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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

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

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    It comes from the alternating series theorem (https://teachingcalculus.com/2013/02/22/error-bounds/) – 2017-02-19
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    @JeanMarie $|a_n|$ is not necessarily strictly decreasing in the beginning though, or is that not a problem? – 2017-02-19
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    @JeanMarie: It's not as simple as that. The Alternating Series theorem doesn't apply here, because the remaining terms are not in general decreasing in absolute value. – 2017-02-19
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    I agree: I wrote too quickly my comment. But, nevertheless, your 2nd sentence isn't exact: there is a certain, rather large, neighbourhood of the origin ($x\in[-\sqrt{2},\sqrt{2}]$) where we can apply it. – 2017-02-20
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    @JeanMarie: That's why I wrote "in general". Also, by what measure is $[-\sqrt 2,\sqrt 2]$ "rather large"? – 2017-02-20
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    I meant this on a practical point of view, because it is almost $[-\pi/2,\pi/2]$, that's all... – 2017-02-20

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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.

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    This can be found in "101 Great Problems In Elementary Mathematics" by H. Dorrie. – 2017-02-20
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    @Jack D'Aurizio Thank you, that makes sense. But I was looking a general form answer, such as $\cos{x}\gt 1-\frac{x^2}{2!}+\cdots + \frac{x^{4n}}{(4n)!}-\frac{x^{4n+2}}{(4n+2)!}$. So does the application of this trick imply that the inequality is true for all n where n is 2 mod 4? – 2017-02-20
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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.

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    $|a_n|$ eventually decreases but it's not necessarily the case in the beginning. So does that make a difference? – 2017-02-19
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    @jxie20 It follows that $\cos(x)\ge-1$ and your partial sum is eventually decreasing. – 2017-02-19
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    You're right: implicitly, I consider the case when $0\le x\le\frac\pi2 $, which displays all the values in $[0,1]$. – 2017-02-19
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    @jxie20: Yes, it certainly does make a difference. This answer is invalid. – 2017-02-19
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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 $f^{v}(x)0$ when $x>0.$

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

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

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

So $f^i(x)$ is strictly increasing , with $f^i(x)0$ 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$.

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    Isn't $f^{vi}(x)= \cos{x}-1$ ? – 2017-02-20
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    @jxie20.Thanks . I wrote the def'n of f(x) incorrectly. – 2017-02-20