The equation of the normal to the curve y = sin x at (0, 0) is :

The correct answer is Option (2) → $x+y= 0$

$y=\sin x$

$y'=\cos x$

$\left.y'\right]_{(0,0)}=1$

So slope of normal = $\frac{-1}{\left.y'\right]_{(0,0)}}=-1$

so eq: $y-0=-1(x-0)$

$⇒x+y= 0 $