# sine inequality

$$a>0.$$ $$\sin{a}\leq\frac{a}{\sqrt{1+\frac{a^2}{k}}}$$ Find the minimum of $$k$$. It’s obvious that we only need to prove the inequality holds when $$a\in\left(0 , \frac{\pi}{2}\right)$$.

I can prove that the inequality holds when$$a \in \left(0, \frac{\pi}{4}\right)$$ and $$k=4$$. But I am not sure that $$k=4$$ is the minimum and also works when $$a \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$$.

When $$a \in \left(0, \frac{\pi}{4}\right)$$:

$$\sin{a} \leq a \leq \tan{a}$$

Hence $$\cos{a} \leq \frac{\sin{a}}{a} \leq \cos{\frac{a}{2}}$$

$$\cos^2{\frac{a}{2}}=1-\sin^2{\frac{a}{2}} \leq 1-\frac{a^2}{2}$$

$$\Rightarrow \cos^2{\frac{a}{2}} \leq \frac{1}{1+\frac{a^2}{4}}$$

Hence $$\sin{a}\leq\frac{a}{\sqrt{1+\frac{a^2}{4}}}$$

Any ideas?

• Welcome! Is $0\cdot\sin a=0$ right? – manooooh Feb 16 at 15:04
• I’m not sure what you mean. Can you please show me more clearly? – qsa Feb 16 at 15:15
• Yes, sorry. You wrote $0.\sin a$, which is equal to $0$ for all values of $a$. It seems that you did not want to write that expression, so I am asking if you wanted to write that. – manooooh Feb 16 at 15:17
• k=3is the answer. Use a graphing tool and explore why! – John. P Feb 16 at 15:23
• Isolate k and look at the resulting function of a. – marty cohen Feb 16 at 15:29

Now that an answer has been accepted, I'll spell out the method I hinted at in a comment, leaving only a couple of details to fill in. (I haven't even checked the details myself, so it would be unwise to take my word for them!)

For all $$a \in \left(0, \frac{\pi}{2}\right)$$, the terms of the alternating series $$\sin a = a - \frac{a^3}{6} + \frac{a^5}{120} - \cdots$$ decrease in absolute value. For instance, $$\left(\frac{\pi}{2}\right)^2 < 6$$. (Check the other terms.) Therefore: $$$$\label{3115111:eq:1}\tag{1} a - \frac{a^3}{6} < \sin a < a - \frac{a^3}{6} + \frac{a^5}{120} \quad \left(0 < a < \frac{\pi}{2} \right).$$$$

The right hand side of the desired inequality also has a convergent series expansion, by the generalised binomial theorem: $$$$\label{3115111:eq:2}\tag{2} a\left(1 + \frac{a^2}{k}\right)^{-1/2} = a - \frac{a^3}{2k} + \frac{3a^5}{8k^2} - \frac{5a^7}{16k^3} + \cdots \quad (0 < a < \sqrt{k}).$$$$ One must check - I haven't! - that the terms of this series, too, alternate in sign and decrease in absolute value. Then: $$a - \frac{a^3}{2k} < \frac{a}{\sqrt{1+\frac{a^2}{k}}} < a - \frac{a^3}{2k} + \frac{3a^5}{8k^2} \quad (0 < a < \sqrt{k}).$$ If $$k < 3$$, then for small enough $$a$$, we will have: $$\frac{3a^5}{8k^2} < \frac{a^3}{2k} - \frac{a^3}{6},$$ whence: $$\frac{a}{\sqrt{1+\frac{a^2}{k}}} < \sin a,$$ i.e. the inequality is false for those values of $$a$$.

So, we must have $$k \geqslant 3$$.

From \eqref{3115111:eq:2}: $$\frac{a}{\sqrt{1+\frac{a^2}{3}}} > a - \frac{a^3}{6} + \frac{a^5}{24} - \frac{5a^7}{432} \quad (0 < a < \sqrt3).$$ The desired inequality follows from this in conjunction with \eqref{3115111:eq:1}, so long as: $$\frac{a^5}{24} - \frac{5a^7}{432} > \frac{a^5}{120}.$$ This reduces to: $$a^2 < \frac{72}{25}, \text{ i.e. } a < \frac{6\sqrt2}{5}.$$ This holds - just! - for all $$a \in \left(0, \frac{\pi}{2}\right)$$. Therefore, the desired inequality is true when $$k = 3$$.

• Thank you for your answer sincerely! – qsa Feb 17 at 1:58

My solution might not be that elegant, but using calculus is a way to analyze graphs. Let $$f(x)=\sin x - \frac{x}{\sqrt{1+\frac{x^2}{k}}}$$. Then, $$\frac{d}{dx}f(x)= \cos x- \frac{1}{(\frac{k+x^2}{k})^{\frac{3}{2}}}$$ We want $$f'(x)\le 0$$ when $$x \ge 0$$, and since $$f(0)=f'(0)=0, f''(x)$$ should be negative. $$f''(x)=\frac{3x}{k(\frac{k+x^2}{k})^{\frac{5}{2}}}-\sin x, f''(0)=0$$ In same way, $$f'''(x)$$should be negative around $$x=0$$. $$f'''(x)=\frac{3(k-4x^2)}{(k+x^2)^2(\frac{k+x^2}{k})^{\frac{3}{2}}}-\cos x , f'''(0)=\frac{3}{k}-1$$ Therefore $$\frac{3}{k}-1\le0, k\ge3$$

• Thank you for your answer sincerely! – qsa Feb 17 at 1:58