Evaluating $\cos\frac{2\pi}{2021}+\cos\frac{4\pi}{2021}+\cdots+\cos\frac{2020\pi}{2021}$

Question

Evaluate $$\cos\left(\frac{2\pi}{2021}\right)+\cos\left(\frac{4\pi}{2021}\right)+\cdots+\cos\left(\frac{2020\pi}{2021}\right)$$

Answer 1

We can directly use the sum of cosines of angles in an ap formula $$\sum_{k=0}^{n-1}\cos\left(a+kd\right)=\frac{\left(\cos\left(a+\frac{\left(n-1\right)d}{2}\right)\sin\left(\frac{nd}{2}\right)\right)}{\sin\left(\frac{d}{2}\right)}$$

With $$a=2\pi/2021; d=2\pi/2021, n=1010$$

We get the sum as $\begin{array}{l}\frac{\cos\left(\frac{1011\pi}{2021}\right)\sin\left(\frac{1010\pi}{2021}\right)}{\sin\left(\frac{\pi}{2021}\right)}\end{array}$

Now use the identity $\cos \left(A\right)\sin \left(B\right)=\frac{\left(\sin \left(A+B\right)-\sin \left(A-B\right)\right)}{2}$

The above expression evaluates to $\begin{array}{l}\frac{\left(\sin \pi -\frac{\sin \pi }{2021}\right)}{2\sin \left(\frac{\pi }{2021}\right)}\\\end{array}$

Which is equal to $$-1/2$$

Answer 2

Let $t=\cos 2x+\cos 4x+...\cos 2nx$, $x=\frac{\pi}{2n+1}$. It's easy to see: $$2\sin x\cos kx=\sin(k+1)x-\sin(k-1)x.$$ Then \begin{align*} 2t\sin x&=2\sin x(\cos 2x+\cos 4x+...+\cos 2nx)\\ &=[\sin 3x-\sin x]+[\sin 5x-\sin 3x]+...+[\sin(2n+1)x\\&-\sin(2n-1)x]\\ &=\sin(2n+1)x-\sin x\\ &=-\sin x. \end{align*} Therefore $t=-\frac{1}{2}$. Finally $$\cos\left(\frac{2\pi}{2021}\right)+\cos\left(\frac{4\pi}{2021}\right)+...+\cos\left(\frac{2020\pi}{2021}\right)=-\frac{1}{2}.$$