Newton method for finding closest point

Question

A function $\ y=\sin(\frac x 2) $ is given. How to determine a point which is the closest to $(1,0)$, using Newton method?

Answer 1

Hint: To do this you should minimize the distance of the point with the curve which is: $$\min_x\left((x-1)^2 + \sin^2\left(\frac{x}{2}\right)\right)$$

Also, to minimize this, you should find where the above derivative would be zero. There, you can use newton method.

$$2(x-1) + \sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) = 0 \Rightarrow 2(x-1) + 0.5\times sin(x) = 0$$