# 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.

• Certainly, $P(x) = \frac12$ will do Dec 18, 2019 at 15:54
• There's a whole family of those solutions @MatthewTowers: $ax^2 + \frac{1-a}{2}$ will work. Dec 18, 2019 at 15:58
• Apparently solved here: artofproblemsolving.com/community/c6h129406p733854 : The general solution is $(x^2-1/2) R((x^2-1/2)^2) + 1/2$ for an arbitrary polynomial $R$. Dec 18, 2019 at 16:08
• @MartinR: amazing. How did you figure it out? Is there any search engine to check if a problem appears somewhere in the AoPS? Dec 18, 2019 at 16:19
• @W-t-P: Found with Approach0 Dec 18, 2019 at 16:36

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 $$(*)$$.

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