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If $P(x)$ is a polynomial for all $x$ belongs to $[0,π/2]$ such that $P(\cos^2x)=P(\sin^2x)$ for all $x$ belongs to $[0,π/2]$ which of the following are true:

  1. $P(x)$ is an even function

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

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

I tried evaluating it as $$P(\sin^2x) = P(\cos^2x)$$ $$P(\sin^2x) = P(1-\sin^2x)$$ Let $\sin^2x = y $ The equation becomes $P(y)=P(1-y)$ for all $x$ belongs to $[0,1]$.

Now I don't know how to further find the properties of this polynomial. Any help is appreciated :)

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    $\begingroup$ $P(y)=P(1-y)$ for all $y \in [0,1]$ implies $P(y)=P(1-y)$ for all $y \in \mathbb R$. $\endgroup$ – lhf Nov 13 '17 at 13:47
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Since$$(\forall x\in\mathbb{R}):\sin^2(x)+\cos^2(x)=1,$$you know that$$(\forall x\in\mathbb{R}):P(x)=P(1-x).$$Let $Q(x)=P\left(x+\frac12\right)$. Then\begin{align}Q(-x)&=P\left(-x+\frac12\right)\\&=P\left(1-\left(-x+\frac12\right)\right)\\&=P\left(x+\frac12\right)\\&=Q(x).\end{align}So, $Q(x)$ is even, and therefore $Q(x)=R(x^2)$ for some polynomial $R(x)$. But then$$P(x)=Q\left(x-\frac12\right)=R\left(\left(x-\frac12\right)^2\right),$$and this is equivalent to the third item from your list.

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  • $\begingroup$ #2 is also true. $\endgroup$ – lhf Nov 13 '17 at 14:01
  • $\begingroup$ @lhf Indeed it is. $\endgroup$ – José Carlos Santos Nov 13 '17 at 14:01
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$P(y)=P(1-y)$ for all $y \in [0,1]$ implies $P(y)=P(1-y)$ for all $y \in \mathbb R$.

  1. False: $P(x)=x(1-x)$ satisfies the condition but is not an even function

  2. True: If the degree of $P$ is $n$ and its leading coefficient is $a$, then the leading coefficient of $P(1-y)$ is $(-1)^n a$.

  3. Has been answered nicely by José.

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