Unsympathetic Trigonometric Arithmetic

$\begin{aligned} \sum_{j=1}^{44} \cos{(j^\circ)} & = \sum_{j=46} ^{89} \sin{(j^\circ)} = \sum_{j=1}^{44} \sin{(45+j)^\circ} \\ & = \sum_{j=1}^{44} [\sin{(45^\circ)}\cos{(j^\circ)} + \cos{(45^\circ)}\sin{(j^\circ)}] \\ & = \frac{1}{\sqrt{2}} \sum_{j=1}^{44} [\cos{(j^\circ)} + \sin{(j^\circ)}] \\ \Rightarrow \left(1 -\frac{1}{\sqrt{2}} \right) \sum_{j=1}^{44} \cos{(j^\circ)} & = \frac{1}{\sqrt{2}} \sum_{j=1}^{44} \sin{(j^\circ)} \\ \sum_{j=1}^{44} \cos{(j^\circ)} & = \left( \frac{1}{\sqrt{2}-1} \right) \sum_{j=1}^{44} \sin{(j^\circ)} \\ \frac{\displaystyle \sum_{j=1}^{44} \cos{(j^\circ)} }{\displaystyle \sum_{k=1}^{44} \sin{(k^\circ)}} & = 1+\sqrt{2} \\ \Rightarrow 2\cot^{-1} \left( \frac{\displaystyle \sum_{j=1}^{44} \cos{(j^\circ)} }{\displaystyle \sum_{k=1}^{44} \sin{(k^\circ)}} \right) & = 2\cot^{-1} \left( 1+\sqrt{2} \right) \\ & = 2 \tan^{-1} \left( \frac{1}{1+\sqrt{2}} \right) \\ & = 2 \tan^{-1} \left(\color{#3D99F6}{\sqrt{2}-1} \right) \quad \quad \color{#3D99F6}{[\text{See Note}]} \\ & = 2 \left( \color{#3D99F6}{22.5^\circ} \right) = \boxed{45}^\circ \end{aligned}$

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$\color{#3D99F6}{\text{Note:}}$

Let $t = \tan{22.5^\circ}$

$\begin{aligned} \Rightarrow \tan{45^\circ} = 1 & = \frac{2t}{1-t^2} \\ \Rightarrow t^2 + 2t -1 & = 0 \Rightarrow t = \tan{\color{#3D99F6}{22.5^ \circ}} & = \frac{-2+\sqrt{8}}{2} = \color{#3D99F6}{\sqrt{2} - 1} \end{aligned}$

Let $w=e^{1^\circ i}+e^{2^\circ i}+\cdots+e^{n^\circ i}$ , then $w=\dfrac{e^{(n+1)^\circ i}-e^{1^\circ i}}{e^{1^\circ i}-1}$ using the formula for the sum of a geometric progression.

Now, we can manipulate it to obtain $w=\dfrac{e^{(1+n/2)^\circ i}}{e^{1^\circ i/2}} \cdot \dfrac{e^{n i^\circ /2}-e^{-n i^\circ/2}}{e^{1^\circ i/2}-e^{-1^\circ i/2}}$

Use $e^{\theta i}-1/e^{\theta i}=2i \sin \theta$ :

$w=e^{(n+1)^\circ i/2} \cdot \dfrac{2i \sin \frac{n^\circ}{2}}{2i \sin \frac{1^\circ}{2}}$

$w=\left[\cos\left(\dfrac{(n+1)^\circ}{2}\right)+i\sin\left(\dfrac{(n+1)^\circ}{2}\right)\right] \cdot \dfrac{\sin \frac{n^\circ}{2}}{\sin \frac{1^\circ}{2}}$

That is the derivation of an special case for the summation of sines and cosines of angles in arithmetic progression.

Clearly, $\displaystyle \sum_{j=1}^{n} \cos(j^\circ)=\text{Re}(w)$ and $\displaystyle \sum_{k=1}^{n} \sin(k^\circ)=\text{Im}(w)$ , so:

$\dfrac{\displaystyle \sum_{j=1}^{n} \cos(j^\circ)}{\displaystyle \sum_{k=1}^{n} \sin(k^\circ)}=\cot \left(\dfrac{(n+1)^\circ}{2}\right)$

The expression above is valid only when $\sin\left(\dfrac{n^\circ}{2}\right) \neq 0$ , in this problem $n=44$ , so we can apply the formula:

$x^\circ=2 \text{arccot}\left(\cot\left(\dfrac{(44+1)^\circ}{2}\right)\right)=2\left(\dfrac{45^\circ}{2}\right)$

Hence, $x=\boxed{45}$ .