Maximum value of sin(x/2)

Question

I am trying to solve a fixed point problem and I have the $\sin(x/2)$ in my function. After checking that my function is defined in a closed interval which is convex and also maps elements onto itself. I am trying to define the interval in which I will solve the fixed point problem.

Here is the function: $$F(x,y)= (\sin(x/2)\cos(y-1) + 1/4, \cos(x+1)\sin(y/2) - 1/2)$$in $[-1,1]\times[-1,1].$

So normally $|\sin(a)|\le1$ and I think it remains true regardless of $a=x/2.$ But apparently, in the defined intervals above, $\sin(x/2)\in[-1/2,1/2].$

Can someone help with why please? I am kinda lost here

Answer 1

If one does not want to accept the graphical solution, as suggested in the comments, the inequality $|\sin(\alpha)| \leq |\alpha|$, $\forall \alpha \in \mathbb{R}$ (which can be established e.g. using Taylor's formula) might be useful here.

Answer 2

The first thing to notice that that it is still true that $|\sin(\frac{x}{2})| \leq 1$ in the domain of interest. However, what's happened is that we're in a situation where this isn't a tight bound, i.e. there are values $a$ such that $|\sin(\frac{x}{2})| \leq a < 1$ over the domain.

We can use the extreme value theorem (or a variant of it), which tells us that the function takes its global maximum and minimum values either (a) where the derivative equals zero or doesn't exist, or (b) at the boundary of its domain. The places where $\sin(\frac{x}{2})$ would have zero derivative are all outside the interval $[-1, 1]$ (they happen when $x = \pm \pi, \pm 3 \pi, \ldots$), so the maximum value has to happen on the bounary, i.e. at either $x = -1$ or $x = 1$.

We can then see that this means that on the interval $[-1, 1]$, $\sin(\frac{x}{2})$ has a minimum of $-\sin(\frac{1}{2}) \approx -0.48$, and a maximum of $\sin(\frac{1}{2}) \approx 0.48$.

This is fine, though - as I said at the beginning, this doesn't invalidate the requirement that $|\sin(\frac{x}{2})| \leq 1$, because all that's happened is that we've shown $|\sin(\frac{x}{2})| \leq \sin(\frac{1}{2}) < 1$.