A continuous function $f: [-1, 1] \to R$ satisfies that
$$ f(2x^2 - 1) = 2xf(x) $$
Furthermore, another function defined $g: G \to R$ with $G = \{t \in R: t \ne nπ, n \in Z\}$ as
$$ g(t) = \dfrac{f(\cos{t})}{\sin{t}} $$
With these two functions provided, I would like to know
- f(-1) and f(1)
- Show that function $g$ is an odd function
- For every $t \in G$, show that $g(t) = g(t/2)$
I am trying to figure out what is the main point of such a question, and as far as I know about functions, I tried the first and second ones as follows
1. $$ \begin{align*} & 2x^2 -1 = -1, \quad x = 0 \\\ & f(-1) = 2 \cdot 0 \cdot f(-1) = 0 \end{align*} $$
$$ \begin{align*} & 2x^2 -1 = 1, \quad x = \pm1 \\\ & f(1) = 2 \cdot f(1) = -2 \cdot f(1), \quad f(1) = 0 \end{align*} $$
2. $$ \begin{align*} & g(-t) = \dfrac{f(\cos(-t))}{\sin(-t)} = -\dfrac{f(\cos(t))}{\sin(t)} \\ & 2x^2 - 1 = \cos{x} (?) \end{align*} $$
Actually, I think I do not fully understand this question, and I think what I did is probably nonsense. What I want to know is how can I solve them, and what concepts are these questions asking, it looks like this is a composite function, yet I am not sure where to start or to delve into related concepts.
With the advice from Klaus, the answers for 1. and 2. are already described in the OP.
As for the last question, it can be derived from
$$ \begin{align*} g(t/2) & = \dfrac{f(\cos(t/2))}{\sin(t/2)} = \dfrac{f(2\cos^2(t/2) - 1)}{2\sin(t/2)\cos(t/2)} = \dfrac{f(\cos{t})}{\sin{t}} \end{align*} $$