Is $f: [5,8] \rightarrow \mathbb{R}$ defined by $f(x) = \cos^3(x + 1)$ uniformly continuous?

Question

Let $f: [5,8] \rightarrow \mathbb{R}$ be defined by $f(x) = \cos^3(x + 1)$. Then $f$ is uniformly continuous if for every $\epsilon > 0$, there exists a $\delta > 0$ such that $x,y \in [5,8]$ and $|x-y| < \delta$ implies $|f(x)-f(y)| < \epsilon$.

I am attempting to find a $\delta > 0$ to show that $|f(x)-f(y)| < \epsilon$.

Suppose $|x-y| < \delta = ?$ and $x,y \in [5,8]$. Then $|f(x)-f(y)| = |\cos^3(x + 1)-\cos^3(y + 1)|$.

Using the trigonometric identity $\cos^3a = \frac{\cos 3a + 3\cos a}{4}$, I could write $|f(x)-f(y)|$ as

$\frac{|\cos(3x+3) + 3\cos(x+1) - \cos(3y+3) - 3\cos(y+1)|}{4}$ = $\frac{|\cos(3x+3) - \cos(3y+3) + 3[\cos(x+1) - \cos(y+1)]|}{4}$

But up to here I am a bit stuck since I am not sure how to show that this is less than $\epsilon$, using the fact that $|x-y| < \delta$.

Answer 1

Hint 1: If you've seen compactness, you can just appeal to that, which will make the problem easier.

Hint 2: If not, rather than using the trig identity, use $(f^3 - g^3) = (f - g)(f^2 + gf + g^2)$.

Answer 2

Why not simply use the Mean value inequality? $$|f(x)-f(y)|\le M\,|x-y|,$$ where $M$ is an upper bound for $\;|f'(c)|=3\cos^2\mkern-3muc\mkern3mu|\sin c| $.