Ratio between trigonometric sums: $\sum_{n=1}^{44} \cos n^\circ/\sum_{n=1}^{44} \sin n^\circ$ What is the value of this trigonometric sum ratio: $$\frac{\displaystyle\sum_{n=1}^{44} \cos n^\circ}{\displaystyle \sum_{n=1}^{44} \sin n^\circ} = \quad ?$$ 
The answer is given as $$\frac{\displaystyle\sum_{n=1}^{44} \cos n^\circ}{\displaystyle \sum_{n=1}^{44} \sin n^\circ}  \approx \displaystyle \frac{\displaystyle\int_{0}^{45}\cos n^\circ dn}{\displaystyle\int_{0}^{45}\sin n^\circ dn} = \sqrt{2}+1$$
Using the fact $$\displaystyle  \sum_{n = 1}^{44}\cos\left(\frac{\pi}{180}\cdot n\right)\approx\int_0^{45}\cos\left(\frac{\pi}{180}\cdot x\right)\, dx $$
My question is that I did not understand the last line of this solution.
Please explain to me in detail. Thanks.
 A: Note: $\cos{n^{\circ}} = \cos{\frac{\pi}{180} n}$
so the above sums can be expressed in terms of geometric series, i.e.
$$\sum_{n=1}^{44} e^{i \pi n/180} = \frac{e^{i \pi 45/180}-e^{i \pi/180}}{1-e^{i \pi/180}} = \frac{e^{i \pi/4}-e^{i \pi/180}}{1-e^{i \pi/180}}$$
This simplifies to:
$$\frac{(e^{i \pi/4}-e^{i \pi/180})(1-e^{-i \pi/180})}{4 \sin^2{\pi/360}} = \frac{e^{i 23\pi/180} i 2 \sin{(22 \pi/180)} e^{-i \pi/360} i 2 \sin{(\pi/360)}}{4 \sin^2{\pi/360}} $$
which in turn becomes
$$-e^{i 45 \pi/360} \frac{\sin{(22 \pi/180)}}{\sin{(\pi/360)}}$$
We seek the ratio of the real part of this expression to the imaginary part.  The result is
$$\frac{\displaystyle\sum_{n=1}^{44} \cos n^\circ}{\displaystyle \sum_{n=1}^{44} \sin n^\circ} =  \cot{(\pi/8)} = \sqrt{2}+1$$
A: The last line in the argument you give could say
$$
\sum_{n=1}^{44} \cos\left(\frac{\pi}{180}n\right)\,\Delta n \approx \int_1^{44} \cos n^\circ\, dn.
$$
Thus the Riemann sum approximates the integral.  The value of $\Delta n$ in this case is $1$, and if it were anything but $1$, it would still cancel from the numerator and the denominator.
Maybe what you didn't follow is that $n^\circ = n\cdot\dfrac{\pi}{180}\text{ radians}$?
The identity is ultimately reducible to the known tangent half-angle formula
$$
\frac{\sin\alpha+\sin\beta}{\cos\alpha+\cos\beta}=\tan\frac{\alpha+\beta}{2}
$$
and the rule of algebra that says that if
$$
\frac a b=\frac c d,
$$
then this common value is equal to
$$
\frac{a+c}{b+d}.
$$
Just iterate that a bunch of times, until you're done.
Thus
$$
\frac{\sin1^\circ+\sin44^\circ}{\cos1^\circ+\cos44^\circ} = \tan 22.5^\circ
$$
and
$$
\frac{\sin2^\circ+\sin43^\circ}{\cos2^\circ+\cos43^\circ} = \tan 22.5^\circ
$$
so
$$
\frac{\sin1^\circ+\sin2^\circ+\sin43^\circ+\sin44^\circ}{\cos1^\circ+\cos2^\circ+\cos43^\circ+\cos44^\circ} = \tan 22.5^\circ
$$
and so on.
Now let's look at $\tan 22.5^\circ$.  If $\alpha=0$ then the tangent half-angle formula given above becomes
$$
\frac{\sin\beta}{1+\cos\beta}=\tan\frac\beta2.
$$
So
$$
\tan\frac{45^\circ}{2} = \frac{\sin45^\circ}{1+\cos45^\circ} = \frac{\sqrt{2}/2}{1+(\sqrt{2}/2)} = \frac{\sqrt{2}}{2+\sqrt{2}} = \frac{1}{\sqrt{2}+1}.
$$
In the last step we divided the top and bottom by $\sqrt{2}$.
What you have is the reciprocal of this.
Postscript four years later: In my answer I explained why the answer that was "given" was right, but I forgot to mention that $$ \frac{\displaystyle\sum_{n=1}^{44} \cos n^\circ}{\displaystyle \sum_{n=1}^{44} \sin n^\circ} = \sqrt{2}+1 \text{ exactly, not just approximately.} $$ The reason why the equality is exact is in my answer, but the explicit statement that it is exact is not.
