A trigonometry problem...

[EnmitySpecter]

I am having difficulty with a problem that reads as follows:

A circle with radius r has central angle theta measured in radians. Prove that the area A of the shaded region is A = 1/2 * r^2 (theta - sin of theta).

It comes with a picture of a sector of a circle, and in a triangle POD with O origin, it would be split by line PD, with the non-triangle portion as the shaded region.

I don't really know how to solve it. I'm stuck on this idea of "1/2 * r^2 - 1/2 bh" but that gets me nowhere, since in order to find b or h I have to split the angle and then whatever I find is useless. I don't know how sin of theta can even come into the equation.

Help me please.

 

[galactus]

You could use ratios.

Let A=area of sector and \(\displaystyle {\theta}\)=central angle of the sector.

\(\displaystyle \L\\\frac{area\ of\ sector}{area\ of\ circle}=\frac{central\ angle\ of\ sector}{central\ angle\ of\ circle}\)

Which equals: \(\displaystyle \L\\\frac{A}{{\pi}r^{2}}=\frac{{\theta}}{2{\pi}}\)

Solving for A, we get \(\displaystyle \L\\\frac{1}{2}r^{2}{\theta}=area\ of\ sector\)

The area of a isosceles triangle is given by \(\displaystyle \L\\\frac{1}{2}r^{2}sin{\theta}\)
This proof can be found in a trig book or do a google search.

The area of the segment of a circle is given by the area of the sector minus the isosceles triangle with area \(\displaystyle \L\\\frac{1}{2}r^{2}sin{\theta}\)

So, we have \(\displaystyle \L\\\frac{1}{2}r^{2}{\theta}-\frac{1}{2}r^{2}sin{\theta}\)

Factoring gives: \(\displaystyle \L\\\frac{1}{2}r^{2}({\theta}-sin{\theta})\)

 

[daon]

It is known (at least Im assuming so in your class) that \(\displaystyle \L\\ A_a= \frac{\theta}{2}r^2\) for a portion of a circle.

The area of a triangle is \(\displaystyle \L\\ A_t= \frac{1}{2} bh\)

b= \(\displaystyle r\) and h= \(\displaystyle rsin\theta\)

\(\displaystyle \L\\ A_t=\frac{1}{2}r^2 sin\theta\)

The area of the shaded region will be

\(\displaystyle \L\\ A_a-A_t\) = \(\displaystyle \L\\ \frac{1}{2}r^2[\theta - sin\theta]\)

edit. I guess I was a little late :!:

But anyway, I am a little confused. Is the triangle penetrating the circumference of the circle, or is it interior? If the triangle is interior, then the base of the tringle is \(\displaystyle rcos\theta\) and not r. If this is the case then the area of the shaded region is:
\(\displaystyle \L\\ \frac{1}{2}r^2[\theta - sin\theta cos\theta]\)