pairs of polynomials $P,Q$ with $P(\sin x)+Q(\cos x)=1$: say, from $\sin 3x=3\sin x-4\sin^3 x$ and $\cos 3x=4\cos^3 x-3\cos x$ it follows that 
  $$ (3\sin x-4\sin^3 x)^2+(4\cos^3 x-3\cos x)^2=1. $$
My motivation for this question is a pure curiosity; hope it counts. 
 A: 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 $(*)$.
A: 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 $Q(t)=16t^3-3t$, $P(\sin^2x)=\sin^2x(3-4\sin^2x)^2$.
