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
$\displaystyle\lim_{\theta\to 0^+}\frac{\sin\theta}{\theta}= 1$
Answer
Hint: WORK IN RADIANS!
$a)$ $$\text{Area of }\Delta OAD=\dfrac{1}{2}\cdot OD\cdot AD\\
\text{Area of sector }OAC=\dfrac{\theta}{360}\pi (OA)^2\\
\text{Area of }\Delta OBC=\dfrac{1}{2}\cdot OC\cdot BC$$
See that
$$\text{Area of }\Delta OAD<\text{Area of sector }OAC<\text{Area of }\Delta OBC\\
\implies \dfrac{1}{2}\cdot \cos\theta\cdot \sin\theta<\dfrac{\theta}{360}\pi (OA)^2<\dfrac{1}{2}\cdot 1\cdot BC$$
Now,
$$DC=1-\cos\theta\\
BC=\tan\theta$$
$b)$ Then, after doing $a)$, use the fact that
$$\dfrac{1}{2}\sin\theta\cos\theta<\theta/2\\
\theta/2<\dfrac{1}{2}\tan\theta$$
Then use the squeeze theorem. The limit follows.
No comments:
Post a Comment