Indian Statistical Institute B.Math & B.Stat : Trigonometry

Find the value of the sum $\cos \frac{2 \pi}{1000} + \cos \frac{4 \pi}{1000} + \dots + \cos \frac{1998 \pi}{1000}$. Let $z = \cos \frac{\pi}{1000} + i \sin \frac{\pi}{1000} = \cos \theta + i \sin \theta$ where $\theta = \frac{\pi}{1000}$ . It is easy to see that $z \neq 1,-1$. Consider the sum $1 +z^2+z^4+ \dots + z^{1998}$, $z \neq 1,-1$. Putting $w = z^2$ the sum reduces to $1 +w+w^2+ \dots + w^{999}$, $w \neq 1$. Now, $1 +w+w^2+ \dots + w^{999} = \frac{w^{1000}-1}{w-1}$ Substituting back $w$ we have the following identity $1 +z^2+z^4+ \dots + z^{1998} = \frac{z^{2000}-1}{z^2-1}$,$z \neq 1,-1$. Using $De-Moivre's$ theorem we have $z^n = \cos n \theta + i \sin n \theta$ for $n \in \mathbb{N}$. Substituting back in the above identity we have, $\big(1+ \cos 2 \theta + \cos 4 \theta + \dots + \cos 1998 \theta \big) + i \big(1+ \sin 2 \theta + \sin 4 \theta + \dots + \sin 1998 \theta \big) = \frac{\cos 2000 \theta + i \sin 2000 \theta -1}{\cos 2 \theta + i \sin 2 \theta -1}$ Equating the real part from both side we have. $1+ \cos 2 \theta + \cos 4 \theta + \dots + \cos 1998 \theta = Re \bigg( \frac{\cos 2000 \theta + i \sin 2000 \theta -1}{\cos 2 \theta + i \sin 2 \theta -1} \bigg) = Re \bigg( \frac{\cos 2 \pi + i \sin 2 \pi -1 }{\cos 2 \theta + i \sin 2 \theta -1}\bigg)$, since $\theta = \frac{\pi}{1000}$. Therefore $1+ \cos 2 \theta + \cos 4 \theta + \dots + \cos 1998 \theta = Re (0) = 0 \implies \cos 2 \theta + \cos 4 \theta + \dots + \cos 1998 \theta = -1$. $\implies \cos \frac{2 \pi}{1000} + \cos \frac{4 \pi}{1000} + \dots + \cos \frac{1998 \pi}{1000} = -1$