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.
 A: 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 $
A: Consider the problem 
$$\max_{\mathbf{y}\in \mathbb{R}^2} \mathbf{y} \cdot \mathbf{c}$$ subject to $\mathbf{b}\cdot \mathbf{y} \le 0$ and $\|\mathbf{y}\|=1$, where $\mathbf{b}=(-a,1)$, $\mathbf{c}=(-a^3,1)$. Since $\mathbf{b} \cdot \mathbf{c} \ge 0$ it's not too hard to see   that the optimal argument is given by $\mathbf{y}^*=(\cos x,\sin x) = (-1,-a)/\sqrt{1+a^2}$. (Drawing a picture of the feasible region helps.) So then we want to check that:
$$\mathbf{y}^*\cdot \mathbf{c} = \frac{a^3-a}{\sqrt{1+a^2}} \le \frac{1}{3}\sqrt{1+a^6}$$
Since $a \ge 1$ this is equivalent to:
$$9 a^2(a^2-1)^2 \le (1+a^2)(1+a^6).$$
Which then by collecting terms and factoring, is equivalent to:
$$a^8-8a^6+18a^4-8a^2+1=(a^4-4a^2+1)^2\ge 0.$$
A: Since $\sin(x)-a\cos(x)\le0$,
$$
\sin(x)-a^3\cos(x)\le(a-a^3)\cos(x)\tag{1}
$$
Since $a\in\mathbb{N}$, $a-a^3\le0$. If $\cos(x)\ge0$, then $(1)$ implies that $\sin(x)-a^3\cos(x)\le0$. So assume wlog that $\cos(x)\le0$. Therefore, $\sin(x)\le a\cos(x)\le0$, and thus,
$$
a\le\tan(x)\tag{2}
$$
Since $x$ is in the third quadrant, $(2)$ implies that
$$
\sin(x)-a^3\cos(x)\le\frac{a^3-a}{\sqrt{1+a^2}}\tag{3}
$$
Maximizing
$$
f(a)=\dfrac{a^3-a}{\sqrt{1+a^2}}\dfrac1{\sqrt{1+a^6}}\tag{4}
$$
with
$$
f'(a)=-\dfrac{a^4+1}{\sqrt{(a^6+1)^3(a^2+1)}}(a^4-4a^2+1)\tag{5}
$$
gives $f'(a)=0$ at $a=\sqrt{2+\sqrt3}\implies f(a)\le\frac13$.
Therefore, putting $(3)$ and $(4)$ together yields
$$
\sin(x)-a^3\cos(x)\le\frac13\sqrt{1+a^6}\tag{6}
$$
However, since $a\in\mathbb{N}$ and $f(2)=\dfrac{6}{5\sqrt{13}}$, we can improve the $\dfrac13$ to $\dfrac{6}{5\sqrt{13}}$. That is, with the restriction that $a\in\mathbb{N}$, we have a slightly stronger inequality:
$$
\sin(x)-a^3\cos(x)\le\frac{6}{5\sqrt{13}}\sqrt{1+a^6}\tag{7}
$$

For $x\le1000000$, the closest we get to equality in $(7)$ is $x=11270$ and $a=2$. For those values,
$$
\sin(x)-a\cos(x)=-0.00000638535498889561
$$
and
$$
\frac{\sin(x)-a^3\cos(x)}{\sqrt{1+a^6}}=0.33281742491402203493
$$
where
$$
\frac{6}{5\sqrt{13}}=0.33282011773513747321
$$
