Continuity of $\cos^2(ax)$ on the unit interval I need to prove that $g(x)=\cos^2(ax)$ is continuous on the unit interval $x=[0,1]$. No bounds of $a$ were given. Any thoughts?
 A: First we prove that $\cos(ax)$ is continuous. Let $\varepsilon > 0$ and suppose $\left| {y - {y_0}} \right| < \delta$. Note that
$$\left| {\cos (y) - \cos ({y_0})} \right| = \left| {- 2\sin \left( {\frac{{y + {y_0}}}{2}} \right)\sin \left( {\frac{{y - {y_0}}}{2}} \right)} \right| < 2\left( {\frac{{y - {y_0}}}{2}} \right).$$
Therefore $\left| {\cos (y) - \cos ({y_0})} \right| < \delta $, so choose $\varepsilon = \delta$ i.e. $\left| {\cos (y) - \cos ({y_0})} \right| < \varepsilon$. We showed that for every $\varepsilon  > 0$, there exists some $\delta  > 0$ such that $\left| {y - {y_0}} \right| < \delta$ implies $\left| {f(y) - f({y_0})} \right| < \varepsilon$. So $\cos(y)$ is continuous.
A: The more general result - that $\cos^2(\alpha x)$ is continuous everywhere, for any $\alpha$ - can be deduced from this outline. (Do it!)
(1) We know - or we can prove - that $\lim_{x \to 0} \cos(x) = 1$, and $\lim_{x \to 0} \sin(x) = 0$. Now, note that as $x \to x_0$, we have $(x - x_0) \to 0$. Thus, letting $(x - x_0) = h$, we get $$\lim_{x \to x_0} \cos(x) = \lim_{h \to 0} \cos(x_0 + h).$$
Use this - and the addition identity for $\cos(x)$ - to prove that $\cos(x)$ is continuous at all real $x$.
(2) Recall that if $\lim_{x \to a} f(x) = L$, and $g$ is continuous at $L$, then 
$$ \lim_{x \to a} g(f(x)) = g(\lim_{x \to a} f(x)) = g(L).$$ 
Apply this principle as needed to conclude that $\cos^2(\alpha x)$ is continuous. 
