trigonometric identity involving product of sines of roots of unity


Let n>1 be a positive integer, and ζn=e2iπ/n, a primitive nth root of unityMathworldPlanetmath.

The purpose of this article is to prove

Theorem 1.

Let m=n2. Then

k=1msin2(πkn)=k=1n-1sin(πkn)=n2n-1 (1)

The theorem follows easily from the following simple lemma:

Lemma 2.

Let n>1 be a positive integer. Then

k=1n-1(1-ζnk)=n
Proof.

We have xn-1=k=1n(x-ζnk). Dividing both sides by x-1 gives

xn-1x-1=1+x+x2+xn-1=k=1n-1(x-ζnk)

Substitute x=1 to get the result. ∎

Proof of Theorem 1.

Using the definition of ζn and the half-angle formulas, we have

1-ζnk =1-cos(2πkn)-isin(2πkn)
=2sin2(πkn)-2isin(πkn)cos(πkn)
=2sin(πkn)(sin(πkn)-icos(πkn))

Note that |sinθ-icosθ|=sin2θ+cos2θ=1, so taking absolute valuesMathworldPlanetmathPlanetmathPlanetmath, we get

|1-ζnk|=2|sin(πkn)|

Now, for 1kn-1, sin(πkn)>0 so is equal to its absolute value. Thus (using, for n even, the fact that sinπ2=1),

k=1msin2(πkn) =k=1msin(πkn)sin(π-πkn)=k=1msin(πkn)sin(π(n-k)n)
=k=1n-1sin(πkn)=k=1n-1|sin(πkn)|
=12n-1|k=1n-1(1-ζnk)|=12n-1|n|
=n2n-1

(Thanks to dh2718 for greatly simplifying the original proof.)

Title trigonometric identity involving product of sines of roots of unity
Canonical name TrigonometricIdentityInvolvingProductOfSinesOfRootsOfUnity
Date of creation 2013-03-22 19:00:03
Last modified on 2013-03-22 19:00:03
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 11
Author rm50 (10146)
Entry type Theorem
Classification msc 26A09
Classification msc 33B10