Ratio between trigonometric sums: $\sum_{n=1}^{44} \cos n^\circ/\sum_{n=1}^{44} \sin n^\circ$

Question

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.

Answer 1

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.

Answer 2

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

$$\frac{\displaystyle\sum_{n=1}^{44} \cos n^\circ}{\displaystyle \sum_{n=1}^{44} \sin n^\circ} = \cot{(\pi/8)} = \sqrt{2}+1$$