How to comment on degree and whether polynomial function is even or odd given a particular condition for it? Let $P(x)$ be a polynomial with real coefficients such that $P(\sin^2x) = P(\cos^2x)$ for all x in interval [0,π/2] Find which of the following statements are true ?


*

*$P(x)$ is an even function

*$P(x)$ can be expressed as a polynomial in $(2x - 1)^2 $

*$P(x)$ is a polynomial of even degree.


I tried taking a general poynomial and tried equating bpth sides. I am confused whether to replace x by π/2 - x or replace it by 1 - x while equating and how to move further. Or is there any other simpler approach ?
 A: Firstly, we see that setting $t=\sin^2(x)$ for $x \in [0, \pi/2]$ then $t \in [0, 1]$ and the condition above becomes $P(t) = P(1-t)$ for all $t \in [0, 1]$. 


*

*False. Let $P(x) = (x - \frac{1}{2})^2$.

*If we consider the polynomial $P(x) - P(1-x)$ the above condition tells us that it is $0$ for all $x \in [0,1]$. Since this polynomial has infinitely many roots it must be that $P(x) = P(1-x)$ for all $x$. Now let $Q$ be a polynomial such that $P(x) = Q(2x - 1)$ (you should think about why we always can do this). Our condition then implies that
$$
Q(x) = P\left(\frac{x+1}{2}\right) = P\left(\frac{1-x}{2} \right) = Q(-x)
$$
and thus Q is even. Since $Q$ is an even polynomial it must only contain even powers and therefore we can write $Q(x) = R(x^2)$ and thus $P(x) = R((2x-1)^2)$.

*True. This follows from 2.
A: Let
$$y=x-\frac{1}{2}$$
and
$$P(x)=Q(y)$$
Then because
$$P(x)=P(1-x)$$
for $0 \le x \le 1$, we have
$$Q(y)=Q(-y)$$
for $-\frac{1}{2}\le y \le \frac{1}{2}$.
Hence $Q(y)$ has only even powers in $y$.
Therefore
$$P(x)=Q(y)=R(y^2)=R((x-1/2)^2)=S((2x-1)^2)$$
