# $\sin 9^{\circ}$ or $\tan 8^{\circ}$ which one is bigger?

$\sin 9^{\circ}$ or $\tan 8^{\circ}$ which one is bigger ?

someone ask me that , and said without using calculator !!
now my question is ,how to find which is bigger ?
Is there a logical way to find ?
I s there a mathematical method to show which is greater ?
I am thankful for your guide , hint or solution

• A brute force solution: Calculate enough terms of the Taylor series of $\sin$ and $\cos$ so that the error is negligible. Then calculate $\tan 8^\circ = \sin 8^\circ/\cos 8^\circ$. – Henricus V. May 21 '17 at 14:42

I would check the first two terms of the Taylor series. $\sin 9^\circ \approx \frac \pi{20}-\frac {\pi^3}{6 \cdot 20^3}, \tan 8^\circ \approx \frac {2\pi}{45}+\frac {8\pi^3}{3 \cdot 45^3}$, so $$\sin 9^\circ -\tan 8^\circ\approx \frac \pi{20}-\frac {\pi^3}{6 \cdot 20^3}-\frac {2\pi}{45}-\frac {8\pi^3}{3 \cdot 45^3}\\\approx \frac \pi{180}-\frac{(3^6+2^{10})\pi^3}{2^73^75^3}\\ \approx \frac \pi{180}(1-\frac {1753}{2^43^55}) \\ \approx \frac \pi{180}(1-\frac {1753}{17440})\\ \gt 0$$ where I used $\pi^2 \approx 10$. Alpha agrees, but I didn't check until I was done.

When $0<x<1$ then$$\sin x>x-{x^3\over6},\qquad\tan x={\sin x\over \cos x}<{x\over 1-{x^2\over2}}\ ,$$ by the theorem on alternating series. Let $t:=1^\circ={2\pi\over 360}<{1\over50}$. Then $${\sin(9t)\over\tan(8t)}>{9t\over 8t}\left(1-{81t^2\over6}\right)\left(1-{64t^2\over2}\right)>{9\over8}\left(1-{273\over6}t^2\right)>1\ .$$

In terms of radian, $9^\circ = \frac{\pi}{20} \approx 0.157$ is reasonable small. We can use Taylor series expansion to estimate the value of $\sin$ and $\tan$. For small $\theta$, we have

$$\sin\theta \approx \theta - \frac{\theta^3}{6} \quad\text{ and }\quad \tan\theta = \frac{\sin\theta}{\cos\theta} \approx \frac{\theta - \frac{\theta^3}{6}}{1 - \frac{\theta^2}{2}} \approx \theta + \frac{\theta^3}{3}$$ This implies $$\tan^{-1}\theta \approx \theta - \frac{\theta^3}{3}\quad\text{ and }\quad \tan^{-1}(\sin\theta) \approx \theta - \frac{\theta^3}{6} - \frac{\theta^3}{3} = \theta\left(1 - \frac{\theta^2}{2}\right)$$

In order for $\tan\phi$ equals to $\sin 9^\circ$, $\phi$ should be around $$9^\circ \times \left( 1 - \frac{0.157^2}{2}\right) \approx 9^\circ \times 0.98\;(\text{ or } 0.99???)$$ No matter what the actual value of last factor is, the $\phi$ required to make $\tan\phi = \sin 9^\circ$ is much closer to $9^\circ$ than $8^\circ$. This means $\tan 8^\circ \le \sin 9^\circ$.

I'll show a theoretical way to get at this answer without these Taylor/Maclaurin series featured in the other answers, though this would take a lot of time to do by hand.

### Setup

Let $s=\sin x^\circ$ and note that for $x<\dfrac\pi2$, we have $\cos x^\circ=\sqrt{1-s^2}$. Then by using the sum formulas repeatedly, we find that: $$\tag{1}\sin \left(9x^\circ\right)=256s^9-576s^7+432s^5-120s^3+9s$$ $$\tag{2}\tan \left(8x^\circ\right)=\dfrac{\left(-128s^7+192s^5-80s^3+8s\right)\sqrt{1-s^2}}{128s^8-256s^6+160s^4-32s^2+1}$$

To compare $\sin 9^\circ$ and $\tan 8^\circ$, we need a bound on $\sin 1^\circ$.

### Estimating $\sin 1^\circ$

One known value is $\sin 15^\circ=\dfrac{\sqrt3-1}{2\sqrt2}$. The sum formulas give us that $\sin \left(15x^\circ\right)$ is
$$-16384 s^{15}+61440 s^{13}-92160 s^{11}+70400 s^9-28800 s^7+6048 s^5-560 s^3+15 s$$ Evaluating that polynomial at $s=\dfrac{1}{20}$ yields $\dfrac{1363735274101499}{2000000000000000}>\dfrac12>\dfrac{\sqrt3-1}{2\sqrt2}$. Therefore, $0<\sin1^\circ<\dfrac{1}{20}$ for sure. (In reality, $\sin1^\circ$ is more like $1/50$.)

### Conclusion

Evaluating the polynomials from (1) and (2) at $s=\sin x^\circ=\dfrac{1}{20}$ yield $\sin \left(9x^\circ\right)=\dfrac{870269101}{2000000000}$ and $\tan\left(8x^\circ\right)=\dfrac{3900599\sqrt{399}}{184199201}<\dfrac{3900599*20}{184199201}<\dfrac{870269101}{2000000000}$.

It remains to check why this really means $\sin 9^\circ$ is greater than $\tan8^\circ$. Since $\tan$ has an asymptote at $90^\circ$, $\tan{8x^\circ}$ must overtake $\sin{9x^\circ}$ sometime before $x=90/8$, so the expression has an asymptote at $s=\sin\left(\dfrac{90^\circ}{8}\right)=\dfrac12\sqrt{2-\sqrt{2+\sqrt2}}\approx\dfrac15$. By considering concavity, the $\tan$ function should overtake the $\sin$ function exactly once before this, and since it hasn't happened by $s=\dfrac{1}{20}>\sin 1^\circ$, $\boxed{\sin 9^\circ>\tan8^\circ}$ after all.

### Pictures

Here is a graph of the two functions of $s$. $\tan$ overtakes $\sin$ before $s=.06$.

If you had large enough drawing materials, you might be able to compare the values directly with a construction like this:

This question can be resolved by expressing these values in the their expansions, to wit

$$\sin x=x-\frac{x^6}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+...\\ \tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+...$$

This should put you on the path to determine which of the two is the larger. To lowest order, the sine is clearly larger, so you then look at the second term. The rest will be higher order terms that do not affect the outcome. (It's probably true of the second term as well.)

• The last term in the sine expansion should be negative – Ross Millikan May 21 '17 at 14:52
• You're right, but the series is infinite, so there really is no last term... – Obinna Nwakwue May 21 '17 at 15:02
• @ObinnaNwakwue: I meant the last displayed term – Ross Millikan May 21 '17 at 15:13
• I know, but you're right anyways. – Obinna Nwakwue May 21 '17 at 15:19
• @RossMillikan You've a keen eye, thanks. I've corrected the equation. – Cye Waldman May 21 '17 at 15:28