why is $\cos{x} \gt 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}$

Question

For $x>0$, why is $$\cos{x}\gt1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}$$ and $$\sin{x}\lt x-\frac{x^3}{3!}+\frac{x^5}{5!}.$$

Or in general, how do I know that

$$\cos{x}\gt 1-\frac{x^2}{2!}+\cdots + \frac{x^{4n}}{(4n)!}-\frac{x^{4n+2}}{(4n+2)!}$$

Answer 1

If $0\le x\le\dfrac\pi 2$, this results from Leibniz'criterion for alternating series: if an alternating series $\sum a_n$ is such that $\lvert a_n\rvert$ decreases to $0$ , then it converges. Furthermore, if $L$ is the sum of the series, $$\biggl\lvert\sum_{n=0}^N a_n-L\biggr\rvert\le\lvert a_{n+1}\rvert ,$$ and $(\sum_{n=0}^N a_n)-L$ and $a_{n+1}$ have the same sign.

Answer 2

For any $t>0$ we have $\sin t< t$, hence by integrating both sides over $(0,x)$ we get that for any $x>0$ the inequality $1-\cos x < \frac{x^2}{2}$ holds. If we consider $\cos t>1-\frac{t^2}{2}$ and integrate both sides over $(0,x)$ we get $\sin x > x-\frac{x^3}{6}$. From $\sin t > t-\frac{t^3}{6}$ it follows that $1-\cos x > \frac{x^2}{2!}-\frac{x^4}{4!}$, and by performing the same trick twice we get the claim.

Answer 3

Let $f(x)=-(1-x^2/2!+x^4/4!-x^6/6!)+\cos x.$

For Roman numeral $W$ let $f^W$ denote the $W$-th derivative of $f$.

We have $f^{vi}(x)=1-\cos x>0$ except when $x/2\pi \in \mathbb Z$.

So $f^{v}(x)$ is strictly increasing, with $f^{v}(x)<f^v(0)=0 $ when $x<0$, and $f^v(x)>0$ when $x>0.$

So $f^{iv}(x)>f^{iv}(0)=0$ for $x\ne 0.$

So $f^{iii}(x)$ is strictly increasing, with $f^{iii}(x)<f^{iii}(0)=0$ when $x<0$ , and $f^{iii}(x)>0$ when $x>0.$

So $f^{ii}(x)>f^{ii}(0)=0$ for $x\ne 0$.

So $f^i(x)$ is strictly increasing , with $f^i(x)<f^i(0)=0$ when $x<0,$ and $f^i(x)>0$ when $x>0.$

Therefore $f(x)>f(0)=0$ when $x\ne 0 $ .

$(\bullet) $This will of course generalize to $ (\;\sum_{j=0}^{2n+1}(-1)^jx^{2j}/(2j)!\;)-\cos x$.