#### Answer

$$ y=\frac{\sqrt{2}}{2} x+\frac{\sqrt{2}}{2}\left(1-\frac{\pi}{4}\right).$$

#### Work Step by Step

Since $ y=\sin x $, then $ y'=\cos x $ and hence the slope at $ x=\frac{\pi}{4}$ is $ m=\cos \frac{\pi}{4}=\frac{\sqrt{2}}{2}.$ Now, the tangent line equation is given by
$$ y=\frac{\sqrt{2}}{2} x+c.$$
Since the curve and line coincide at $ x=\frac{\pi}{4}$, then
$$\sin \frac{\pi}{4}= \frac{\sqrt{2}}{2}\frac{\pi}{4}+c \Longrightarrow c=\frac{\sqrt{2}}{2}\left(1-\frac{\pi}{4}\right) .$$
Hence the equation of the tangent line is given by
$$ y=\frac{\sqrt{2}}{2} x+\frac{\sqrt{2}}{2}\left(1-\frac{\pi}{4}\right).$$