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.

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 $\color{red}{\text{0.00003 %}}$.
