The equation to the normal to the curve

Given that $y=\sin x$

Slope of the tangent $\frac{d y}{d x}=\cos x$

Slope at origin $=\cos 0=1$

Equation of normal:

$\left(y-y_{1}\right)=\frac{-1}{\text { Slope of tangent }}\left(x-x_{1}\right)$

$\Rightarrow(\mathrm{y}-0)=\frac{-1}{1}(\mathrm{x}-0)$

$\Rightarrow \mathrm{y}+\mathrm{x}=0$

$\Rightarrow y+x=0$

Hence option C is correct.