Sum of cosines

Geometry Level 5

A = cos ⁡ ( 13 π 20 ) + cos ⁡ ( 21 π 20 ) + cos ⁡ ( 29 π 20 ) + cos ⁡ ( 37 π 20 ) A=\cos\left(\frac{13\pi}{20}\right) +\cos\left(\frac{21\pi}{20}\right) +\cos\left(\frac{29\pi}{20}\right) +\cos\left(\frac{37\pi}{20}\right)

Find the value of ⌊ 1000 A ⌋ \lfloor 1000A \rfloor .


Notation: ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function .


The answer is -708.

Chan Lye Lee by Chan Lye Lee , 44, Singapore

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3 solutions

A = cos ⁡ ( 13 π 20 ) + cos ⁡ ( 21 π 20 ) + cos ⁡ ( 29 π 20 ) + cos ⁡ ( 37 π 20 ) = cos ⁡ ( 2 π 5 + π 4 ) + cos ⁡ ( 4 π 5 + π 4 ) + cos ⁡ ( 6 π 5 + π 4 ) + cos ⁡ ( 8 π 5 + π 4 ) = 1 2 ( cos ⁡ ( 2 π 5 ) − sin ⁡ ( 2 π 5 ) + cos ⁡ ( 4 π 5 ) − sin ⁡ ( 4 π 5 ) + cos ⁡ ( 6 π 5 ) − sin ⁡ ( 6 π 5 ) + cos ⁡ ( 8 π 5 ) − sin ⁡ ( 8 π 5 ) ) = 1 2 ( cos ⁡ ( 2 π 5 ) + cos ⁡ ( 4 π 5 ) + cos ⁡ ( 6 π 5 ) + cos ⁡ ( 8 π 5 ) − sin ⁡ ( 2 π 5 ) − sin ⁡ ( 4 π 5 ) − sin ⁡ ( 6 π 5 ) − sin ⁡ ( 8 π 5 ) ) = 1 2 ( cos ⁡ ( 2 π 5 ) + cos ⁡ ( 4 π 5 ) − cos ⁡ ( π 5 ) − cos ⁡ ( 3 π 5 ) − sin ⁡ ( 3 π 5 ) − sin ⁡ ( π 5 ) + sin ⁡ ( π 5 ) + sin ⁡ ( 3 π 5 ) ) See proof below. = 1 2 ( − 1 2 − 1 2 ) = − 1 2 ≈ − 0.7071 \begin{aligned} A & = \cos \left(\frac {13\pi}{20}\right) + \cos \left(\frac {21\pi}{20}\right) + \cos \left(\frac {29\pi}{20}\right) + \cos \left(\frac {37\pi}{20}\right) \\ & = \cos \left(\frac {2\pi}5 + \frac \pi 4\right) + \cos \left(\frac {4\pi}5 + \frac \pi 4\right) + \cos \left(\frac {6\pi}5 + \frac \pi 4\right) + \cos \left(\frac {8\pi}5 + \frac \pi 4\right) \\ & = \frac 1{\sqrt 2} \left( \cos \left(\frac {2\pi}5 \right) - \sin \left(\frac {2\pi}5 \right) + \cos \left(\frac {4\pi}5\right) - \sin \left(\frac {4\pi}5 \right) + \cos \left(\frac {6\pi}5\right) - \sin \left(\frac {6\pi}5 \right) + \cos \left(\frac {8\pi}5\right) - \sin \left(\frac {8\pi}5 \right) \right) \\ & = \frac 1{\sqrt 2} \left( \cos \left(\frac {2\pi}5 \right) + \cos \left(\frac {4\pi}5\right) + \cos \left(\frac {6\pi}5\right) + \cos \left(\frac {8\pi}5\right) - \sin \left(\frac {2\pi}5 \right) - \sin \left(\frac {4\pi}5 \right) - \sin \left(\frac {6\pi}5 \right) - \sin \left(\frac {8\pi}5 \right) \right) \\ & = \frac 1{\sqrt 2} \left({\color{#3D99F6}\cos \left(\frac {2\pi}5 \right) + \cos \left(\frac {4\pi}5\right)}{\color{#D61F06} - \cos \left(\frac {\pi}5\right) - \cos \left(\frac {3\pi}5\right)} - \sin \left(\frac {3\pi}5 \right) - \sin \left(\frac {\pi}5 \right) + \sin \left(\frac {\pi}5 \right) + \sin \left(\frac {3\pi}5 \right) \right) & \small \color{#3D99F6} \text{See proof below.} \\ & = \frac 1{\sqrt 2} \left({\color{#3D99F6}-\frac 12}{\color{#D61F06} - \frac 12} \right) = - \frac 1{\sqrt 2} \approx - 0.7071 \end{aligned}

Therefore, ⌊ 1000 A ⌋ = − 708 \lfloor 1000A\rfloor = \boxed{-708} .

Reference: Proof for ∑ k = 0 n − 1 cos ⁡ ( 2 k + 1 2 n + 1 ) = 1 2 \displaystyle \sum_{k=0}^{n-1} \cos \left(\frac {2k+1}{2n+1}\right) = \frac 12

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Chan Lye Lee
Aug 2, 2018

Note that the 5 roots of z 5 = 1 + i 2 \displaystyle z^5 = \frac{1+i}{\sqrt{2}} are e i ( 8 k + 5 ) π 20 \displaystyle e^{i \frac{(8k+5)\pi}{20}} where k = 0 , 1 , 2 , 3 , 4 k=0,1,2,3,4 .

The sum of roots is 0, by Vieta formula, which implies that the real part of the sum is also 0. Hence ∑ k = 0 4 Re ( e i ( 8 k + 5 ) π 20 ) = 0 \displaystyle \sum_{k=0}^4 \text{Re} \left(e^{i \frac{(8k+5)\pi}{20}}\right) =0 . Now

cos ⁡ ( 5 π 20 ) + cos ⁡ ( 13 π 20 ) + cos ⁡ ( 21 π 20 ) + cos ⁡ ( 29 π 20 ) + cos ⁡ ( 37 π 20 ) = 0 \cos\left(\frac{5\pi}{20}\right) + \cos\left(\frac{13\pi}{20}\right) +\cos\left(\frac{21\pi}{20}\right) +\cos\left(\frac{29\pi}{20}\right) +\cos\left(\frac{37\pi}{20}\right)=0

So, A = cos ⁡ ( 13 π 20 ) + cos ⁡ ( 21 π 20 ) + cos ⁡ ( 29 π 20 ) + cos ⁡ ( 37 π 20 ) = − cos ⁡ ( 5 π 20 ) = − 1 2 ≈ − 0.7071 A= \cos\left(\frac{13\pi}{20}\right) +\cos\left(\frac{21\pi}{20}\right) +\cos\left(\frac{29\pi}{20}\right) +\cos\left(\frac{37\pi}{20}\right)=-\cos\left(\frac{5\pi}{20}\right) =-\frac{1}{\sqrt{2}} \approx -0.7071

Finally, ⌊ 1000 A ⌋ = ⌊ − 707.1 ⌋ = − 708 \lfloor 1000A \rfloor = \lfloor -707.1 \rfloor = \boxed{-708} .

4 Helpful 1 Interesting 1 Brilliant 0 Confused
Naren Bhandari
Aug 2, 2018

A = cos ⁡ ( 13 π 20 ) + cos ⁡ ( 21 π 20 ) + cos ⁡ ( 29 π 20 ) + cos ⁡ ( 37 π 20 ) = 2 cos ⁡ ( 50 π 40 ) ⋅ cos ⁡ ( 24 π 40 ) + 2 cos ⁡ ( 50 π 40 ) ⋅ cos ⁡ ( 8 π 40 ) = 2 cos ⁡ 5 π 4 { cos ⁡ ( 3 π 5 ) + cos ⁡ ( π 5 ) } = 2 2 cos ⁡ ( 5 π 4 ) = cos ⁡ ( π + π 4 ) = − 1 2 = − 0.7071067812 \begin{aligned} A & =\cos\left(\frac{13\pi}{20}\right) +\cos\left(\frac{21\pi}{20}\right) +\cos\left(\frac{29\pi}{20}\right) +\cos\left(\frac{37\pi}{20}\right)\\ & = 2\cos \left(\dfrac{50\pi }{40}\right) \cdot\cos \left(\dfrac{24\pi }{40}\right) + 2\cos \left(\dfrac{50\pi }{40}\right) \cdot \cos \left(\dfrac{8\pi }{40}\right) \\ & = 2\cos \dfrac{5\pi }{4}\left\{ \cos \left(\dfrac{3\pi }{5}\right)+\cos \left(\dfrac{\pi }{5}\right) \right\} \\ & = \dfrac{2}{2} \cos \left(\dfrac{5\pi}{4}\right) = \cos \left(\pi + \dfrac{\pi}{4}\right) =- \dfrac{1}{\sqrt 2} = - 0.7071067812 \end{aligned} Therefore, ⌊ 1000 A ⌋ = ⌊ − 707.1067812 ⌋ = − 708 \left\lfloor 1000A \right\rfloor\ = \left\lfloor- 707.1067812 \right\rfloor= \boxed{-708 }

Note cos ⁡ ( π 5 ) = 1 − 2 sin ⁡ 2 ( π 10 ) = 1 − 2 ( 5 − 1 4 ) 2 = 5 + 1 4 = ϕ 2 cos ⁡ ( 3 π 5 ) = − sin ⁡ ( π 2 + π 10 ) = − sin ⁡ ( π 10 ) = − ( 5 − 1 4 ) = 1 − 5 4 = − ϕ − 1 2 cos ⁡ ( π 5 ) + cos ⁡ ( 3 π 5 ) = 5 + 1 4 + 1 − 5 4 = 1 2 \cos \left(\dfrac{\pi}{5} \right) = 1 - 2 \sin ^2 \left(\dfrac{\pi}{10}\right) = 1 - 2\left(\dfrac{\sqrt 5 -1}{4}\right)^2 = \dfrac{\sqrt 5 +1}{4} = \dfrac{\phi}{2} \\ \cos \left(\dfrac{3\pi}{5}\right) =- \sin \left(\dfrac{\pi}{2} + \dfrac{\pi}{10}\right)= - \sin \left(\dfrac{\pi}{10}\right) = -\left(\dfrac{\sqrt 5-1 }{4 }\right) = \dfrac{1- \sqrt 5}{4}=-\dfrac{\phi^{-1}}{2} \\ \cos \left( \dfrac{\pi}{5} \right) + \cos \left(\dfrac{3\pi}{5}\right) = \dfrac{\sqrt 5+1}{4} + \dfrac{1- \sqrt 5 }{4}=\dfrac{1}{2}

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