If the area of the region bounded by y=cos x, | vedclass

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(C) The area of the region bounded by $y=\sin x$ and $y=\cos x$ from $x=\frac{\pi}{4}$ to $x=\pi$ is given by the integral of the upper curve minus th

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$\frac{\sqrt{2}-1}{2 \sqrt{2}}$

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(C) The area of the region bounded by $y=\sin x$ and $y=\cos x$ from $x=\frac{\pi}{4}$ to $x=\pi$ is given by the integral of the upper curve minus the lower curve. In the interval $[\frac{\pi}{4}, \frac{\pi}{2}]$,$\sin x \ge \cos x$,and in $[\frac{\pi}{2}, \pi]$,$\sin x \ge \cos x$ (since $\cos x$ is negative). Thus,the total area is $\int_{\frac{\pi}{4}}^{\pi} (\sin x - \cos x) dx$.According to the problem,the line $x=a$ bisects this area,so:$\int_{\frac{\pi}{4}}^{a} (\sin x - \cos x) dx = \int_{a}^{\pi} (\sin x - \cos x) dx$Evaluating the integral:$[-\cos x - \sin x]_{\frac{\pi}{4}}^{a} = [-\cos x - \sin x]_{a}^{\pi}$$(-\cos a - \sin a) - (-\cos \frac{\pi}{4} - \sin \frac{\pi}{4}) = (-\cos \pi - \sin \pi) - (-\cos a - \sin a)$$-\cos a - \sin a + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = -(-1) - 0 + \cos a + \sin a$$-\cos a - \sin a + \frac{2}{\sqrt{2}} = 1 + \cos a + \sin a$$\sqrt{2} - 1 = 2(\sin a + \cos a)$$\sin a + \cos a = \frac{\sqrt{2}-1}{2}$Divide both sides by $\sqrt{2}$:$\frac{1}{\sqrt{2}} \sin a + \frac{1}{\sqrt{2}} \cos a = \frac{\sqrt{2}-1}{2 \sqrt{2}}$$\sin a \cos \frac{\pi}{4} + \cos a \sin \frac{\pi}{4} = \frac{\sqrt{2}-1}{2 \sqrt{2}}$$\sin \left(a + \frac{\pi}{4}\right) = \frac{\sqrt{2}-1}{2 \sqrt{2}}$

If the area of the region bounded by $y=\cos x$,$y=\sin x$,$x=\frac{\pi}{4}$ and $x=\pi$ is bisected by the line $x=a$,then $\sin \left(a+\frac{\pi}{4}\right)=$

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