Approximating $\tan61^\circ$ using a Taylor polynomial centered at $\frac \pi 3$ : how to proceed? Here's what I have so far...
I wrote a general approximation of $f(x)=\tan(x)$ , which then simplified a bit to this:
$$\tan \left(\frac{61π}{180}\right) + \sec^2\left(\frac{61π}{180}\right)\left(\frac{π}{180}\right) + \tan\left(\frac{61π}{180}\right) \sec^2\left(\frac{61π}{180}\right)\left(\frac{π}{180}\right)^2 $$
Thing is, I'm not seeing anything obvious to do next... any hints/suggestions on how to proceed in my approximation?
Thanks in advance!
 A: There are many ways to approximate (even very accurately) functions close to a point.
The simplest is Taylor expansion; in the case of the tangent, assuming $b<<a$, the expansion is $$\tan(a+b)=\tan (a)+ \left(\tan ^2(a)+1\right)b+ \left(\tan ^3(a)+\tan (a)\right)b^2+$$ $$
   \left(\tan ^4(a)+\frac{4 \tan ^2(a)}{3}+\frac{1}{3}\right)b^3+ \left(\tan
   ^5(a)+\frac{5 \tan ^3(a)}{3}+\frac{2 \tan (a)}{3}\right)b^4+$$ $$ \left(\tan ^6(a)+2
   \tan ^4(a)+\frac{17 \tan ^2(a)}{15}+\frac{2}{15}\right)b^5+O\left(b^6\right)$$ Applied to $a=\frac \pi 3$, it gives $$\tan(\frac \pi 3+b)=\sqrt{3}+4 b+4 \sqrt{3} b^2+\frac{40 b^3}{3}+\frac{44 b^4}{\sqrt{3}}+\frac{728
   b^5}{15}+O\left(b^6\right)$$ Using $b=\frac \pi {180}$ and the successive orders the approximate value is $$1.80186397765$$ $$1.80397442904$$ $$1.80404531673$$ $$1.80404767396$$ $$1.80404775256$$ while, for twelve significant digits, the exact value should be $$1.80404775527$$
Edit
The following is just for your curiosity
Another way is to use Pade approximant (these are ratios of polynomials); built at $a=\frac \pi 3$, the simplest would be $$P_{(1,1)}(x)=\frac{(x-\frac{\pi }{3})+\sqrt{3}}{1-\sqrt{3} \left(x-\frac{\pi }{3}\right)}$$ which gives for $x=\frac {61\pi} {180}$ $$1.80404021672$$ Similarly $$P_{(2,2)}(x)=\frac{-\frac{\left(x-\frac{\pi }{3}\right)^2}{\sqrt{3}}+(x-\frac{\pi
   }{3})+\sqrt{3}}{-\frac{1}{3} \left(x-\frac{\pi }{3}\right)^2-\sqrt{3}
   \left(x-\frac{\pi }{3}\right)+1}$$ which gives $$1.80404775512$$
A: $$f(x)=\tan(x)\\x=60^\circ=\frac{\pi}{3} \\h=\Delta x=1^\circ=\frac{\pi}{180}$$now 
$$ f(x+h)=f(x)+hf'(x)+\frac{h^2}{2!}f''(x)+\frac{h^3}{3!}f'''(x)+...\\f(\frac{\pi}{3}+\frac{\pi}{180})=f(\frac{\pi}{3})+hf'(\frac{\pi}{3})+\frac{h^2}{2!}f''(\frac{\pi}{3})+\frac{h^3}{3!}f'''(\frac{\pi}{3})+...\\f(\frac{\pi}{3})+(\frac{\pi}{180})f'(\frac{\pi}{3})+\frac{(\frac{\pi}{180})^2}{2!}f''(\frac{\pi}{3})+\frac{(\frac{\pi}{180})^3}{3!}f'''(\frac{\pi}{3})+...$$
$$=\sqrt{3} +\frac{\pi}{180}(1+(\sqrt{3})^2)+ \frac{(\frac{\pi}{180})^2}{2!}2\tan(\frac{\pi}{3})\sec^2(\frac{\pi}{3})+...$$
