If $\sin(\theta') \le (1+\alpha)\sin(\theta)$ then $\theta' \le (1+\sqrt\alpha)\theta$ I need to show that if $\sin(\theta') \le (1+\alpha)\sin(\theta)$ then $\theta' \le (1+\sqrt\alpha)\theta$, for small $\alpha$ and $\theta$ and $\theta'$ are between $0$ and $π/2$
I tried to use the series expansions for sine, replacing $x$ with $(1+\sqrt\alpha)x$ , but see no way to collect all the extra $\alpha$ terms together to get the required form. 
 A: We can formulate the problem as follows: Prove that for all $\theta', \theta \in [0, \pi/2]$, there exists an $A > 0$ such that whenever $0 < \alpha \le A$ and $\sin(\theta') \le (1+\alpha)\sin(\theta)$, we have that $\theta' \le (1+\sqrt{\alpha})\theta$.
First, note the inequality is trivial when $\theta' = \theta$, so we may assume WLOG that $\theta' \neq \theta$ (non-equality). Next, note that the first inequality implies that $\sin(\theta') - \sin(\theta) \le \alpha \sin(\theta)$. Mean Value Theorem gives us $\cos(\theta^*)(\theta'-\theta) \le \alpha \sin(\theta)$ for some $\theta^*$ strictly between $\theta'$ and $\theta$ (strict in-betweeness). Hence, $\theta' \le \alpha \frac{\sin(\theta)}{\cos(\theta^*)}+\theta$, which is well defined since $\cos(\theta^*) > 0$, because $\theta^* \in (0, \pi/2)$ by the strict in-betweeness and non-equality established earlier. It suffices to prove this last bound is $\le \theta + \sqrt{\alpha}\theta$, or
\begin{align*}
\sqrt{\alpha} \le \cos(\theta^*) \frac{\theta}{\sin(\theta)}
\end{align*}
Take $\sqrt{A} = \cos(\theta^*) \frac{\theta}{\sin(\theta)}$, and we are done. 
