<s>[INST] Prove, that $\sin x- a^3\cos x\leq \frac 1 3 \sqrt{1+a^6}$ Let a and $x$ be natural numbers with the property that $\sin x\leq a\cos x$.
Prove that $\sin x- a^3  \cos x\leq \frac 1 3 \sqrt{1+a^6}$.
Again, I'm looking for a second solution. I don't know how to use LaTex and my solution is hard to write here. It would be very nice to see a different solution from mine.
 [/INST] Not sure if it is correct.
\begin{align*}
\sin x- a^3\cos x &= \sqrt{1 + (-a^3)^2} \left( \frac{1}{\sqrt{1 + a^6}} \sin(x) - \frac{a^3}{\sqrt{1 + a^6}} \cos(x)\right )\\ 
 &= \sqrt{1+a^6} \left( \cos(\phi) \sin(x) - \sin(\phi) \cos(x))  \right )\\ 
 &= \sqrt{1 +a^6} \sin \left( x - \phi  \right )\\
\end{align*}
Where $\displaystyle \phi = \arctan(a^3)$. Given, constraint $\sin(x) \le a \cos(x) \implies x \le \arctan(a) \le \frac \pi 2  $
\begin{align*}
\sin(x-\arctan(a^3)) &\le \sin (\arctan(a) - \arctan(a^3)) \\ 
 &= \sin \left(\arctan \left( \frac{a - a^3}{1 + a^4 } \right ) \right )\\ 
 &= \frac{a - a^3}{ \sqrt{ (a -a^3)^2 + (1+a^4)}} \\ 
\end{align*}
From calculus, that thing seems to have maximum value of $\displaystyle \frac 1 3 $
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