We seek to show that the inequality $|\sin(n\theta)| \leqslant n \sin(\theta)$ is satisfied for all $n \in \mathbb{N}^*$ and all $\theta \in \left[0, \frac{\pi}{2}\right]$.
a) Show that $\sin(n\theta) \leqslant n \sin(\theta)$ for all $n \in \mathbb{N}^*$ and all $\theta \in \left[0, \frac{\pi}{2n}\right]$.
b) Show that, for all $\theta \in \left[0, \frac{\pi}{2}\right]$, we have $\sin(\theta) \geqslant \frac{2}{\pi} \theta$.
c) Deduce that:
$$\forall \theta \in \left[\frac{\pi}{2n}, \frac{\pi}{2}\right], \quad 1 \leqslant n \sin(\theta)$$
d) Conclude.
e) For which values of $\theta \in \left[0, \frac{\pi}{2}\right]$ do we have $|\sin(n\theta)| = n \sin(\theta)$?