Summation or Product with trigonometric function

by fungarwai, May 24, 2019, 1:49 PM

Summation with trigonometric function

$\sum_{k=0}^{n-1} \cos(a+kb)
=\frac{1}{2\sin\frac{b}{2}}\left(\sin(a-\frac{b}{2}+nb)-\sin(a-\frac{b}{2})\right)$

$\sum_{k=0}^{n-1} \sin(a+kb)
=\frac{1}{2\sin\frac{b}{2}}\left(\cos(a-\frac{b}{2})-\cos(a-\frac{b}{2}+nb))\right)$

Proof

Product with trigonometric function

$\prod_{k=0}^{n-1}\cos 2^k\theta = \frac{\sin2^n \theta}{2^n\sin\theta}$

Proof

$\prod_{k=0}^{n-1}\sin\left(x+\frac{k\pi}{n}\right) = \frac{\sin nx}{2^{n-1}}$

Proof

Equations for trigonometric function roots

For $x=\tan\theta, \theta=\theta_0+\frac{k\pi}{2n}, k=0,1,2,\dots,2n-1$

$\sum_{r=0}^n (-1)^r \binom{2n}{2r}x^{2r}
-\cot 2n\theta_0\sum_{r=0}^{n-1} (-1)^r \binom{2n}{2r+1}x^{2r+1}=0$

For $x=\tan\theta, \theta=\theta_0+\frac{k\pi}{2n+1}, k=0,1,2,\dots,2n$

$\tan (2n+1)\theta_0\sum_{r=0}^n (-1)^r \binom{2n+1}{2r}x^{2r}
-\sum_{r=0}^n (-1)^r \binom{2n+1}{2r+1}x^{2r+1}=0$

Proof

Example

$x=\cos^2 \theta, \theta=\frac{k\pi}{2n}, k=1,2,\dots, n-1,
\sum_{r=0}^{n-1}(-1)^r \binom{2n-1-r}{r}\frac{x^{n-1-r}}{2^{2r}}=0$

$x=\cos^2 \theta, \theta=\frac{k\pi}{2n+1}, k=1,2,\dots, n,
\sum_{r=0}^n (-1)^r \binom{2n-r}{r} \frac{x^{n-r}}{2^{2r}}=0$

$x=\sec^2 \theta, \theta=\frac{k\pi}{2n}, k=1,2,\dots, n-1,
\sum_{r=0}^{n-1}(-1)^r \binom{2n-1-r}{r} 2^{2(n-1-r)} x^r=0$

$x=\sec^2 \theta, \theta=\frac{k\pi}{2n+1}, k=1,2,\dots, n,
\sum_{r=0}^n (-1)^r \binom{2n-r}{r} 2^{2(n-r)}x^r=0$

Proof

Example

For $x=\tan^2 \theta, \theta=\frac{k\pi}{2n}, k=1,2,\dots,n-1$, $\sum_{r=0}^{n-1} (-1)^{n-r}\binom{2n}{2r+1}x^r=0$

For $x=\tan^2 \theta, \theta=\frac{k\pi}{2n+1}, k=1,2,\dots,n$, $\sum_{r=0}^n (-1)^{n-r}\binom{2n+1}{2r+1}x^r=0$

For $x=\cot^2 \theta, \theta=\frac{k\pi}{2n}, k=1,2,\dots,n-1$, $\sum_{r=0}^{n-1} (-1)^r \binom{2n}{2r+1}x^{n-1-r}=0$

For $x=\cot^2 \theta, \theta=\frac{k\pi}{2n+1}, k=1,2,\dots,n$, $\sum_{r=0}^n (-1)^r \binom{2n+1}{2r+1}x^{n-r}=0$

Proof

Example


$\begin{array}{|c|c|c|c|c|c|c|} 
\hline
\displaystyle f(\theta) & \sin\theta & \cos\theta & \tan\theta &
\csc\theta & \sec\theta & \cot\theta \\
\hline
\displaystyle \sum_{k=0}^{2n-1} f(\theta_0+\frac{k\pi}{2n}) & 
\displaystyle \frac{\cos(\frac{\pi}{4n}-\theta_0)}{\sin\frac{\pi}{4n}} & 
\displaystyle \frac{\sin(\frac{\pi}{4n}-\theta_0)}{\sin\frac{\pi}{4n}} &
\displaystyle -2n\cot 2n\theta_0 &
\displaystyle ~ & 
\displaystyle ~ & 
\displaystyle ~\\
\hline
\displaystyle \sum_{k=0}^{2n} f(\theta_0+\frac{k\pi}{2n+1}) & 
\displaystyle \frac{\cos(\frac{\pi}{2(2n+1)}-\theta_0)}{\sin\frac{\pi}{2(2n+1)}} & 
\displaystyle \frac{\sin(\frac{\pi}{2(2n+1)}-\theta_0)}{\sin\frac{\pi}{2(2n+1)}} &
\displaystyle (2n+1)\tan (2n+1)\theta_0 &
\displaystyle ~ & 
\displaystyle ~ & 
\displaystyle ~\\
\hline
\displaystyle \sum_{k=1}^{n-1} [f(\frac{k\pi}{2n})]^2 & 
\displaystyle \frac{n-1}{2} & 
\displaystyle \frac{n-1}{2} &
\displaystyle \frac{(n-1)(2n-1)}{3} &
\displaystyle \frac{2(n-1)(n+1)}{3} & 
\displaystyle \frac{2(n+1)(n-1)}{3} & 
\displaystyle \frac{(n-1)(2n-1)}{3}\\
\hline
\displaystyle \sum_{k=1}^{n} [f(\frac{k\pi}{2n+1})]^2 & 
\displaystyle \frac{2n+1}{4} & 
\displaystyle \frac{2n-1}{4} &
\displaystyle n(2n+1) &
\displaystyle \frac{2n(n+1)}{3} & 
\displaystyle 2n(n+1) & 
\displaystyle \frac{n(2n-1)}{3}\\
\hline
\displaystyle \prod_{k=1}^{n-1} f(\frac{k\pi}{2n}) & 
\displaystyle \frac{\sqrt{n}}{2^{n-1}} & 
\displaystyle \frac{\sqrt{n}}{2^{n-1}} &
\displaystyle 1 &
\displaystyle \frac{2^{n-1}}{\sqrt{n}} & 
\displaystyle \frac{2^{n-1}}{\sqrt{n}} & 
\displaystyle 1\\
\hline
\displaystyle \prod_{k=1}^{n} f(\frac{k\pi}{2n+1}) & 
\displaystyle \frac{\sqrt{2n+1}}{2^n} & 
\displaystyle \frac{1}{2^n} &
\displaystyle \sqrt{2n+1} &
\displaystyle \frac{2^n}{\sqrt{2n+1}} & 
\displaystyle 2^n & 
\displaystyle \frac{1}{\sqrt{2n+1}}\\
\hline
\end{array}$
This post has been edited 15 times. Last edited by fungarwai, Jun 9, 2020, 8:01 AM

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