## Monday, May 25, 2015

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

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.

(This question showed up here(math-teacher-guide) but I did not include its solution in that slide show).

My solution:

Normally for simplifying trigonometric question such as this one, there is one thing that's so worth noticing:

1. The terms could be paired up so the sum of them always yields to the same number.

In this case, we can rewrite the given expression so both numerator and denominator carry more terms for better visualization effect:

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

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

$=\dfrac{(\cos 1^{\circ}+\cos 44^{\circ})+(\cos 2^{\circ}+\cos 43^{\circ})+(\cos 3^{\circ}+\cos 42^{\circ})\cdots+(\cos 22^{\circ}+\cos 23^{\circ})}{(\sin 1^{\circ}+\sin 44^{\circ})+(\sin 2^{\circ}+\sin 43^{\circ})+(\sin 3^{\circ}+\sin 42^{\circ})\cdots+(\sin 22^{\circ}+\sin 23^{\circ})}$

Ahem, we're not done yet, we need the sum-to-product formula to simplify the expression after grouping further:

Sum-to-product formulas tell us:

1. $\cos A+\cos B=2\cos \dfrac{A+B}{2}\cos \dfrac{A-B}{2}$

1. $\sin A+\sin B=2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}$

Therefore, we get:

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

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

$=\dfrac{(\cos 1^{\circ}+\cos 44^{\circ})+(\cos 2^{\circ}+\cos 43^{\circ})+(\cos 3^{\circ}+\cos 42^{\circ})+\cdots+(\cos 22^{\circ}+\cos 23^{\circ})}{(\sin 1^{\circ}+\sin 44^{\circ})+(\sin 2^{\circ}+\sin 43^{\circ})+(\sin 3^{\circ}+\sin 42^{\circ})\cdots+(\sin 22^{\circ}+\sin 23^{\circ})}$

$=\dfrac{2\cos \dfrac{45^{\circ}}{2}\cos \dfrac{43^{\circ}}{2}+2\cos \dfrac{45^{\circ}}{2}\cos \dfrac{41^{\circ}}{2}+2\cos \dfrac{45^{\circ}}{2}\cos \dfrac{39^{\circ}}{2}+\cdots+2\cos \dfrac{45^{\circ}}{2}\cos \dfrac{1}{2}}{2\sin \dfrac{45^{\circ}}{2}\cos \dfrac{43^{\circ}}{2}+2\sin \dfrac{45^{\circ}}{2}\cos \dfrac{41^{\circ}}{2}+2\sin \dfrac{45^{\circ}}{2}\cos \dfrac{39^{\circ}}{2}+\cdots+2\sin \dfrac{45^{\circ}}{2}\cos \dfrac{1^{\circ}}{2}}$

$=\dfrac{2\cos \dfrac{45^{\circ}}{2}\left(\cos \dfrac{43^{\circ}}{2}+\cos \dfrac{41^{\circ}}{2}+\cos \dfrac{39^{\circ}}{2}+\cdots+\cos \dfrac{1^{\circ}}{2}\right)}{2\sin \dfrac{45^{\circ}}{2}\left(\cos \dfrac{43^{\circ}}{2}+\cos \dfrac{41^{\circ}}{2}+\cos \dfrac{39^{\circ}}{2}+\cdots+\cos \dfrac{1^{\circ}}{2}\right)}$

$=\dfrac{\cancel{2}\cos \dfrac{45^{\circ}}{2}\cancel{\left(\cos \dfrac{43^{\circ}}{2}+\cos \dfrac{41^{\circ}}{2}+\cos \dfrac{39^{\circ}}{2}+\cdots+\cos \dfrac{1^{\circ}}{2}\right)}}{\cancel{2}\sin \dfrac{45^{\circ}}{2}\cancel{\left(\cos \dfrac{43^{\circ}}{2}+\cos \dfrac{41^{\circ}}{2}+\cos \dfrac{39^{\circ}}{2}+\cdots+\cos \dfrac{1^{\circ}}{2}\right)}}$

$=\dfrac{1}{\tan \dfrac{45^{\circ}}{2}}$

Since we know $\tan 45^{\circ}=1$ and $\tan 2x=\dfrac{2\tan x}{1-\tan^2 x}$, we get the value for $\tan \dfrac{45}{2}$ as follows:

$\tan 2\left(\dfrac{45^{\circ}}{2}\right)=\dfrac{2\tan \dfrac{45^{\circ}}{2}}{1-\tan^2 \dfrac{45^{\circ}}{2}}$

$\tan \cancel{2}\left(\dfrac{45^{\circ}}{\cancel{2}}\right)=\dfrac{2\tan \dfrac{45^{\circ}}{2}}{1-\tan^2 \dfrac{45^{\circ}}{2}}$

$\tan 45^{\circ}=\dfrac{2\tan \dfrac{45^{\circ}}{2}}{1-\tan^2 \dfrac{45^{\circ}}{2}}$

$1(1-\tan^2 \dfrac{45^{\circ}}{2})=2\tan \dfrac{45^{\circ}}{2}$

$\tan^2 \dfrac{45^{\circ}}{2}+2\tan \dfrac{45^{\circ}}{2}-1=0$

Solve for $\tan \dfrac{45^{\circ}}{2}$ we get:

$\tan \dfrac{45^{\circ}}{2}=\dfrac{-2\pm\sqrt{2^2-4(1)(-1)}}{2}=-1\pm \sqrt{2}=-1+\sqrt{2}(>0)$

Therefore we obtain:

\begin{align*}\dfrac{\cos 1^{\circ}+\cos 2^{\circ}+\cdots+\cos 44^{\circ}}{\sin 1^{\circ}+\sin 2^{\circ}+\cdots+\sin 44^{\circ}}&=\dfrac{1}{\tan \dfrac{45^{\circ}}{2}}\\&=\dfrac{1}{-1+\sqrt{2}}\\&=\dfrac{-1-\sqrt{2}}{(-1+\sqrt{2})(-1-\sqrt{2})}\\&=1+\sqrt{2}\end{align*}