# Trigonometric Summation

$$\sum_{j=0}^7 \cos^{16} \left(\frac {j\pi}8\right)$$

Is it possible to derive directly a nice closed form for this summation?

The answer is $1.57129$ or $\frac {25744}{16384}$ or $\frac {2\cdot \left[\binom {16}0+\binom {16}8+\binom{16}{16}\right]}{2^{14}}$ which was where the summation in this question emerged from. The objective is to use the summation in the question to compute the summation of the binomial coefficients.

• $$2\cos^2\left\{(2m+1)\dfrac\pi8\right\}=1+\sqrt2$$ Now $\cos^2\dfrac{7\pi}8=\cos^2\dfrac\pi8,\cos^2\dfrac{3\pi}8=\cos^2\dfrac{5\pi}8$ are the roots of $$8d^2-8d+1=0$$ – lab bhattacharjee Nov 24 '16 at 18:40
• @labbhattacharjee - Did you mean $1\pm \frac 1{\sqrt{2}}$? – hypergeometric Nov 26 '16 at 1:56

As $\cos(\pi-y)=-\cos y$

$\cos\dfrac{7\pi}8=-\cos\dfrac\pi8,$

$\cos\dfrac{6\pi}8=-\cos\dfrac{2\pi}8=-\dfrac1{\sqrt2}$

$\cos\dfrac{5\pi}8=-\cos\dfrac{3\pi}8$

$$\implies\sum_{r=0}^7\cos^{16}\dfrac{r\pi}8=\cos^{16}\dfrac{0\cdot\pi}8+\cos^{16}\dfrac{4\cdot\pi}8+2\sum_{r=1}^3\cos^{16}\dfrac{r\pi}8$$

$$=1+0+2\left(\dfrac1{\sqrt2}\right)^{16}+2\left(\cos^{16}\dfrac{\pi}8+\cos^{16}\dfrac{3\pi}8\right)$$

As $\dfrac{3\pi}8+\dfrac{\pi}8=\dfrac\pi2,\cos\dfrac{3\pi}8=\sin\dfrac{\pi}8$

$$\implies\cos^{16}\dfrac{\pi}8+\cos^{16}\dfrac{3\pi}8=\cos^{16}\dfrac{\pi}8+\sin^{16}\dfrac{\pi}8$$

Now $\cos^2\dfrac{\pi}8\sin^2\dfrac{\pi}8=\dfrac{\sin^2\dfrac\pi4}4=\dfrac18$

$\cos^4\dfrac{\pi}8+\sin^4\dfrac{\pi}8=1-2\cos^2\dfrac{\pi}8\sin^2\dfrac{\pi}8=1-2\cdot\dfrac18=\dfrac34$

$\cos^8\dfrac{\pi}8+\sin^8\dfrac{\pi}8=\left(\cos^4\dfrac{\pi}8+\sin^4\dfrac{\pi}8\right)^2-2\left(\cos^2\dfrac{\pi}8\sin^2\dfrac{\pi}8\right)^2=\left(\dfrac34\right)^2-2\left(\dfrac18\right)^2=\cdots=\dfrac{17}{32}$

$\cos^{16}\dfrac{\pi}8+\sin^{16}\dfrac{\pi}8=\left(\cos^8\dfrac{\pi}8+\sin^8\dfrac{\pi}8\right)^2-2\left(\cos^2\dfrac{\pi}8\sin^2\dfrac{\pi}8\right)^4=\cdots=\dfrac{577}{1024}$

$$\implies\sum_{r=0}^7\cos^{16}\dfrac{r\pi}8=1+\dfrac2{2^8}+\dfrac{577}{1024}\approx1.5712890625$$

• Very nice, thanks. I had earlier posted a similar solution here. Do you think there is a nice closed form for general parameters, e.g. $\displaystyle \sum_{r=0}^{n-1}\cos^{2n}\left(\frac {r\pi}n\right)$, or perhaps only where $n=2^m (m\in \mathbb Z)$? – hypergeometric Nov 26 '16 at 16:04
• @hypergeometric, More generally, $$\sum_{r=0}^{n-1}\cos^{2k}\dfrac{r\pi}n$$ – lab bhattacharjee Nov 26 '16 at 16:07
• @hypergeometric, Initially, I was planning to form a Polynomial equation in cosine. The current approach is probably not ready for generalization. – lab bhattacharjee Nov 26 '16 at 16:09
• Yes, and thanks - corrected. – hypergeometric Nov 26 '16 at 16:09