How to show that $\Psi: E \rightarrow E$, $\Psi(f) = \sin(f(t))$ is continuous and differentiable? 
Let $E = \mathcal{C}([0,1],\mathbb{R})$ a Banach space of continuous fonctions mapping from $[0,1]$ to \mathbb{R}, with the norm $||f|| = \underset{t \in [0,1]}{\sup}|f(t)|$. Let $\Psi : E \rightarrow E$ defined as $\Psi(f) = \sin(f(t)), \forall f \in E, \forall t \in [0,1] $.
  1) Verify that $\Psi$ is continuous
  2) Show that $\Psi$ is Frechet differentiable on $E$, and find its differential.

I trying to solve this exercise, but I struggle a lot. I think that to show its continuity, you have to show that $\Psi$ is a composite function of two continuous functions. Once you manage to describe $\Psi$ with the composition of two functions, answering the second question might be more easier.
The issue I get with the composition, is that my intuition first tells me that $$\Psi(f) = g \circ u $$ where $u:E \rightarrow \mathbb{R}$ such that $u(f) = f(t)$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ such that $g(x)=sin(x)$ Now this is obviously false as $\Psi$ maps to $E$ and not $\mathbb{R}$.
Can someone help me to find the two functions that fit?
 A: Consider $f,f_0\in E$ and let $g=\Psi(f)-\Psi(f_0)$. Note that for $x\in [0,1]$, $$|g(x)|=|\sin(f(x))-\sin(f_0(x))|\leq |f(x)-f_0(x)|$$
since $\sin$ is $1$-Lipschitz (see this proof)
Thus $|g(x)|\leq \|f-f_0\|$ for all $x$, hence $\|g\|\leq \|f-f_0\|$ that is $\|\Psi(f)-\Psi(f_0)\|\leq \|f-f_0\|$. Hence $\Psi$ is $1-$Lipschitz, thus continuous.

Fix some $f\in E$. Let $h\in E$ and $x\in [0,1]$. Note that $$\Psi(f+h)(x)-\Psi(f)(x) = \cos(f(x))h(x)+ \cos(f(x))(\sin(h(x))-h(x))+\sin(f(x))(\cos(h(x))-1)$$
which rewrites functionally as $$\Psi(f+h)-\Psi(f) = h\cdot \cos\circ f + (\sin\circ h -h ) \cdot \cos\circ f+(\cos\circ h -1) \cdot \sin\circ f$$
A candidate for the differential of $\Psi$ at $f$ is $h\mapsto h\cdot \cos\circ f$ which is obviously linear and continuous.
It suffices to prove that $(\sin\circ h -h )= o(h)$ and $\cos\circ h -1 = o(h)$. This is a simple consequence of the inequalities $$|\sin(x)-x|\leq \frac{x^2}{2}$$
$$|\cos(x)-1|\leq \frac{x^2}{2}$$
which you can prove using Taylor theorem with integral remainder.
A: Fix a function $f\in E$, and expand $\sin(f+h)$. Notice that if you put $A=\cos f$ then 
$$\lim_{||h||\to 0}\frac{||\cos f\sin h+\sin f\cos h-\sin f-h\cos f||}{||h||}=0,$$
via some easy calculations. So the Frechet derivative is $\cos f$ at every point $f$. This also imply the continuity of your map at any point of $E$.
