Let $t > 0$. I am trying to show that then $\frac{\sin(t\theta)}{\sin(\theta)}\leq t$. I am under the assumption that this shouldn't be too hard, which leads me to believe that I must be missing something obvious as I am stuck with the proof. One observation is that you can show that $\lim_{\theta\to 0}\frac{\sin(t\theta)}{\sin(\theta)} = t$ with L'Hopital's rule. So it would suffice to show that for no $\theta\in\mathbb{R}$, $\frac{\sin(t\theta)}{\sin(\theta)} > t$.
Proving that $\frac{\sin(t\theta)}{\sin(\theta)}\leq t$ for $t > 0$
real-analysistrigonometry
Best Answer
This is a nice problem, but the statement needs to be modified slightly: $$ \frac{\sin t\theta}{\sin\theta} \le t \text{ for all integers } t>0. $$ To prove this, let $\omega = e^{i\theta}$ so that $\sin \theta = (\omega - \omega^{-1})/(2i)$. Then $$ \frac{\sin t\theta}{\sin\theta} = \frac{\omega^t - \omega^{-t}}{\omega - \omega^{-1}} = \omega^{t-1} + \omega^{t-3} + \cdots + \omega^{3-t} + \omega^{1-t}, $$ which one can verify by multiplying each side by $\omega - \omega^{-1}$. Because there are $t$ terms in the expansion above, and because $|\omega^j| = 1$ for any integer $j$, $$ \left|\frac{\sin t\theta}{\sin\theta} \right| \le |\omega^{t-1}| + |\omega^{t-3}| + \cdots + |\omega^{3-t}| + |\omega^{1-t}| = t, $$ from which the result follows (giving a lower bound of $-t$ as well).