Exactly How Does This Proof Mean That The Cosine Function Is Continuous The question is: Prove that cosine is a continuous function.
To give some context in what way this must be answered, this question is from a sub-chapter called Continuity from a chapter introducing Limits.
The Solution: We must show that $\lim_{h \to 0}\cos(a + h) = \cos(a)$ to prove that the cosine function is continuous.
$\lim_{h \to 0}\cos(a + h) = \lim_{h \to 0}(\cos(a)\cos(h)-\sin(a)\sin(h)$
which after applying the limit laws $= \cos(a)(1) - \sin(a)(0) = \cos(a)$
Now how I understand how this shows that the cosine function is continuous, is that a can be any number and the cosine function is defined for any number a, thus the proof shows that the cosine function is continuous. But now why use $\cos(a+h)$?
I feel slightly uncomfortable with the way I understand this proof as I feel it's almost too elaborate to show something that I have taken to be true without question.
Could anybody explain to me in words how this proof shows that the cosine function is continuous?
 A: The reasoning
$$
\lim_{h \to 0}\cos(a + h) = \lim_{h \to 0}(\cos a\cos h -\sin a \sin h )=
\cos a\cos 0-\sin a\sin 0=\cos a
$$
reduces continuity of the cosine function to the continuity of the sine and the cosine at $0$, because only if you already know that the functions are continuous at zero you can say
$$
\lim_{h\to 0}\cos h=\cos 0
\qquad\text{and}\qquad
\lim_{h\to 0}\sin h=\sin 0.
$$
Thus you're right in thinking this doesn't prove continuity of the cosine, if the book hasn't already established continuity of sine and cosine at $0$.
How can you establish continuity of the sine at $0$? You can use the fact that
$$
\sin t < t
$$
for $t>0$, which can be justified with a geometric reasoning similar to the one used for proving that $\lim_{x\to 0}\frac{\sin x}{x}=1$. Now, from the above inequality you can derive
$$
\lvert\sin t\rvert\le |t|
$$
for any $t$, by using $\sin(-t)=-\sin t$. This ends the proof of continuity at $0$ of the sine. (Why?)
For the cosine it's a bit more complicated, try your hand at it.
To turn to the “general” question: you want to prove that
$$
\lim_{x\to a}\cos x=\cos a.
$$
This is perfectly equivalent to proving that
$$
\lim_{h\to 0}\cos(a+h)=\cos a.
$$
Just try solving the necessary inequalities. For instance, if you assume the second fact, fix $\varepsilon>0$; then you know that there exists $\delta>0$ such that, for $0<h<\delta$, $\lvert\cos(a+h)-\cos a\rvert<\varepsilon$. Then, for
$$
0<|x-a|<\delta
$$
you have $\lvert\cos x-\cos a\rvert<\varepsilon$: just set $x-a=h$ and apply the hypothesis. Similarly you get the second limit once you know the first one.
A: We use $\cos(\alpha+h)$ because, if we let $h$ be very small, $\alpha+h$ is very close to $\alpha$.
To show that the cosine function is continuous at $x=a$, we need to show that 
$$\lim_{x \to \alpha} \cos x = \cos \alpha$$
If $h > 0$ then starting at $x=\alpha+h$ means we start "a little" to the right of $x=\alpha$. Letting $h$ tend towards zero means that we creep from $x=\alpha + h$ towards $x=\alpha$. That is exactly what a limit is. You start just to the side of $x=\alpha$ and shuffle closer and closer towards $x=\alpha$.
To prove continuity, we need also to look at what happens when $h < 0$. This means $x=\alpha+h$ is slightly to the left of $x=\alpha$. As $h$ tends towards zero we shuffle from $x=\alpha+h$ towards $x=\alpha$.
If when shuffling from the right $(h>0)$ and from the left $(h<0)$, we end head towards the same value, namely $\cos \alpha$, we say that the cosine function is continuous at $x=\alpha$.
