Too much brute force can kill you!

Relevant wiki: Sum and Difference Trigonometric Formulas - Problem Solving

At first, think about a small case.

Let, $n=5$ . So, the ratio becomes, $\xi=\frac{sin0^\circ+sin\frac{\phi}{5}+sin\frac{2\phi}{5}+sin\frac{3\phi}{5}+sin\frac{4\phi}{5}+sin\phi}{cos0^\circ+cos\frac{\phi}{5}+cos\frac{2\phi}{5}+cos\frac{3\phi}{5}+cos\frac{4\phi}{5}+cos\phi}$ or, $\xi=\frac{sin\phi+sin\frac{\phi}{5}+sin\frac{4\phi}{5}+sin\frac{2\phi}{5}+sin\frac{3\phi}{5}}{1+cos\phi+cos\frac{\phi}{5}+cos\frac{4\phi}{5}+cos\frac{2\phi}{5}+cos\frac{3\phi}{5}}$ Using the formula for compound angles, we have $\xi=\frac{2sin\frac{\phi}{2}cos\frac{\phi}{2}+2sin\frac{\phi}{2}cos\frac{3\phi}{10}+2sin\frac{\phi}{2}cos\frac{\phi}{10}}{2cos^2\frac{\phi}{2}+2cos\frac{\phi}{2}cos\frac{3\phi}{10}+2cos\frac{\phi}{2}cos\frac{\phi}{10}}$ Now, you can factor the common terms [ $sin\frac{\phi}{2}$ in the numerator and $cos\frac{\phi}{2}$ in the denominator] out of the rest. And, you'll see the rest is the same both in the numerator and the denominator. So, they will cancel each other out. Hence, $\xi=\frac{sin\frac{\phi}{2}}{cos\frac{\phi}{2}}=tan\frac{\phi}{2}$ This is true for all $n$ . The general case can be derived in the same way. For, all even $n$ , the term $sin\frac{\frac{n}{2}\phi}{n}$ or $cos\frac{\frac{n}{2}\phi}{n}$ is readily in the form $sin\frac{\phi}{2}$ or $cos\frac{\phi}{2}$ . As you can see, $\xi$ depends only on $\phi$ . Now, putting $\phi=71^\circ$ in the formula for $\xi$ we get $\xi=tan\frac{71^\circ}{2}\approx\boxed{0.713}$

Simple problem. Use Lagrange's identities to generalize the numerator and the denominator and cancel out the common factors. The answer comes out to be $\tan{\phi /2}$ . Plug in $\phi = 71^{\circ}$ to get $\boxed{0.71329}$ as the answer.

In general,

$\sin(a) + \sin(a+d) +\sin(a+2d) +...+\sin(a+(n-1)d) = \frac{\sin \left(a+\frac{(n-1)d}{2} \right)\sin(nd/2)}{\sin(d/2)}$

And,

$\cos(a) + \cos(a+d) +\cos(a+2d) +...+\cos(a+(n-1)d) = \frac{\cos \left(a+\frac{(n-1)d}{2} \right)\sin(nd/2)}{\sin(d/2)}$

Using these,

$\frac{\sin(a) + \sin(a+d) +\sin(a+2d) +...+\sin(a+(n-1)d)}{\cos(a) + \cos(a+d) +\cos(a+2d) +...+\cos(a+(n-1)d)} = \frac{\sin \left(a+\frac{(n-1)d}{2} \right)}{\cos \left(a+\frac{(n-1)d}{2} \right)}=\tan \left(a+\frac{(n-1)d}{2}\right)$

Number of terms are $n+1$ and and we have $a=0^\circ$ and $d=\phi/2$ , hence we end up with $\tan(\phi/2)=\tan(71^\circ/2) \approx \boxed{0.713}$ .

Since in our problem $\frac{\phi}{n}=1$ we can rewrite $\xi$ more simply as

$\large{\xi=\frac{\sin0+\sin1+\dots +\sin71}{\cos0+\cos1+\dots +\cos71}}$

where here and from now on we will take it for granted that the arguments are in degrees.

Using Euler's wonderful formula $\large{e^{i\theta}=\cos\theta + i\sin\theta}$ we can see that the numerator and denominator of $\xi$ are the imaginary and real parts of the series

$\large{x=e^{0}+e^{1i}+e^{2i}+\dots+e^{71i}}$

This is a geometric series with first term equal to one, common ratio $e^{i}$ and 72 terms, so it is easily summed with the g.p. formula to give

$\large{x=\frac{e^{72i}-1}{e^{i}-1}}$

The denominator can be made real by multiplying by its complex conjugate, to give

$\large{x=\frac{(e^{72i}-1)(e^{-i}-1)}{(e^{i}-1)(e^{-i}-1)}=\frac{e^{71i}-e^{72i}-e^{-i}+1}{(e^{i}-1)(e^{-i}-1)}}$

The answer to our problem is the ratio of the imaginary to real parts of this expression. Since we need only their ratio we can ignore the purely real denominator and use Euler's formula again to get

$\large{\xi=\frac{\sin71-\sin72-\sin(-1)}{\cos71-\cos72-\cos(-1)+1}}=\boxed{0.713}$

Easy prob try solving orally try to make a general formula for ap os sine and cos