Sum up trigonometric series [duplicate]

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$$\cos \frac{2π}{2013} +\cos \frac{4π}{2013} +\cdots+\cos \frac{2010π}{2013} + \cos \frac{2012π}{2013}$$

How to sum it up? *Calculator is not allowed.

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3 Answers

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Let $\alpha=\frac{\pi}{2013}$ and let

$$S=cos2\alpha+cos4\alpha+cos6\alpha+\cdots+cos2010\alpha+cos2012\alpha$$ Multiply both sides with $2sin\alpha$ and use

$$2sinAcosB=sin(A+B)+sin(A-B)$$ so we get

$$(2sin\alpha) \:S=2sin\alpha cos2\alpha+2sin\alpha cos4\alpha+2sin\alpha cos6\alpha+\cdots+2sin\alpha cos2010\alpha+2sin\alpha cos2012\alpha$$ $\implies$

$$(2sin\alpha)S=sin3\alpha-sin\alpha+sin5\alpha-sin3\alpha+sin7\alpha-sin5\alpha+\cdots+sin2011\alpha-sin2009\alpha+sin2013\alpha-sin2011\alpha$$ $\implies$

$$(2sin\alpha)S=-sin\alpha$$

$$S=\frac{-1}{2}$$

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okay Ill give you the hint Note the angles are in AP with common difference $d=2\pi/2013$ and $a=2\pi/2013$ so general formula for such series is $\frac{cos(a+\frac{(n-1)d}{2})}{sin\frac{d}{2}}.sin\frac{nd}{2}$ here $a$= first angle $d$=common difference between two angles and $n$= number of terms. Hope it helps you.

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Original Approach: Using the formula for the sum of a geometric series, we get $$ \sum_{k=0}^{n-1}e^{\frac{2\pi i}nk}=\frac{e^{2\pi i}-e^0}{e^{\frac{2\pi i}n}-1}=0\tag{1} $$ and using Euler's Formula$$ \begin{align} \sum_{k=0}^{n-1}e^{\frac{2\pi i}nk} &=\sum_{k=0}^{n-1}\cos\left(\frac{2\pi}nk\right)+i\sum_{k=0}^{n-1}\sin\left(\frac{2\pi}nk\right)\tag{2} \end{align} $$ Substituting $n\mapsto2n+1$ and subtracting the $k=0$ term yields $$ \sum_{k=1}^{2n}\cos\left(\frac{2\pi}{2n+1}k\right)=-1\tag{3} $$ Substituting $k\mapsto2n+1-k$ shows that $$ \sum_{k=1}^n\cos\left(\frac{2\pi}{2n+1}k\right) =\sum_{k=n+1}^{2n}\cos\left(\frac{2\pi}{2n+1}k\right)\tag{4} $$ Combine $(3)$ and $(4)$ to get the desired result.

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Alternate Approach: Using the formula for the sum of a geometric series, we get $$ \begin{align} \sum_{k=1}^ne^{\frac{2\pi i}{2n+1}k} &=\frac{e^{\frac{2\pi i}{2n+1}(n+1)}-e^{\frac{2\pi i}{2n+1}}}{e^{\frac{2\pi i}{2n+1}}-1}\\ &=\frac{e^{\pi i}-e^{\frac{\pi i}{2n+1}}}{e^{\frac{\pi i}{2n+1}}-e^{-\frac{\pi i}{2n+1}}}\\ &=\frac{-1-\cos\left(\frac\pi{2n+1}\right)-i\sin\left(\frac\pi{2n+1}\right)}{2i\sin\left(\frac\pi{2n+1}\right)}\\ &=\frac i2\cot\left(\frac\pi{4n+2}\right)\color{#C00000}{-\frac12}\tag{5} \end{align} $$ and using Euler's Formula$$ \begin{align} \sum_{k=1}^ne^{\frac{2\pi i}{2n+1}k} &=\color{#C00000}{\sum_{k=1}^n\cos\left(\frac{2\pi}{2n+1}k\right)}+i\sum_{k=1}^n\sin\left(\frac{2\pi}{2n+1}k\right)\tag{6} \end{align} $$ The real parts of $(5)$ and $(6)$ give the desired result.

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