The right reciprocal

Geometry Level 3

∑ k = 1 44 cos ⁡ k ∘ ∑ k = 1 44 sin ⁡ k ∘ − ∑ k = 1 44 sin ⁡ k ∘ ∑ k = 1 44 cos ⁡ k ∘ = ? \dfrac{\displaystyle\sum_{k=1}^{44}\cos{k^{\circ}}}{\displaystyle\sum_{k=1}^{44}\sin{k^{\circ}}}-\dfrac{\displaystyle\sum_{k=1}^{44}\sin{k^{\circ}}}{\displaystyle\sum_{k=1}^{44}\cos{k^{\circ}}}=\ ?


The answer is 2.

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Ikkyu San by Ikkyu San , 23, Thailand

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1 solution

Daniel Rabelo
Jun 2, 2015

∑ k = 1 44 sin ⁡ k º = ∑ k = 1 44 sin ⁡ ( 45 º − k º ) = ∑ k = 1 44 2 2 ( cos ⁡ k º − sin ⁡ k º ) = \sum_{k=1}^{44} \sin kº =\sum_{k=1}^{44} \sin (45º-kº)=\sum_{k=1}^{44} \frac{\sqrt2}{2} (\cos kº - \sin kº)=

= 2 2 ∑ k = 1 44 ( cos ⁡ k º − sin ⁡ k º ) =\frac{\sqrt2}{2}\sum_{k=1}^{44} (\cos kº - \sin kº)

then: ∑ k = 1 44 cos ⁡ k º = ( 1 + 2 ) ∑ k = 1 44 sin ⁡ k º \sum_{k=1}^{44} \cos kº =(1+\sqrt2)\sum_{k=1}^{44} \sin kº

then the problem is reduced to:

1 + 2 − 1 1 + 2 = 3 + 2 2 − 1 1 + 2 = 2 ( 1 + 2 ) ( 1 + 2 ) = 2. 1+\sqrt2 - \frac{1}{1+\sqrt2} = \frac{3+2\sqrt2-1}{1+\sqrt2} = 2\frac{(1+\sqrt2)}{(1+\sqrt2)}=2.

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