# Using Taylor's theorem show that $1 - \frac{x^2}{2} < \cos x < 1- \frac{x^2}{2} + \frac{x^4}{24}$

Using Taylor's theorem show that $1-\frac{x^2}{2} < \cos x < 1-\frac{x^2}{2} + \frac{x^4}{24} \, \text{ for all } x \in \mathbb{R}$

I know this is true because of the Alternating series truncation error that alternates in sign, but is there a way to prove it?

• Check $x = 0$, are you sure about strict inequalities? – Mohamad Ali Baydoun May 6 '17 at 14:30
• It's definitely not true for all $x \in \mathbb{R}$: $\cos (2 \pi) = 1 > 1 - (2 \pi)^2 / 2$. – Michael Seifert May 6 '17 at 14:35
• @MichaelSeifert I don't see any problem with that – Simply Beautiful Art May 6 '17 at 14:55
• Alternating series remainder? – Simply Beautiful Art May 6 '17 at 14:55
• Whoops, you're right. That's what I get for commenting before the coffee takes hold. – Michael Seifert May 6 '17 at 16:34

With Taylor theorem (Lagrange version) we have : $$\forall x \in \mathbb R, \exists c\in (0,1), \cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}\cos''''(cx)=1-\frac{x^2}2+\frac{x^4c^4}{24}\cos(cx)$$ $$\cos(cx)\leq 1$$ $$\cos(x)\leq 1 -\frac{x^2}2+\frac{x^4c^4}{24}\leq 1 -\frac{x^2}2+\frac{x^4}{24}$$ You can apply the same kind of reasonning for the other inequality.
We keep integrating the inequality $\cos x\le x$, for $x\ge 0$, in the interval $[0,x]$, and we recursively obtain $$\cos x \le 1\quad\Longrightarrow\quad \sin x \le x\quad\Longrightarrow\quad 1-\cos x \le \frac{x^2}{2!}\quad\Longrightarrow\quad 1- \frac{x^2}{2!}\le\cos x \quad\Longrightarrow\quad x-\frac{x^3}{3!}\le \sin x\quad\Longrightarrow\quad \frac{x^2}{2!}-\frac{x^4}{4!}\le 1-\cos x\quad\Longrightarrow\quad \cos x\le 1-\frac{x^2}{2!}+\frac{x^4}{4!}\quad\Longrightarrow\quad\cdots$$ In general, continuing this we obtain $$\sum_{k=0}^{2n-1}\frac{(-1)^kx^{2k}}{(2k)!}\le\cos x\le \sum_{k=0}^{2n}\frac{(-1)^kx^{2k}}{(2k)!}$$ Note that it holds, not only for $x\ge 0$, but for $x<0$, since all the functions above are even.