Problem finding normal line equation using partial derivative and gradient vector

I'm trying to solve a problem but the answer is different from the book.

So wondering what am i doing wrong. The problem is like this:

Find equations for the lines that are normal to function:

$\displaystyle

y - \sin{(x)} = 1$ at $\displaystyle P_0\left(\pi, 1)

$

The way i did was first find the gradient vector:

$\displaystyle

\nabla f = \left(-\cos{(x)} . i\right) + 1.j = \left(-\cos{(x)} . i\right) + j

$

Then get the derivative $\displaystyle D_uf\left|_{(\pi,1)}$ which is(because we know that slope

of a function is it's derivative at that point):

$\displaystyle

D_uf\left|_{(\pi,1)} = \sqrt{\left(-\cos{(x)}\right)^2 + (1^2)} = \sqrt{1^2 + 1^2} = \sqrt{2}

$

Finally the equation of normal line is:

$\displaystyle

y - 1 = \sqrt{2}\left(x - \pi\right)

$

Is this right or am i wrong? The answer back of my book is $\displaystyle y = x - \pi + 1 $

Why book's answer is different? Any ideas?