Is $\frac{d \sin\theta}{d \theta} \big|_{\theta=0} = 1$ a consequence of the geometrical definition of $\sin\theta \equiv \frac{opp}{hyp}$?

Question

Is $\dfrac{d \sin\theta}{d \theta} \big|_{\theta=0} = 1$ a consequence of the geometrical definition of $\sin\theta \equiv \dfrac{opp}{hyp}$?

I do not know whether $\dfrac{d \sin\theta}{d \theta} \big|_{\theta=0} = 1$ is an additional constraint on the sine function or not.

Very Intersting question. It's my belif that we should need other constraint because there can be many f such that Df(0)=1. — , Dec 11 '20 at 05:47
Does this answer your question? How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$? — Martin R, Dec 11 '20 at 05:54
Since $$\begin{align}\left.\frac{\mathrm{d}}{\mathrm{d}\theta}\sin(\theta)\right|{\theta=0} &=\lim{\theta\to0}\frac{\sin(\theta)-\sin(0)}{\theta-0}\ &=\lim_{\theta\to0}\frac{\sin(\theta)}{\theta}\&=1 \end{align}$$ Martin R's comment does seem relevant. There is also this answer that might be relevant. — robjohn, Dec 11 '20 at 07:38

score 0 · Answer 1 · answered Dec 11 '20 at 05:55

$$\dfrac{d \sin\theta}{d \theta} \big|_{\theta=0} = \lim _{h \to 0} \dfrac{\sin(0+h) - \sin(0)}h \\ \; \\ \; \\= \lim _{h \to 0} \dfrac{\sin(h) }h $$ The geometric definition of $\sin(h)$ allows us to set up the inequality $$| \sin(h)| \le |h|\le |\tan(h)| $$ ($h$ is in radians) now dividing through by $| \sin(h)|$ ... $$1\le \bigg | \dfrac{ h}{ \sin(h)} \bigg | \le |\sec(h)| $$ So by the squeeze theorem we have $$\lim _{h \to 0} \bigg | \dfrac{ h}{ \sin(h)} \bigg | =1$$ Hopefully you can take it from here

Is $\frac{d \sin\theta}{d \theta} \big|_{\theta=0} = 1$ a consequence of the geometrical definition of $\sin\theta \equiv \frac{opp}{hyp}$?

1 Answers1