Determine the value of $\cos(2)$ I'm trying to prove that the function $f(x)=\cos(x)$ has an $x$-intercept in the interval $[0,2]$. To do that I showed that $f(0)=\cos(0)=1>0$. Then I would have to show that $\cos(2)<0$, however how would I workout the value of $\cos(2)$? My professor mentioned something about using the Taylor Polynomial but I'm not sure.
 A: Curiously, in my Analysis I class, $\pi/2$ was defined as the first positive zero of the $\cos$ function, so the existence of such a zero had to be first shown. $\cos$ was defined as the even part of the exponential function, which was defined using a Taylor series.
On the off chance you're in a similar scenario where you can only use the Taylor expansion and no properties of $\pi$:
$$\begin{align}\cos x= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n}&=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots \\
\cos x &< 1-\frac{x^2}{2!}+\frac{x^4}{4!} \\
\cos 2 &< 1-\frac{4}{2!}+\frac{16}{4!}=-\frac{1}{3}<0\end{align}$$
as required.
As David C. Ullrich states in the comments, it's not immediately clear that the inequality I've written above holds. However, using arguments about alternating series, it's not difficult to show true for $x^2 < 30$, which is all that's required here.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\cos\pars{2} & = \cos\pars{{2\pi \over 3} + \bracks{2 - {2\pi \over 3}}}
\\[5mm] & =
\overbrace{\cos\pars{2\pi \over 3}}^{\ds{-\,{1 \over 2}}}\
\overbrace{\cos\pars{6 - 2\pi \over 3}}^{\ds{\approx\ 1}}\ -\
\overbrace{\sin\pars{2\pi \over 3}}^{\ds{\root{3} \over 2}}\
\overbrace{\sin\pars{6 - 2\pi \over 3}}^{\ds{\approx\ {6 - 2\pi \over 3}}}
\\[5mm] & \approx
\bbx{{2\root{3}\pi - 6\root{3} - 3 \over 6} \approx -0.4183}\label{1}\tag{1}
\end{align}

A calculator yields $\ds{\cos\pars{2} = -0.416146836547142\ldots}$.
  The \eqref{1} relative error is $\ds{\color{red}{\large0.5057\ \%}}$.

A: $2$ is between $\frac{\pi}{2}$ and $\pi$, so $\cos(2)$ is certainly negative. For a numerical determination of such constant, one may invoke
$$ \forall x\in\mathbb{R},\qquad \cos(x)=\sum_{n\geq 0}\frac{(-1)^n x^{2n}}{(2n)!}.\tag{A}$$
By evaluating the RHS at $x=2$ and truncating at $n=3$, we get $\cos(2)\approx -\frac{19}{45}$.
There are more efficient methods, however. For instance, 
$$\cos(2)=2\cos^2(1)-1=8\cos^4\left(\frac{1}{2}\right)-8\cos^2\left(\frac{1}{2}\right)+1$$
and by $(A)$ the constant $\cos\frac{1}{2}$ is pretty simple to approximate: it is about $\frac{337}{384}$.
A: Simply notice that $$ \frac{\pi}{2} < \frac{3.2}{2} = 1.6 < 2 < \pi$$
Since $\cos$ is stricly decreasing on the interval $[\frac{\pi}{2}, \pi]$, we conclude $\cos(2) < \cos(\frac{\pi}{2})$.
Alternatively, if you want to use the power series, recall that $\cos(x) = \sum\limits_{n=0}^\infty \frac{(-1)^n (x)^{2n}}{(2n)!} $.
Now, notice that $\frac{-(2)^{4k-2}}{(4k-2)!} + \frac{(2)^{4k}}{(4k)!} < 0$ for $ k \in \mathbb{N}$. 
Thus $\cos(2) < \sum\limits_{n=0}^2 \frac{(-1)^n (2)^{2n}}{(2n)!} = 1 - \frac{4}{2} + \frac{16}{24} < 0$.
A: Just for the fun of it.
Using the same idea as Felix Marin, let us build the $[n,n]$ Padé approximant of $\cos \left(x+\frac{2\pi }{3}\right)$ around $x=0$ and evaluate it for $x=2-\frac{2\pi }{3}$.
For $n=1$
$$-\frac{\frac{1}{2}+\frac{7 }{4 \sqrt{3}} x}{1+\frac{1}{2 \sqrt{3}}x }\implies \cos(2)\approx -\frac{9+21 \sqrt{3}-7 \sqrt{3} \pi }{18+6 \sqrt{3}-2 \sqrt{3} \pi }\approx -0.415961$$ which corresponds to a relative error of $\color{red}{\text{0.04455 %}}$.
For $n=2$
$$-\frac{\frac{1}{2}+\frac{17 }{12 \sqrt{3}}x-\frac{19 }{72}x^2 } { 1-\frac{1}{6 \sqrt{3}}x+\frac{5 }{36}x^2}\implies \cos(2)\approx \frac{90-153 \sqrt{3}-114 \pi +51 \sqrt{3} \pi +19 \pi ^2}{252-18 \sqrt{3}-60 \pi
   +6 \sqrt{3} \pi +10 \pi ^2}\approx -0.416147$$ which corresponds to a relative error of $\color{red}{\text{0.00003 %}}$.
