Bijectivity of sinx on the interval $[-\pi, \pi]$ to $[-1, 1]$.

Question

how can I prove that sinx is a bijective map from the domain $[-\pi, \pi]$ to the co-domain $[-1,1]$. I had no problem proving that using the graphical representation of sinx, but rigorously could not.

Any help

Answer 1

$\sin$ is not a bijective map from $[-\pi, \pi]$ to $[-1, 1]$. If it was, then $\forall a,b \in [-\pi, \pi]$, $\sin(a) = \sin(b) \implies a = b$. However notice that $\sin(-\pi) = \sin(\pi) = 0$ and $-\pi \neq \pi$. Therefore we have a contradiction.

Restrict yourself to $[-\pi/2, \pi/2]$ and then you have bijectivity.

Answer 2

Let $f(x)=x/2$ from $[-π, π]$ to $[-π/2,π/2]$is a bijection. Let $g(x)=sin x $ from $[-π/2,π/2]$ to $[-1,1]$ is a bijection. Taking composition of these two function.

The required function is $h(x)=sin(x/2)$.