Why is $ \left| \frac{\sin(\theta+\epsilon)-\sin(\theta)}{\epsilon} \right| \leq 1$ true?

Question

I am conducting an error analysis of some sensor measurements associated with uncertainties. I ended up with the equation $\mathcal{S}$ of the form below. I numerically simulated the equation and noticed that it is always bounded by $\pm 1$. Nonetheless, I need to prove that mathematically. So my question is why the following inequality holds for any $\theta$ and $\epsilon$:

$$ \left| \mathcal{S} = \frac{\sin(\theta+\epsilon)-\sin(\theta)}{\epsilon} \right| \leq 1. $$

Thanks.

$$\sin a - \sin b=2\sin\left(\frac{a-b}2\right)\cos\left(\frac{a+b}2\right)$$ — Stefan Lafon, Jun 26 '23 at 22:55
Minus the limit notation, this is basically the limit definition of the derivative for sine, which is cosine and is always less than or equal to 1 — H. sapiens rex, Jun 26 '23 at 22:55
Quick beginner guide for asking a well-received question + please avoid "no clue" questions. — Anne Bauval, Jun 26 '23 at 22:56
Duplicate: Show that $|\sin{a}-\sin{b}| \le |a-b| $ for all $a$ and $b$. — Anne Bauval, Jun 27 '23 at 00:21

score 0 · Answer 1 · answered Jun 28 '23 at 09:06

0

Since the function sinus is of classe C^\infty, then by the Lagrange's theorem there exists \lambda belongs to (\theta, \theta + \epsilon) such that

sin(\theta + \epsilon) - \sin(\theta) = \epsilon \sin(\lambda)

and by boundedness of sinus, you obtain the result.

answered Jun 28 '23 at 09:06

Isaac

11

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Jun 28 '23 at 09:09

Why is $ \left| \frac{\sin(\theta+\epsilon)-\sin(\theta)}{\epsilon} \right| \leq 1$ true?

1 Answers1