# 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.

• Use parentheses to write clearly what you mean...or much better: writhe mathematics with LaTeX – DonAntonio Sep 27 '13 at 17:37
• Let me know if I broke the meaning of the question - the previous editor was inconsistent, but at least the body and the title agree now. – Thomas Andrews Sep 27 '13 at 17:43
• Do you really want $a,x$ to be natural numbers? That seems strange for a problem like this. – Thomas Andrews Sep 27 '13 at 17:43
• It's very good now! – user85046 Sep 27 '13 at 17:43
• What is your solution? At the very least, provide some details and sketch out the argument. Otherwise, how do we know what a 'second solution' is? – Calvin Lin Sep 27 '13 at 19:06

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 $\displaystyle \frac 1 3$

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: $$a^8-8a^6+18a^4-8a^2+1=(a^4-4a^2+1)^2\ge 0.$$

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 $$\frac{6}{5\sqrt{13}}=0.33282011773513747321$$

• Case of equality? – user85046 Sep 28 '13 at 10:59
• @CFG: are you asking if we can have equality? Since $x\in\mathbb{N}$, I would doubt that we could find $x\equiv\pi+\tan^{-1}(2)\pmod{2\pi}$ (in the third quadrant), so I would doubt we would ever get equality. – robjohn Sep 28 '13 at 11:18
• Yes, true. Thank you. – user85046 Sep 28 '13 at 11:22