Compare $\arcsin (1)$ and $\tan (1)$

Solution 1:

One can write: $$ \frac{1}{\sqrt{1 + \tan^21}} = \cos1 = 1 - 2\sin^2\frac{1}{2} > \frac{17639}{32768}, $$ because $$ \sin\frac{1}{2} < \frac{1}{2} - \frac{1}{2^3\cdot3!} + \frac{1}{2^5\cdot5!} = \frac{1920 - 80 + 1}{3840} < \frac{1845}{3840} = \frac{123}{256}. $$ On the other hand, using Archimedes's lower bound, $\pi > 3\tfrac{10}{71}$: $$ \frac{1}{\sqrt{1 + \left(\sin^{-1}1\right)^2}} = \frac{1}{\sqrt{1 + \left(\frac{\pi}{2}\right)^2}} < \frac{1}{\sqrt{1 + \left(\frac{223}{142}\right)^2}} = \frac{142}{\sqrt{69893}} < \frac{142}{\sqrt{69696}} = \frac{142}{264} = \frac{71}{132}. $$ So, one can prove that $\tan1 < \sin^{-1}1$ by proving that: $$ \frac{17639}{32768} > \frac{71}{132}, $$ which simplifies to $33 \times 17639 > 71 \times 8192$, that is, $582087 > 581632$ - which is true.

Solution 2:

Problem with the Approach in the Question

Note that $\sin(\tan(1))\lt1$ does not insure that $\tan(1)\lt\arcsin(1)$ because $\arcsin(\sin(\tan(1)))\ne\tan(1)$ if $\tan(1)\gt\frac\pi2$.

Another Approach

Let $\alpha=\arctan\left(\frac\pi2\right)$. Since $\tan(\alpha)=\frac\pi2\lt\sqrt3=\tan\left(\frac\pi3\right)$, we know that $\alpha\lt\frac\pi3$.

Furthermore, $$ \frac\pi3-\alpha\le\tan\left(\frac\pi3-\alpha\right)=\frac{\sqrt3-\frac\pi2}{1+\sqrt3\frac\pi2}\tag1 $$ Thus, using the underestimate $\frac{333}{106}$ for $\pi$ and the overestimate $\frac{26}{15}$ for $\sqrt3$, we get $$ \begin{align} \alpha &\ge\frac\pi3-\frac{\sqrt3-\frac\pi2}{1+\sqrt3\frac\pi2}\\ &\ge\frac{111}{106}-\frac{\frac{26}{15}-\frac{333}{212}}{1+\frac{26}{15}\cdot\frac{333}{212}}\\ &=\frac{314804}{313707}\\[9pt] &\gt1\tag2 \end{align} $$ Therefore, $$ \bbox[5px,border:2px solid #C0A000]{\tan(1)\lt\tan(\alpha)=\arcsin(1)}\tag3 $$

A Note About The Inequality

In the inequality $(2)$, we use the fact that $\frac{a-b}{1+ab}$ is increasing in $a$ and decreasing in $b$, for $a,b\ge0$. One way to see this is to recall that this expression came from $$ \tan(\alpha-\beta)=\frac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\tan(\beta)} $$ and let $\alpha=\arctan(a)$ and $\beta=\arctan(b)$. Since $\alpha,\beta\in\left[0,\frac\pi2\right)$, we know that $\alpha-\beta\in\left(-\frac\pi2,\frac\pi2\right)$, in which interval, $\tan$ is an increasing function.

Solution 3:

I try to calculate $\tan 1 <1.57<\frac{\pi}{2}$

$\tan x =x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+\frac{62x^9}{2835}+o(x^9)$

$|\tan x-x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+\frac{62x^9}{2835}|<\frac{62x^9}{2835},x\in[0,1]$

So:$\tan 1<1+\frac{1}{3}+\frac{2}{15}+\frac{17}{315}+\frac{62}{2835}+\frac{62}{2835}<1.565<1.57<\frac{\pi}{2}=\arcsin 1$

Edits: It seems that there just need four terms by under @Ian reminding

$\tan x =x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+\frac{62\xi^9}{2835},\xi\in[0,1]$

$\tan 1<1+\frac{1}{3}+\frac{2}{15}+\frac{17}{315}+\frac{62}{2835}<1.56<1.57<\frac{\pi}{2}=\arcsin 1$

Solution 4:

$\arcsin(1)=\frac{\pi}{2}$ while from the Weierstrass product for the cosine function we have $$ \tan(1) = \sum_{n\geq 0}\frac{8}{(2n+1)^2 \pi^2-4 }=2\sum_{n\geq 0}\left[\frac{1}{(2n+1)\pi-2}-\frac{1}{(2n+1)\pi+2}\right]$$ such that an effective integral representation of $\tan(1)$ through an almost-Gaussian integral is $$\tan(1)=\int_{\mathbb{R}}\frac{\sinh(2x)}{\sinh(\pi x)}\,dx.$$ As an alternative, by expanding $\frac{8}{(2n+1)^2\pi^2-4}$ as a geometric series we get $$ \tan(1) = 2 \sum_{m\geq 1}\frac{(4^m-1)\zeta(2m)}{\pi^{2m}} $$ which is equivalent to the previous integral representation via $\zeta(2m)=\frac{1}{(2m-1)!}\int_{0}^{+\infty}\frac{z^{2m-1}}{e^z-1}\,dz$.
In order to prove that $\tan(1)<\frac{\pi}{2}$ it is enough to show that $\tan\left(\frac{1}{2}\right)<\frac{\pi}{2+\sqrt{\pi^2+4}}$, since $\tan(2z)=\frac{2\tan z}{1-\tan^2 z}$. $\tan\left(\frac{1}{2}\right)$ has a fast-convergent series representation $$ \tan\left(\tfrac12\right)=4\sum_{m\geq 1}\frac{\zeta(2m)}{\pi^{2m}}\left(1-\frac{1}{4^m}\right) $$ which allows to state $$ \tan\left(\tfrac12\right) < 4\sum_{m= 1}^{3}\frac{\zeta(2m)}{\pi^{2m}}\left(1-\frac{1}{4^m}\right)+4\,\zeta(8)\sum_{m \geq 4}\frac{1}{\pi^{2m}}\left(1-\frac{1}{4^m}\right)$$ or $$ \tan\left(\tfrac12\right)<\frac{131}{240}+\frac{\pi ^2 \left(85 \pi^2-21\right)}{50400 \left(\pi^2-1\right) \left(4 \pi^2-1\right)}$$ (this is extremely accurate). The proof is finished by exploiting $\frac{227}{23}<\pi^2<\frac{79}{8}$ which follows from the study of the Beuker-like integrals $$ \iint_{(0,1)^2}\frac{x^m(1-x)^m y^n (1-y)^n}{1-xy}\,dx\,dy.$$

An alternative approach. The Shafer-Fink inequality gives that for any $x\in(0,3/2)$ $$ \tan(x) < \frac{3x+2x\sqrt{9-3x^2}}{9-4x^2} $$ holds, so by the duplication formula for the tangent function the sharper inequality $$ \tan(x) < \frac{4 x\left(9-x^2\right) \left(3+\sqrt{36-3 x^2}\right)}{324-117 x^2+7 x^4-6 x^2 \sqrt{36-3 x^2}} $$ holds too. This gives $$ \tan(1) < \frac{32 \left(3+\sqrt{33}\right)}{214-6 \sqrt{33}}=\frac{4}{697} \left(105+29 \sqrt{33}\right) $$ so it is enough to show that $\pi>\frac{32(3+\sqrt{33})}{107-3\sqrt{33}}$, which is a pretty loose inequality.