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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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The correct answer is:
$\frac{\sqrt{2}-1}{2 \sqrt{2}}$
According to the question,
$\int_{\frac{\pi}{4}}^a(\sin x-\cos x) d x=\int_a^\pi(\sin x-\cos x) d x$
$\begin{aligned} {[-\cos x-\sin x]_{\frac{\pi}{4}}^a } & =[-\cos x-\sin x]_a^\pi \\ -\cos a & -\sin a+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=1+\cos a+\sin a \\ \sqrt{2}-1 & =2(\cos a+\sin a) \\ \frac{\sqrt{2}-1}{2} & =\sin a+\cos a \\ \frac{\sqrt{2}-1}{2 \sqrt{2}} & =\frac{1}{\sqrt{2}} \sin a+\frac{1}{\sqrt{2}} \cos a \\ \frac{\sqrt{2}-1}{2 \sqrt{2}} & =\sin a \cdot \cos \frac{\pi}{4}+\cos a \sin \frac{\pi}{4} \\ \frac{\sqrt{2}-1}{2 \sqrt{2}} & =\sin \left(a+\frac{\pi}{4}\right)\end{aligned}$
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