# On a trigonometric inequality by Huygens

The following inequality, ascribed to Huygens, appeared in this post:
$$\begin{equation*} 1-\frac43\,\frac{\sin^3\theta/2}{\theta-\sin\theta} >(1-\cos\theta/2)\Big(\frac35-\frac3{1400}\frac{\pi^2}{n^2}\Big), \tag{10}\label{10} \end{equation*}$$ where $$\theta\in[0,\pi/2]$$ and $$n$$ is an integer $$\ge2$$. Later, this inequality was changed to its opposite. Note that neither the strict inequality \eqref{10} nor its strict opposite hold for $$\theta=0$$. It was also said that "the inequality as stated here is the beginning of the asymptotic expansion of the quotient of the left-hand side by the trigonometric function on the right".

The mysterious role of $$n$$ here has not been explained. However, one has $$\begin{equation*} f(\theta):=\Big(1-\frac43\,\frac{\sin^3\theta/2}{\theta-\sin\theta}\Big) \Big/(1-\cos\theta/2) =\frac35-\frac{3\theta^2}{1400}-\frac{11\theta^4}{336000}+O(\theta^6). \end{equation*}$$ So, it makes sense to ask if $$\begin{equation*} f(\theta)<\frac35-\frac{3\theta^2}{1400} \quad\text{for }\theta\in(0,\pi/2]. \tag{20}\label{20} \end{equation*}$$

It will be shown below that this is indeed true.

$$\renewcommand{\th}{\theta}$$As in the previous answer, use the substitution $$t=\tan\frac\th4,\quad \sin\frac\th2=\frac{2t}{1+t^2}, \quad \cos\frac\th2=\frac{1-t^2}{1+t^2}, \quad \sin\th=\frac{4t(1-t^2)}{(1+t^2)^2}, \quad \th=4\tan^{-1}t$$ in this case to rewrite the inequality \eqref{20} in question as $$\begin{equation*} g(t):=36 \tan^{-1}(t)^3-\frac{36 (1-t^2) t \tan^{-1}(t)^2}{\left(t^2+1\right)^2} +\frac{105 \left(5-t^2\right) \tan^{-1}(t)}{t^2}-\frac{35 \left(3 t^4+22 t^2+15\right)}{t \left(t^2+1\right)^2}<0 \tag{30}\label{30} \end{equation*}$$ for all $$\begin{equation*} t\in(0,t_*],\quad t_*:=\tan\frac\pi8=\sqrt{2}-1. \end{equation*}$$ Let $$\begin{equation*} g_1(t):=g'(t)\frac{\left(t^2+1\right)^3}{t^4+6 t^2+1} \\ =72 \tan ^{-1}(t)^2 -\frac{6 \left(163 t^6+537 t^4+525 t^2+175\right) \tan ^{-1}(t)}{t^3 \left(t^4+6 t^2+1\right)} +\frac{70 \left(33 t^4+40 t^2+15\right)}{t^2 \left(t^4+6 t^2+1\right)}, \end{equation*}$$ \begin{equation*} \begin{aligned} g_2(t)&:=g_1'(t)\frac{t^4 \left(t^2+1\right) \left(t^4+6 t^2+1\right)^2}{187 t^{12}+1084 t^{10}+6903 t^8+15784 t^6+15937 t^4+6300 t^2+525} \\ &=6 \tan ^{-1}(t)-\frac{2 t \left(2799 t^{10}+12455 t^8+34970 t^6+41826 t^4+18375 t^2+1575\right)}{187 t^{12}+1084 t^{10}+6903 t^8+15784 t^6+15937 t^4+6300 t^2+525}, \end{aligned} \end{equation*} \begin{equation*} \begin{aligned} g_3(t)&:=g_2'(t)\frac{\left(t^2+1\right) \left(187 t^{12}+1084 t^{10}+6903 t^8+15784 t^6+15937 t^4+6300 t^2+525\right)^2}{192 t^8} \\ &=6545 t^{16}+59300 t^{14}+36012 t^{12}-912092 t^{10}-2894890 t^8-3467092 t^6 \\ &-1800788 t^4-362100 t^2-24255<0 \end{aligned} \end{equation*} for $$t\in(0,t_*]$$. Also, $$g_2(0)=g_1(0+)=g(0+)=0$$. So, \eqref{30} immediately follows. $$\quad\Box$$
The question states that it makes sense to ask if $$$$f(\theta)<\frac35-\frac{3\theta^2}{1400} \quad\text{for }\theta\in(0,\pi/2].$$$$
However, this is the wrong question. The right question is to ask if $$\begin{equation*} f(\theta) with $$$$c=\frac35-\frac{3(\pi/2)^2}{1400},$$$$
which happens to be false: $$$$f(1)-c=0.003112,$$$$ $$$$f(\pi/2)-c=-0.000201.$$$$