Problem source

Which of the following statements are true
and which are false? Justify your answers.
\begin{questionparts}
\item $a^{\ln b}=b^{\ln a}$ for all $a,b>0$.
\item $\cos(\sin\theta)=\sin(\cos\theta)$ for all real $\theta$.
\item There exists  a polynomial $\mathrm{P}$ such that
$|\mathrm{P}(\theta)-\cos\theta|\leqslant 10^{-6}$
for all real $\theta$.
\item $x^{4}+3+x^{-4}\geqslant 5$ for all $x>0$.
\end{questionparts}

Solution source

\begin{questionparts}
\item True. \begin{align*}
&& \ln a \cdot \ln b &= \ln b \cdot \ln a \\
\Leftrightarrow && \exp ( \ln a \cdot \ln b) &= \exp ( \ln b \cdot \ln a) \\
\Leftrightarrow && \exp ( \ln a )^{\ln b} &= \exp ( \ln b )^{\ln a} \\
\Leftrightarrow && a^{\ln b} &= b^{\ln a} \\
\end{align*}
\item False. Consider $\theta = 0$. We'd need $\cos 0 = 1 =  \sin 1$, but $0 < 1 < \frac{\pi}{2}$ so $\sin 1 \neq 1$

\item False. If the polynomial has positive degree, then as $n \to \infty$, $\P(x) \to \pm \infty$, in particular it must be well outside the interval $[-1,1]$. Therefore it can't be within $10^{-6}$ of $\cos \theta$ which is confined to that interval. The only polynomial which is restricted to that range are constants, but then $|\cos 0 - c| \leq 10^{-6}$ and $|\cos \pi - c| \leq 10^{-6}$ $2 = |1-(-1)| \leq |1-c| + |-1-c| \leq 2\cdot 10^{-6}$ contradiction.

\item True.
\begin{align*}
&& (x^2-x^{-2})^2 &\geq 0 \\
\Leftrightarrow && x^4-2+x^{-4} &\geq0 \\
\Leftrightarrow && x^4+3+x^{-4} &\geq 5 \\
\end{align*}
\end{questionparts}