Limit and Ratio of two figure

kunny

(1) For $\theta > 0$ , prove that $\frac {1}{6}\theta ^ 3 - \frac {1}{120}\theta ^ 5 < \theta - \sin \theta < \frac {1}{6}\theta ^ 3.$

(2) Given a circle $C$ with the center the point $O$ and radius 1. For $0 < \theta <\pi$ , take two points $P,\ Q$ such that $\angle{POQ} = \theta$ . Let $R$ be the intersection point of two tangent lines at $P,\ Q$ on the circle $C$ . Denote by $S_1$ the area of the part bounded by the line segment $PQ$ and the minor arc $PQ$ and denote by $S_2$ the area of the triangle ${PQR}$ . Find $\lim_{\theta \rightarrow + 0} \frac {S_1}{S_2}$

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Today's calculation of Integral Digest

earldbest

for (1)

let $f(x)=x- \sin x + \frac{1}{120}x^{5} -\frac{1}{6}x^{3}$ , $g(x)= x- \sin x -\frac{1}{6}x^{3}$ .

take $f''''(\theta)$ and $g''(\theta)$ , the rest follows Mr. Green

for (2)

consider $\theta < \pi$ .

then $\frac{S_{1}}{S_{2}}=\frac{\theta - \sin \theta}{\sin \theta}$ .

therefore the answer is $0$ . Mr. Green

**Posted:** Sat Nov 07, 2009 8:35 am

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Ninja

42 is the answer to the universe.

kunny

What's the result of $S_2$ ?

**Posted:** Sat Nov 07, 2009 9:10 am

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Today's calculation of Integral Digest