have trouble with this limit question 
a) By considering the areas of the triangle OAD, the sector OAC and the triangle OBC,
show that
$(\cos \theta)(\sin \theta) < \theta < \frac{\sin\theta}{\cos\theta}$
I find out:
Area of OAD=$\frac{1}{2}OD\cdot AD \cdot \sin \theta$
Area of OAC=$\frac{1}{2}OC^2 \theta$
Area of OBC=$\frac{1}{2}OC\cdot BC \cdot\sin\theta$
Now I'm stuck at how to apply this to prove
How to prove?
(b) Use (a) and the Squeeze Theorem to show that
$\displaystyle\lim_{\theta\to 0^+}\frac{\sin\theta}{\theta}= 1$

The geometry has been discussed in comments. So we have 
$$\frac{1}{2}\cos\theta\lt \frac{1}{2}\theta.$$
Multiply through by $2$. We get $\cos\theta\sin\theta\lt \theta$.
Divide both sides by $\theta\cos\theta$. We get
$$\frac{\sin\theta}{\theta}\lt \frac{1}{\cos\theta}.\tag{1}$$
We also got from the geometry that
$$\frac{1}{2}\theta\lt \frac{1}{2}\frac{\sin \theta}{\cos\theta}.$$ Multiplying by $2$, and rearranging, we get $\theta\cos\theta\lt \sin\theta$. Divide through by $\theta$. We get
$$\cos\theta\lt\frac{\sin\theta}{\theta}.\tag{2}$$
Now putting (1) and (2) together, we get
$$\cos\theta\lt \frac{\sin\theta}{\theta}\lt \frac{1}{\cos\theta}.\tag{3}$$
Finally, let $\theta\to 0$. The left wall and the right wall in (3) both approach $1$, and poor $\frac{\sin\theta}{\theta}$ is therefore forced to approach $1$. 
