- #1
lfdahl
Gold Member
MHB
- 749
- 0
Prove the identity
\[\sum_{j=1}^{n-1}\csc^2\left ( \frac{j\pi}{n} \right ) = \frac{n^2-1}{3 }.\]
\[\sum_{j=1}^{n-1}\csc^2\left ( \frac{j\pi}{n} \right ) = \frac{n^2-1}{3 }.\]
Thankyou, June29, for a nice and thorough solution! (Cool)June29 said:By De Moivre's theorem followed by the binomial theorem:
$$ \begin{aligned} & ~~\cos(nx)+i\sin(nx) = \cos^n(x)\left(1+i\tan(x)\right)^n \\& \implies \cos(nx) = \Re(\mathcal{RHS}) =\cos^n(x)\sum_{k ~\text{even}}\binom{n}{k}(-1)^{k/2}\tan^{k}(x) \\& \implies \displaystyle \sin(nx) = \Im(\mathcal{RHS}) =\cos^n(x)\sum_{k ~\text{odd}}\binom{n}{k}(-1)^{\frac{1}{2}k(k+1)+1}\tan^{k}(x) \\& \implies \tan(nx) = \frac{\sum_{k ~\text{odd}}\binom{n}{k}(-1)^{\frac{1}{2}k(k+1)+1}\tan^{k}(x)}{\sum_{k ~\text{even}}\binom{n}{k}(-1)^{k/2}\tan^{k}(x)} \end{aligned}$$The $\frac{1}{2}k(k+1)+1$ is not deep - it's just there to ensure alternating signs.
The $\mathcal{LHS}$ gives $\tan(nx) = 0 \implies x = \frac{j\pi}{n} $ for $1 \le j \le n-1$ & from $\mathcal{RHS}$
$$\sum_{k ~\text{odd}}\binom{n}{k}(-1)^{\frac{1}{2}k(k+1)+1}\tan^{k}(x) =0$$
Multiplying out by $\cot^n(x)$ we get
$$\sum_{k ~\text{odd}}\binom{n}{k}(-1)^{\frac{1}{2}k(k+1)+1}\cot^{k}(x) =0$$
So $\cot(x) = t$ where $\displaystyle x = \frac{j\pi}{n}, ~~ 1 \le j \le n-1$ satisfies
$$\sum_{k ~\text{odd}}\binom{n}{k}(-1)^{\frac{1}{2}k(k+1)+1} t^{k} =0$$
By Vieta's formulas, the sum of the squares of the roots is
$$\begin{aligned} \sum_{j=1}^{n-1}\cot^2\left(\frac{j\pi}{n}\right) &= \bigg(\sum_{j=1}^{n-1}\cot\left(\frac{j\pi}{n}\right)\bigg)^2-2\bigg(\sum_{j > j' \ge 1}^{n-1}\cot\left(\frac{j\pi}{n}\right)\cot\left(\frac{j'\pi}{n}\right)\bigg) \\& = 2\binom{n}{3}\bigg/\binom{n}{1} = \frac{1}{3} (n-2) (n-1).\end{aligned}$$
Since $\cot^2{x}+1 = \csc^2{x}$ our desired conclusion is reached:
$$\begin{aligned} \sum_{j=1}^{n-1}\csc^2\left(\frac{j\pi}{n}\right) & = \sum_{j=1}^{n-1}\bigg(1+\cot^2\left(\frac{j\pi}{n}\right)\bigg) \\& = n-1+\sum_{j=1}^{n-1}\cot^2\left(\frac{j\pi}{n}\right) \\& = n-1+\frac{1}{3} (n-2) (n-1) \\& = \frac{1}{3}(n^2-1). \end{aligned} $$
A trigonometric sum identity is an equation that relates the sum of two or more trigonometric functions to another trigonometric function.
The most commonly used trigonometric sum identity is the Pythagorean identity, which states that sin^2(x) + cos^2(x) = 1.
To prove a trigonometric sum identity, you can use basic trigonometric identities, algebraic manipulation, and the properties of sine, cosine, and tangent.
Trigonometric sum identities are important because they allow us to simplify and solve complex trigonometric equations, and they have many applications in fields such as physics, engineering, and navigation.
One example of a trigonometric sum identity is the double-angle identity for cosine, which states that cos(2x) = cos^2(x) - sin^2(x).