Answer
(a) $y=2\pi x-2\pi$
(b) $y=-\frac{1}{2\pi}x+\frac{1}{2\pi}$
Work Step by Step
To verify that the given point is on the curve, plug-in the coordinates to both sides of the equation: $LHS=0$, $RHS=2 sin(\pi-0)=0=LHS$
(a) Given the equation $y=2sin(\pi x-y)$, differentiate both sides with respect to $x$: $y'=2cos(\pi x-y)(\pi -y')$, which gives $y'=\frac{2\pi cos(\pi x-y)}{1+2cos(\pi x-y)}$
Thus the slope of the tangent line at the given point is $m=y'=\frac{2\pi cos(\pi -0)}{1+2cos(\pi -0)}=\frac{-2\pi }{1-2}=2\pi$ and the equation for the tangent line is $y-0=2\pi (x-1)$ or $y=2\pi x-2\pi$
(b) The slope of the line normal to the tangent line is $n=-\frac{1}{m}=-\frac{1}{2\pi}$, and the line equation is $y-0=-\frac{1}{2\pi}(x-1)$ or $y=-\frac{1}{2\pi}x+\frac{1}{2\pi}$
