$\sin 9^{\circ}$ 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: 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.
A: 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 $\tan 8^\circ \le \sin 9^\circ$.
A: 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
$${\sin(9t)\over\tan(8t)}>{9t\over 8t}\left(1-{81t^2\over6}\right)\left(1-{64t^2\over2}\right)>{9\over8}\left(1-{273\over6}t^2\right)>1\ .$$
A: 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.)
