# Bounding $\cos x$ using Taylor's theorem

To show that $$\cos x\leq 1-\frac{1}{5}x^2 \quad\text{for}\ -\pi\leq x\leq\pi,$$ I set $\cos x=1-\frac{x^2}{2}\phi(x)$, so that $\phi(x)=\frac{2}{x^2}(1-\cos x)=\frac{4\sin(x^2/2)}{x^2}$, and then showed that $\phi(x)\geq\phi(\pi)=\phi(-\pi)=\frac{4}{\pi^2}\geq\frac{2}{5}$ for $-\pi\leq x\leq\pi$.

But perhaps there's a simpler way of showing this by applying Taylor's theorem directly to $\cos x$? This is supposed to be "a routine application of Taylor's theorem with remainder."

• I don't think it is so trivial, since $1-\frac12 x^2 + \frac{1}{4!}x^4$ becomes larger than $1-\frac15 x^2$ near $x=\pi$. – Siminore Nov 6 '14 at 12:38

Let $$f(x)=\cos x-1+x^2/5$$ Now $$f(\pi)=\pi^2/5-2=(\pi^2-10)/5<0;f(0)=0$$ Now $f(x)$ is decreasing from $0$ to some point then increasing upto zero, so it must be less than zero all this time, also since it is even the rest doesn't needs to be proved.
• It is indeed simple to prove it this way, but this is not an application of Taylor's theorem to $\cos x$. – Aubrey Nov 11 '14 at 19:02