Answer
$$y=-x+\pi$$
Work Step by Step
Finding the derivative of the curve:
$y'=\frac{d\sin(\sin{x})}{dx}$
Using the chain rule:
$y'=\frac{d\sin(\sin{x})}{d\sin{x}}
\times \frac{d\sin{x}}{dx}$
$=\cos(\sin{x}) \times\cos{x}$
Finding the slope ($m$) to the tangent line at ($\pi$,0):
$m=y'(\pi)$
$=\cos(\sin{\pi}) \times\cos{\pi}=\cos(0)\times(-1)=1\times(-1)=-1$
From the point-slope form of a linear equation:
$y-0=-1(x-\pi)$
Thus, an equation of the tangent line would be:
$y=-x+\pi$
