Second Solution to evaluate $\dfrac{\cos 1^{\circ}+\cos 2^{\circ}+\cdots+\cos 44^{\circ}}{\sin 1^{\circ}+\sin 2^{\circ}+\cdots+\sin 44^{\circ}}$

Express $\dfrac{\cos 1^{\circ}+\cos 2^{\circ}+\cdots+\cos 44^{\circ}}{\sin 1^{\circ}+\sin 2^{\circ}+\cdots+\sin 44^{\circ}}$ in the form $a+b\sqrt{c}$, where $a,\,b,\,c$ are positive integers.

Answer:

Other solution:

It's always better to rewrite the given expression so both numerator and denominator carry more terms for better visualization effect:

 $\dfrac{\cos 1^{\circ}+\cos 2^{\circ}+\cdots+\cos 44^{\circ}}{\sin 1^{\circ}+\sin 2^{\circ}+\cdots+\sin 44^{\circ}}$

$=\dfrac{\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+\cdots+\cos 42^{\circ}++\cos 43^{\circ}+\cos 44^{\circ}}{\sin 1^{\circ}+\sin 2^{\circ}++\sin 3^{\circ}\cdots+\sin 42^{\circ}+\sin 43^{\circ}+\sin 44^{\circ}}$

You might notice for each of the angle $1^{\circ},\,2^{\circ},\,\cdots, 44^{\circ}$, there exist the pairs of sine and cosine function with those angles and you can't help but curious about it...

You might wonder what happens if you add those pairs up:

$\begin{align*}\sin 1^{\circ}+\cos 1^{\circ}&=\sqrt{2}\sin (1^{\circ}+45^{\circ})\\&=\sqrt{2}\sin (46^{\circ})\\&=\sqrt{2}\sin (90-44)^{\circ}\\&=\sqrt{2}\cos 44^{\circ}\end{align*}$

$\begin{align*}\sin 2^{\circ}+\cos 2^{\circ}&=\sqrt{2}\sin (2^{\circ}+45^{\circ})\\&=\sqrt{2}\sin (47^{\circ})\\&=\sqrt{2}\sin (90-43)^{\circ}\\&=\sqrt{2}\cos 43^{\circ}\end{align*}$

$\begin{align*}\sin 3^{\circ}+\cos 3^{\circ}&=\sqrt{2}\sin (3^{\circ}+45^{\circ})\\&=\sqrt{2}\sin (48^{\circ})\\&=\sqrt{2}\sin (90-42)^{\circ}\\&=\sqrt{2}\cos 42^{\circ}\end{align*}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\vdots$

$\begin{align*}\sin 44^{\circ}+\cos 44^{\circ}&=\sqrt{2}\sin (44^{\circ}+45^{\circ})\\&=\sqrt{2}\sin (89^{\circ})\\&=\sqrt{2}\sin (90-1)^{\circ}\\&=\sqrt{2}\cos 1^{\circ}\end{align*}$

Ah! This works!!! This actually works when we only toyed with our idea and observation! This is why I always have enjoyed math to the most. Curiosity is an essential ingredient in wanting to learn, and wonder and start to experimenting with different approaches is the key to success.

What remains to be done is to add those equations up (by grouping all the sine and cosine terms separately) and collect like terms:

$(\sin 1^{\circ}+\sin 2^{\circ}+\sin 3^{\circ}+\cdots+\sin 44^{\circ})+(\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+\cdots+\cos 44^{\circ})$

$=\sqrt{2}(\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+\cdots+\cos 44^{\circ})$

$\sin 1^{\circ}+\sin 2^{\circ}+\sin 3^{\circ}+\cdots+\sin 44^{\circ}=(\sqrt{2}-1)(\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+\cdots+\cos 44^{\circ})$

Therefore

$\begin{align*}\dfrac{\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+\cdots+\cos 44^{\circ}}{\sin 1^{\circ}+\sin 2^{\circ}+\sin 3^{\circ}+\cdots+\sin 44^{\circ}}&=\dfrac{1}{\sqrt{2}-1}\\&=\dfrac{1(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}\\&=\sqrt{2}+1\end{align*}$

and we're done.