Proof $\operatorname{Sin}(x)$ is continuous using addition formula 
Show that the sine function $\operatorname{Sin}: \mathbb{R}\to\mathbb{R}, x\to\sin(x)$ is continuous.

I know how to do other proofs but the hint given is: Write $\sin(x)$ as $\sin(a+(x-a))$.
I know that this then is equal to $\sin(a)\cos(x-a)+\sin(x-a)\cos(a)$, but I don't know where to go from here.
 A: If one knows that, as $h \to 0$,
$$
\sin h \to 0, \qquad \cos h\to1,
$$ then using what you have found
$$
\sin x=\sin(a)\cos(x-a)+\sin(x-a)\cos(a)
$$ gives, as $x \to a$,
$$
\sin x \to \sin a,
$$ since both functions $\sin$ and $\cos$ are bounded by $1$ over $\mathbb{R}$.
A: Aside from addition formulas, we will assume following facts of sine and cosine:


*

*$\sin\theta$ is monotonic increasing from $0$ to $1$ on $[0,\frac{\pi}{2}]$.

*$\cos\theta$ is monotonic decreasing from $1$ to $0$ on $[0,\frac{\pi}{2}]$.


For any $\theta \in (0,\frac12)$, take an integer $N \ge 2$ such that $N\theta \le 1 < (N+1)\theta$. We have
$$\sin(N\theta) = \sum_{k=1}^N(\sin(k\theta) - \sin((k-1)\theta)
= \sum_{k=1}^N 2\sin\frac{\theta}{2}\cos\left(\left(k-\frac12\right)\theta\right)$$
On the LHS, $N\theta \in (0,1]$ and $\sin\theta$ increasing there implies $\sin(N\theta) \le \sin(1)$.
On the RHS, all $\left(k - \frac12\right)\theta \in (0,1]$ and $\cos\theta$ decreasing there implies $\cos\left(\left(k - \frac12\right)\theta\right) \ge \cos(1)$.  
Combine these, we find
$$ 
\sin(1) \ge 2N\sin\frac{\theta}{2}\cos(1)
\quad\implies\quad
2\sin\frac{\theta}{2} \le \frac{\tan(1)}{N} = \theta\frac{\tan(1)}{N\theta}
\le \theta\frac{\tan(1)}{1 - \theta} \le (2\tan(1))\theta
$$
The last two inequalities are true because $N\theta \le 1 < (N+1)\theta$ and $\theta < \frac12$. 
For any $\epsilon > 0$, take $\delta = \min\left(\frac{\epsilon}{2\tan(1)}, \frac12\right)$.
For any $x < y$ with $|x - y| < \delta$, above result implies
$$|\sin x - \sin y| = 2\left|\sin\left(\frac{x-y}{2}\right)\right|\left|\cos\left(\frac{x+y}{2}\right)\right|
\le 2 \left|\sin\left(\frac{x-y}{2}\right)\right| \le 2\tan(1)|x-y| \le \epsilon$$
So $\sin x$ is not only continuous, it is uniformly continuous.
A: Prove f(x)=sin(x) is continuous at every point x=c.
Use theorem 1: Function f is continuous at c if and only if $\lim_{h \to 0}f(c+h) = f(c)$
Use theorem 2: $\lim_{x \to 0}sin(x) = 0$
Use theorem 3: $\lim_{x \to 0}cos(x) = 1$
Use Trig identity: $\sin (h+c) = sin(h)cos(c) + cos(h)sin(c)$   
Proof:
$\lim_{h \to 0}\sin (h+c) = \lim_{h \to 0}\left [sin(h)cos(c) + cos(h)sin(c)  \right ]$ by Trig identity.
Sum and product limit laws allow us to simplify to
$\lim_{h \to 0}\sin (h+c) = (\lim_{h \to 0}sin(h))(\lim_{h \to 0}cos(c)) + (\lim_{h \to 0}cos(h))(\lim_{h \to 0}sin(c))$
Theorems 2 and 3 allow us to simplify to
$\lim_{h \to 0}\sin (h+c) = (0)(\lim_{h \to 0}cos(c)) + (1)(\lim_{h \to 0}sin(c))$
$\lim_{h \to 0}\sin (h+c) = (\lim_{h \to 0}sin(c))$
$\lim_{h \to 0}\sin (h+c) = sin(c)$  
Note that the previous equation is true for any value of c.
Therefore, by theorem 1 above, f(x)=sin(x) is continuous at every point x=c.
