The function $f(x)$ is defined, for $|x|\leqslant1$ by $$f(x)=\frac{\sqrt 3}{2}x+\frac{1}{2}\sqrt{1-x^2}.$$ Find an expression for $$f^n(x)=\underbrace{f \circ f \circ \cdots \circ f(x)}_{\text{n times}},$$where $n\in\mathbb{Z^+}$.
Now what I did was to first find $f^2(x)$ and my intention was to find $f^3(x)$ and another few cases so as to recognise a pattern. However, $f^2(x)$ is actually very complicated and does not simplify too much.
I am looking for hints on how to approach the problem and not complete solutions.
Thanks in advance.
If you note that $x$ may be written as $\cos{\phi}$, then
$$f(\cos{\phi}) = \cos{\left(\phi-\frac{\pi}{6}\right)}$$
On the next iteration,
$$f[f(\cos{\phi})] = f\left [ \cos{\left(\phi-\frac{\pi}{6}\right)} \right ] = \cos{\left(\phi-\frac{2 \pi}{6}\right)} $$
I hope you see the pattern...
EDIT
This answer is not totally correct - it is merely formally correct, as pointed out by @achille hui. The problem lies with the evenness of the cosine. The iteration for $\phi=0$ produces an incorrect result. In fact, it is incorrect for $0 \le \phi \le \pi/6$. What to do?
In the $k$th iteration, consider the sign of $\phi-k \pi/6$:
$$f^{(k)}(\cos{\phi}) = \begin{cases}\cos{\left[\phi - (1-(-1)^k)\frac{\pi}{12} \right]} & 0 \le \phi \le k \frac{\pi}{6} \\ \cos{\left(\phi-k \frac{\pi}{6}\right)} & k \frac{\pi}{6} \lt \phi \le \pi \end{cases}$$
Repeat the process for $k > 6$.