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(C) The points of intersection of $y = \sin x$ and $y = \cos x$ are $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.In the interval $\left[\frac{\pi}{4},

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$2\sqrt{2}$

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(C) The points of intersection of $y = \sin x$ and $y = \cos x$ are $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.In the interval $\left[\frac{\pi}{4}, \frac{5\pi}{4}ight]$,$\sin x \geq \cos x$.Therefore,the area is given by the integral:$	ext{Area} = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\sin x - \cos x) \, dx$$= [-\cos x - \sin x]_{\frac{\pi}{4}}^{\frac{5\pi}{4}}$$= -[(\cos x + \sin x)]_{\frac{\pi}{4}}^{\frac{5\pi}{4}}$$= -\left[ \left( \cos\frac{5\pi}{4} + \sin\frac{5\pi}{4} ight) - \left( \cos\frac{\pi}{4} + \sin\frac{\pi}{4} ight) ight]$$= -\left[ \left( -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} ight) - \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} ight) ight]$$= -\left[ -\frac{2}{\sqrt{2}} - \frac{2}{\sqrt{2}} ight]$$= -[ -\sqrt{2} - \sqrt{2} ] = -(-2\sqrt{2}) = 2\sqrt{2}$

The area of one region included between the $sine$ and $cosine$ curves is given by

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