$\sin 9^{\circ}$ 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 

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 $\tan 8^\circ \le \sin 9^\circ$.
