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

Question

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?

Answer 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.

Answer 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 $\tan(1)\gt\frac\pi2$.

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

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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.

Answer 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

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

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

Answer 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 $$ \iint_{(0,1)^2}\frac{x^m(1-x)^m y^n (1-y)^n}{1-xy}\,dx\,dy.$$

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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.

Answer 5

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.

Answer 6

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 $\frac\pi2>\tan(1.00238)>\tan(1)$.

Answer 7

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 $6\cdot50=300\lt301=7\cdot43$.)