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The famous identity $\sin^2 x+\cos^2x =1$ can be written as follows:

The polynomials $P(x)=x^2$ and $Q(x)=1-x^2$ satisfy $$P(\sin x)= Q(\cos x),\quad \text{for all }x\in\mathbb R$$

What are other such pairs of polynomials. In other words, what is the sufficient and essential condition for two real polynomials $P(x)$ and $Q(x)$ to satisfy $P(\sin x)= Q(\cos x)$ for all $x$?

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    $\begingroup$ Down voted for substantially changing the question. $\endgroup$ – Git Gud Jan 25 '14 at 19:43
  • $\begingroup$ If $a \ne 0$ then it would have to work for $x=0$ and $x=\pi$, at both of which $\sin x = 0$, but $\cos 0 = 1$ and $\cos \pi = -1$. So it only works fwhen $a=0$ and b can be any arbitrary constant. $\endgroup$ – NovaDenizen Jan 25 '14 at 20:01
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    $\begingroup$ This question is changed back and forth between two different versions... $\endgroup$ – N. S. Jan 25 '14 at 20:11
  • $\begingroup$ -1 for changing question back/forth again. $\endgroup$ – achille hui Jan 25 '14 at 21:21
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Proposition: $P(\sin(x))=Q(\cos(x))$ if and only if there exists a polynomial $R$ so that

$$P(X)=R(X^2) \, \mbox{and} \, Q(X)=R(1-X^2) \,.$$

Proof:

Step 1: $P(X)$ is an even polynomial:

$$P(-\sin(x))=P(\sin(-x))=Q(\cos(-x))=Q(\cos(x))=P(\sin(x)) \,.$$

Thus, for all $x$, $\sin(x)$ is a root of $P(X)-P(-X)$. Thus $P(X)-P(-X)$ has infinitely many roots, which completes the step 1.

Step 2: $P(X)=R(X^2)$ for some $R$. This is obvious, a $P$ even implies $$P(X)=a_{2n}X^{2n}+a_{2n-2}X^{2n-2}+..+a_2X^2+a_0$$ And we can set $$P(X)=a_{2n}X^{n}+a_{2n-2}X^{n-1}+..+a_2X+a_0$$

Step 3: $Q(X)=R(1-X^2)$

$$Q(\cos(x))=P(\sin(x))=R(\sin^2(x))=R(1-\cos^2(x))$$

Thus, for all $x, \cos(x)$ is root of $Q(X)-R(1-X^2)$. Hence $Q(X)-R(1-X^2)$ has infinitely many roots.

This proves the direct implication. The Converse is trivial to check.

Comment Given $P(X)$ and $Q(X)$ it is easy to check if the proposition holds. Indeed given $P(x)$ and $Q(X)$, you check first that $P(x)$ is even, then construct $R(x)$ as in the proof and finally test the equality $R(1-x^2)=Q(x)$...

Also note that $Q$ must also be even. This follows immediately. from the proposition, but can also be proven exactly like in step 1, just replace $x$ by $\pi-x$ in the given condition.

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  • $\begingroup$ For potential downvoters: This is the answer to the question as it was when I saw it..Now it is a different question... $\endgroup$ – N. S. Jan 25 '14 at 20:10
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This answer applies to an old version of the question.

We have that $1-\sin^2(x) = \cos^2(x)$ for all real x. The left-hand side is a polynomial in $\sin(x)$, and the right hand side is a polynomial in $\cos(x)$.

I'll also point out that, if $R(x)$ is any polynomial, and $P(\sin(x)) = Q(\cos(x))$, then $R(P(\sin(x)) = R(Q(\cos(x))$. (This is unnecessary for existence, of course, but I think it's interesting to notice).

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