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Possible Duplicate:
Solving $2x - \sin 2x = \pi/2$ for $0 < x < \pi/2$

I need a little nudge for this problem. I have figured out everything but the last bit(albeit the most important!).

Two equal circles of radii $a$ intersect and the common chord subtends an angle of $2\theta$ radians at either center. Find an expression for the area of the region common to the two circles.

If this area is equal to half the area of either circle, estimate the value of $\theta$ which satisfies this equation

Using the sector area formula, I found the area of the minor segment = $\dfrac{1}{2}a^2(2\theta - \sin 2\theta)$

Hence area of overlapping region = $a^2(2\theta - \sin 2\theta)$

Since this area is half of either circle, $a^2(2\theta - \sin 2\theta) = \dfrac{1}{2}\pi a^2$

Simplifying further, I got,

$\sin 2\theta = 2\theta - \dfrac{\pi}{2}$

This is how far I have gotten. I don't know how to solve this equation. Can you guys help? Thanks again.

Community
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mathguy80
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  • Hint: Let $\phi = 2\theta - \pi/2$, and express the equation in terms of $\phi$, after simplifying $\sin (\phi + \pi/2)$. You still get an equation that you can't solve, but you can draw a graph to estimate the solution. (Or use the first two terms of the Taylor expansion.) – TonyK Jul 01 '11 at 10:03
  • Nice substitution, makes for an easier graph. Thanks. – mathguy80 Jul 01 '11 at 12:40

2 Answers2

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I think the point of the question is that you are asked to estimate the value. I suggest that you plot the two curves $\sin 2\theta$ and $2\theta - \dfrac{\pi}{2}$ and estimate where they intersect.

Tpofofn
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Well your work so far implies:

$\theta = \dfrac{\sin 2\theta}{2}+ \dfrac{\pi}{4}$

And you know that $\sin 2\theta$ is bounded.

You are asked to estimate ... so could you use an iterative method, or a graphical method, or (likely messier) expand the series for $\sin 2\theta$ ...

Mark Bennet
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  • So, continuing the thought a little, you can see that $\sin 2\theta$ is positive, for the range of possibilities for $\theta$ given by the bound. – Mark Bennet Jul 01 '11 at 10:43
  • I did the graphical solution. The iterative method opened up some new ideas I need to learn, Thanks. – mathguy80 Jul 01 '11 at 12:39