Holiday Problem 3

We have, $x+1= \frac{\sum_{n=1}^{44}( \sin n^\circ + \cos n^\circ)}{\sum_{n=1}^{44} \sin n^\circ}$ $= \sqrt{2}\frac{\sum_{n=1}^{44}( \frac{1}{\sqrt{2}}\sin n^\circ +\frac{1}{\sqrt{2}} \cos n^\circ)}{\sum_{n=1}^{44} \sin n^\circ}$ $=\sqrt{2} \frac{\sum_{n=1}^{44} \cos (45^\circ-n^\circ)}{\sum_{n=1}^{44} \sin n^\circ}$ $=\sqrt{2} \frac{\sum_{n=1}^{44} \cos (n^\circ)}{\sum_{n=1}^{44} \sin n^\circ}=\sqrt{2}x$ Thus, $x=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$

Consider the sum $\sum_{n = 1}^{44} \text{cis } n^\circ$ . The fraction is given by the real part divided by the imaginary part.

The sum can be written $- 1 + \sum_{n = 0}^{44} \text{cis } n^\circ = - 1 + \dfrac {\text{cis } 45^\circ - 1}{\text{cis } 1^\circ - 1}$ (by De Moivre's Theorem with geometric series)

$= - 1 + \dfrac {\frac {\sqrt {2}}{2} - 1 + \dfrac {i \sqrt {2}}{2}}{\text{cis } 1^\circ - 1} = - 1 + \dfrac {\left( \dfrac {\sqrt {2}}{2} - 1 + \dfrac {i \sqrt {2}}{2} \right) (\text{cis } ( - 1^\circ) - 1)}{(\cos 1^\circ - 1)^2 + \sin^2 1^\circ}$ (after multiplying by complex conjugate)

$= - 1 + \dfrac {\left( \frac {\sqrt {2}}{2} - 1 \right) (\cos 1^\circ - 1) + \dfrac {\sqrt {2}}{2}\sin 1^\circ + i\left( \left(1 - \dfrac {\sqrt {2}}{2} \right) \sin 1^\circ + \dfrac {\sqrt {2}}{2} (\cos 1^\circ - 1)\right)}{2(1 - \cos 1^\circ)}$

$= - \dfrac {1}{2} - \dfrac {\sqrt {2}}{4} - \dfrac {i\sqrt {2}}{4} + \dfrac {\sin 1^\circ \left( \dfrac {\sqrt {2}}{2} + i\left( 1 - \dfrac {\sqrt {2}}{2} \right) \right)}{2(1 - \cos 1^\circ)}$

Using the tangent half-angle formula, this becomes $\left( - \dfrac {1}{2} + \dfrac {\sqrt {2}}{4}[\cot (1/2^\circ) - 1] \right) + i\left( \dfrac {1}{2}\cot (1/2^\circ) - \dfrac {\sqrt {2}}{4}[\cot (1/2^\circ) + 1] \right)$ .

Dividing the two parts and multiplying each part by 4, the fraction is $\dfrac { - 2 + \sqrt {2}[\cot (1/2^\circ) - 1]}{2\cot (1/2^\circ) - \sqrt {2}[\cot (1/2^\circ) + 1]}$ .

Although an exact value for $\cot (1/2^\circ)$ in terms of radicals will be difficult, this is easily known: it is really large!

So treat it as though it were $\infty$ . The fraction is approximated by $\dfrac {\sqrt {2}}{2 - \sqrt {2}} = \dfrac {\sqrt {2}(2 + \sqrt {2})}{2} = 1 + \sqrt {2}\Rightarrow \lfloor 100(1+\sqrt2)\rfloor=\boxed{241}$

by summation formula x reduces to cot22.5 =2.414 then we want 100 x= 100 2.414=241.4 so answer 241

It's not rocket science! All you need to solve this question is notice that 1+44 = 2+43 = ...= 22+23. Then, you may want to transform cos a + cos b and sin a + sin b into a product. When you simplify, will will find x=cot(22.5).

Sum is cot (45/2)

expand numerator and denominator and take (first, last);(second, second last)......... terms and apply sum to product rules in numerator and denominator. we get required ratio as cot(22.5)=2.414...... therefore answer=241