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

Which one is greater: $$\arcsin (1)$$ or $$\tan (1)$$?

How to find without using graph or calculator?

I tried using $$\sin(\tan1)\leq1$$, but how do I eliminate the possibility of an equality without using the calculator?

• What method have you tried to use? Where did you get stuck? – TheSimpliFire Aug 9 '18 at 14:32
• How to ask a good question – steven gregory Aug 9 '18 at 14:51
• @Mason Because always $\sin\leq1$. – Michael Rozenberg Aug 9 '18 at 15:07
• Is $arcsin(1)=\frac{\pi}{2}$? – HQR Aug 9 '18 at 15:07
• Very nice problem! Now, good people will come and will close this topic. – Michael Rozenberg Aug 9 '18 at 15:25

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.

• Nice series acceleration trick in passing to the half-angle formula. Also a nice use of a rigorous rational bound for $\pi$. +1. – Ian Aug 9 '18 at 17:56
• Simplifying slightly: $$\cos1 > \frac{17639}{32768} > \frac{17632}{32768} = \frac{551}{1024} > \frac{71}{132},$$ because $33 \times 551 > 71 \times 256$, i.e. $18183 > 18176$. – Calum Gilhooley Aug 10 '18 at 16:16
• N.B. There is a much better answer - to essentially the same question - here. – Calum Gilhooley Mar 4 '19 at 21:18

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, $$\bbox[5px,border:2px solid #C0A000]{\tan(1)\lt\tan(\alpha)=\arcsin(1)}\tag3$$

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.

• Why the downvote? If it is about the estimated values, $\frac{333}{106}$ is a fairly well known continued fraction under-estimate for $\pi$ and $\frac{26}{15}\gt\sqrt3$ can be verified by noting that $26^2=676\gt675=3\cdot15^2$. – robjohn Mar 6 '19 at 18:26

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

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

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

• Technically you need to be a bit more careful about estimating the remainder term. – Ian Aug 9 '18 at 15:43
• I am checking it. @Michael Rozenberg – HQR Aug 9 '18 at 15:45
• @Ian I use lagrange remainder to control it. – HQR Aug 9 '18 at 15:48
• @HQR The point is that the $\xi$ goes into the evaluation of the derivative. In this non-alternating situation you do not have the freedom to evaluate the derivative at $0$, you are forced to evaluate it at the worst possible point which in this case is $\xi=1$. For $\tan$ this is very bad because the derivatives of $\tan$ grow rather fast, even though they do eventually die down (as we know from analyticity). – Ian Aug 9 '18 at 16:13
• I notice with a calculator that the bound $\tan(1) \leq \sqrt{3}+\sec^2(1)(1-\pi/3)$ is good enough to get the result, but I don't know how much tedious grinding of approximation of $\sqrt{3},\sec(1)$ and $\pi$ would be required to prove that. (Presumably quite a bit, because this number is less than $5 \cdot 10^{-4}$ away from $\pi/2$...) – Ian Aug 9 '18 at 16:29

$$\arcsin(1)=\frac{\pi}{2}$$ 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.

Here is another method. The only real disadvantage is that it requires calculating $e^{-1}$ in decimal form, but only two digits are needed, so this can be done by hand easily. It does not require using very large fractions or remembering the series for $\tan(x)$.

As in another answers, we want to find an upper bound for $\sin(1)$ and a lower bound for $\cos(1)$ so that we can show that $\tan(1) < \frac{\pi}{2} = \sin^{-1}(1)$. From there, given the monotonicity of $\sin^{-1}(x)$, the result follows easily.

$\sin(1)$ can be expanded as

$$\sin(1) = \frac{1}{2}\left[ \sin(1) + \sinh(1) \right] + \frac{1}{2} \left[ \sin(1) - \sinh(1) \right]$$

The first half represents the positive terms in the Maclaurin series for $\sin$, and the second term represents the negative terms. On the second half, just keep the first term:

$$\frac{1}{2} \left[ \sin(1) - \sinh(1) \right] = - \sum_{n=0}^\infty \frac{1}{(4n+3)!} \le -\frac{1}{6}$$

Plugging this into the first equation and solving for $\sin(1)$ above, we get

\begin{align} \sin(1) &\le \sinh(1) - \frac{1}{3} \\ &= \frac{e-e^{-1}}{2} - \frac{1}{3} \\ &\le \frac{2.72 - 0.36}{2} - 0.333 \\ &= 0.847 \end{align}

Now consider the Maclaurin series for $\cos(1)$. The terms are alternating and decrease in magnitude, so for a lower bound, we can terminate after any negative term. Keep the first four terms:

\begin{align} \cos(1) & \ge 1 - \frac{1}{2} + \frac{1}{24} - \frac{1}{720} \\ & \ge 1 - \frac{1}{2} + \left( \frac{1}{24} - \frac{1}{600} \right) \\ & = 1 - \frac{1}{2} + \frac{1}{25} \\ &= 1 - 0.5 + 0.04 \\ & = 0.54 \end{align}

This gives us $$\tan(1) \le \frac{0.847}{0.54} < 1.57 < \frac{\pi}{2} = \sin^{-1}(1)$$

Of course, it would have taken me a really long time to come up with this without the use of a calculator for exploration.

Define $g(x)=\sum_{i=0}^4\frac{(-1)^{i+1}}{(2i+1)}\left(\frac1x\right)^{2i+1}$ and $f(x)=\tan^{-1}x-\left(\frac\pi2+g(x)\right)$. We then have the rather delightful derivatives $f'(x)=\frac{-1}{x^{10}(x^2+1)}$ and $g'(x)=\frac{x^{10}+1}{x^{10}(x^2+1)}$, which tells us that $f(x)$ and $g(x)$ are respectively decreasing and increasing. Then, since $\lim_{x\to+\infty}f(x)=0$, we have that $f(x)>0$. As $f\left(\frac\pi2\right)>0$, we have

$$\tan^{-1}\frac\pi2>\frac\pi2+\underbrace{\sum_{i=0}^4\frac{(-1)^{i+1}}{(2i+1)}\left(\frac2\pi\right)^{2i+1}}_{g(\pi/2)}$$

As $g(x)$ is increasing, we have $g(\pi/2)>g(3.14/2)$. Hence, we may put this all together to say

\begin{aligned}\tan^{-1}\frac\pi2&>\frac{3.14}{2}+\sum_{i=0}^4\frac{(-1)^{i+1}}{(2i+1)}\left(\frac2{3.14}\right)^{2i+1}\\\text{RHS}&=\frac{314}{200}-\frac{200}{314}+\ldots-\frac19\left(\frac{200}{314}\right)^9\\&=1.00238\end{aligned}

Finally, as $\tan x$ is increasing, we have that $\frac\pi2>\tan(1.00238)>\tan(1)$.

Since $$\arcsin1=\pi/2$$, it suffices to show $$2\sin1\lt\pi\cos1$$, in order to conclude that $$\tan1\lt\arcsin1$$.

We have

$$2\sin1=2\left(1-{1\over6}+{1\over120}-\cdots \right)\lt2\left(1-{1\over6}+{1\over120}\right)={101\over60}$$

While

$$\pi\cos1=\pi\left(1-{1\over2}+{1\over24}-{1\over720}+\cdots \right)\gt\pi\left(1-{1\over2}+{1\over24}-{1\over720}\right)={389\over720}\pi$$

so it suffices to note that $$\pi\gt1212/389\approx3.1317829$$. To be a little less calculator-dependent, note that

$${1212\over389}\lt{1215\over387}={135\over43}\lt{314\over100}=3.14\lt\pi$$

since $$135\cdot100=13500\lt13502=314\cdot43$$. (And if desired, even this arithmetic can be further simplified by noting that $${135\over43}=3+{6\over43}$$ and $${314\over100}=3+{7\over50}$$, with $${6\over43}\lt{7\over50}$$ since $$6\cdot50=300\lt301=7\cdot43$$.)