$\cos^2(\frac{\pi}{101})+\cos^2(\frac{2\pi}{101})+\cos^2(\frac{3\pi}{101})+...+\cos^2(\frac{100\pi}{101})=?$

Question

Find the value: $$\cos^2\left(\frac{\pi}{101}\right)+\cos^2\left(\frac{2\pi}{101}\right)+\cos^2\left(\frac{3\pi}{101}\right)+\cos^2\left(\frac{4\pi}{101}\right)+\cos^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cdots+\cos^2\left(\frac{99\pi}{101}\right)+\cos^2\left(\frac{100\pi}{101}\right)$$

My attempt:I've tried it by considering the sum $$\sin^2\left(\frac{\pi}{101}\right)+\sin^2\left(\frac{2\pi}{101}\right)+\sin^2\left(\frac{3\pi}{101}\right)+\sin^2\left(\frac{4\pi}{101}\right)+\sin^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cdots+\sin^2\left(\frac{99\pi}{101}\right)+\sin^2\left(\frac{100\pi}{101}\right)$$

along with

$$\cos^2\left(\frac{\pi}{101}\right)+\cos^2\left(\frac{2\pi}{101}\right)+\cos^2\left(\frac{3\pi}{101}\right)+\cos^2\left(\frac{4\pi}{101}\right)+\cos^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cdots +\cos^2\left(\frac{99\pi}{101}\right)+\cos^2\left(\frac{100\pi}{101}\right)$$ which gives $ 100$ as resultant but failed to separate the sum of $$\sin^2\left(\frac{\pi}{101}\right)+\sin^2\left(\frac{2\pi}{101}\right)+\sin^2\left(\frac{3\pi}{101}\right)+\sin^2\left(\frac{4\pi}{101}\right)+\sin^2\left(\frac{5\pi}{101}\right)+\cdots\\ \dots+\sin^2\left(\frac{99\pi}{101}\right)+\sin^2\left(\frac{100\pi}{101}\right)$$ at last.

I tried the next approach by using de Movire's theorem but failed to separate the real and imaginary part.

I've invested a great amount of time in the so it would be better if someone please come up with an answer.

Answer 1

$$\cos\left(\frac{k\pi}{101}\right)= \frac{1}{2} \left(e^{i\frac{k\pi}{101}}+e^{-i\frac{k\pi}{101}} \right) \\ \cos^2\left(\frac{k\pi}{101}\right)= \frac{1}{4} \left(e^{2i\frac{k\pi}{101}}+e^{-2i\frac{k\pi}{101}} +2\right) \\ \sum_{k=1}^{100}\cos^2\left(\frac{k\pi}{101}\right)= \frac{1}{4} \sum_{k=1}^{100}\left(e^{2i\frac{k\pi}{101}}+e^{-2i\frac{k\pi}{101}} +2\right) $$

Now, $$1+\sum_{k=1}^{100}e^{2i\frac{k\pi}{101}}=\sum_{k=0}^{100}\left(e^{2i\frac{\pi}{101}}\right)^k=\frac{1-(e^{2i\frac{\pi}{101}})^{101}}{1-e^{2i\frac{\pi}{101}}}=0 \\ 1+\sum_{k=1}^{100}e^{-2i\frac{k\pi}{101}}=\sum_{k=0}^{100}\left(e^{-2i\frac{\pi}{101}}\right)^k=\frac{1-(e^{-2i\frac{\pi}{101}})^{101}}{1-e^{-2i\frac{\pi}{101}}}=0 $$

Therefore $$\sum_{k=1}^{100}\cos^2\left(\frac{k\pi}{101}\right)= \frac{1}{4} \left(-1-1+200\right) $$

Answer 2

HINT:

Note that

$$\cos^2(x)=\frac{1+\cos(2x)}{2}$$

Then, the problem boils down to evaluating (See This Answer)

$$\sum_{k=1}^{100}\cos(2k\pi/101)=\text{Re}\left(\sum_{k=1}^{100}\left(e^{i2\pi/101}\right)^k\right)$$

The sum $\sum_{k=1}^{100}\cos(2k\pi/101)$ can also be easily evaluated by multiplying by $\frac{\sin(2\pi/101)}{\sin(2\pi/101)}$ and creative telescoping.

The final answer is $\frac12 (100-1)=\frac{99}{2}$.

Answer 3

HINT

Use the fact that $$\cos^2(x) = \dfrac{1}{2}+\dfrac{\cos(2x)}{2}$$

Thus, the result is equal to

$$50 + \dfrac{1}{2}\sum\limits_{k=1}^{100} \cos\left(\dfrac{2\pi k}{101}\right)$$

And the last sum can be proven to be $-1$ using this result: How to prove $\sum_{k=1}^n \cos(\frac{2 \pi k}{n}) = 0$ for any n>1? .

So the answer is $49.5$.

Answer 4

Let $x_{k} = \cos^{2}(k\pi/n)$ where $k = 1, 2, \dots, n$. Then we can see that $z_{k} = 2x_{k} - 1 = \cos(2k\pi/n)$ and hence these are the roots of the equation $P_{n}(z) = 1$ where $P_{n}(z)$ is a polynomial such that $P_{n}(\cos x) = \cos nx$. These polynomials are famous by the name Chebyshev polynomials and satisfy the recurrence $$P_{n + 1}(z) = 2zP_{n}(z) - P_{n - 1}(z)$$ and using the above recurrence we can easily show (via induction) that if $n\geq 2$ then the coefficient of $z^{n - 1}$ in $P_{n}(z)$ is $0$ and hence the sum of roots $z_{k} = \cos(2k\pi/n)$ is $0$. Also note that $z_{n} = 1$ so that $\sum_{k = 1}^{n - 1} z_{k} = -1$ and hence $$\sum_{k = 1}^{n - 1}x_{k} = \frac{n - 2}{2}$$ Putting $n = 101$ we get the desired sum as $99/2$.

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In case the recurrence satisfied by $P_{n} (z) $ seems mysterious one has to note that $$\cos(n+1)x+\cos(n-1)x=2\cos x\cos nx$$ which means that $$P_{n+1}(\cos x) +P_{n-1}(\cos x)=2\cos x P_{n} (\cos x) $$ and replacing $\cos x$ by $z$ we get the desired recurrence.