Prove that $f\left(x\right)=\sin\left(x\right)$ is Continuous. The function $f\left(x\right)=\sin\left(x\right)$ is obviously continuous. But how would you prove this using the $\delta,\varepsilon$ definition of continuity? So given $x\in\mathbb{R}$ and $\varepsilon>0$, how do you determine the $\delta>0$ that guarantees that $\left|x-y\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon$?
 A: This depends on what you're allowed to use. One thing we can say is
$$\sin(x+\delta)-\sin(x)=\sin(x)\cos(\delta)+\sin(\delta)\cos(x)-\sin(x) \\
= \sin(x) (\cos(\delta)-1) + \sin(\delta) \cos(x)$$
Now you can control both terms by making $\delta$ small enough. In particular, you can prove, using only trigonometry, that both terms are no larger in magnitude than $\delta$. Then take $\delta=\varepsilon/2$ and use the triangle inequality to finish the proof.
It actually turns out that the first term is much smaller than $\delta$ when $\delta$ is small, which is why this estimate is worse than the one the mean value theorem would give you.
A: Apply the Mean Value Theorem to $f(t) = \sin t$ on the interval $(x, y)$ to see that $|\sin y - \sin x| \le |y - x|$. We can show that $f(t)$ is uniformly continuous by taking $\delta = \epsilon$. Regular continuity follows as an immediate consequence.
A: You can use $\sin(x + h) = \sin x \cos h + \cos x \sin h$ if you know that $\lim_{h \to 0} \sin h = 0$ and $\lim_{h \to 0} \cos h = 1$.
A: Hint: $$\sin(x+h)-\sin x = 2 \sin\left(\frac{h}{2}\right)\cos\left(x+\frac{h}{2}\right).$$
