Area of region lying inside r(t) = sin(2t) and r(t) = cos(2t)

Dear all,

I am having trouble with the following exercise and would like to ask if the step underlined in red could be please explained.

I understand everything up to that point but don't understand why only $\displaystyle \sin 2\theta $ is considered in the integral. What about $\displaystyle \cos 2\theta$?

Also, what does the $\displaystyle 8 \cdot 2\int^{\pi/8}_{0} \frac{1}{2} \sin^2 2\theta d\theta $ mean? Does it represent the fact that the wedge (highlighted in light green below in my graph) is multiplied by 16 times?

Thank you very much.

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Question: Find the area of the region that lies inside both $\displaystyle r = \sin 2\theta $ and $\displaystyle r = \cos 2\theta $.

**Given solution: http://img830.imageshack.us/img830/9516/is2js10431.jpg**

My graph: http://img838.imageshack.us/img838/4710/isjs10431.jpg