Prove that ∏k=1n−1sin⁡kπn=n2n−1 Using nth root of unity (e2kiπn)n=1

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Prove that   ∏k=1n−1sin⁡kπn=n2n−1  Using nth root of unity  (e2kiπn)n=1

nick1337 · Accepted Answer

Step 1 P=∏k=1n−1sin⁡(kπn) =(2i)1−n∏k=1n−1(eikπn−e−ikπn) =(2i)1−ne−in(n−1)2πn∏k=1n−1(e2ikπn−1) =(−2)1−n∏k=1n−1(ξk−1) =21−n∏k=1n−1(1−ξk) where ξ=e2iπn Now note that xn−1=(x−1)∑k=0n−1 and xn−1=∏k=0n−1(x−ξk)  Cancelling (x−1) we have ∏k=1n−1(x−ξk)=∑k=0n−1xk Substituting x=1 we have ∏k=1n−1(1−ξk)=n ∴P=n21−n Step 2 In order to note that xn−1=∏k=0n−1(x−ξk), note that 1, ξ, ⋯, ξn−1 are roots of xn−1. Therefore by polynomial reminder theorem we have xn−1=Q(x)∏k=0n−1(x−ξk) Comparing degrees we find Q(x) has degree 0. Comparing highest coefficients we conclude Q(x)=1 Step 3 We may instead use the identity |1−e2ikπn==2sin⁡(kπn), k=1, ⋯, n−1 to establish immediately that P=∏k=1n−1sin⁡(kπn)=21−n∏k=1n−1|1−e2ikπn|, and =21−n|∏k=1n−1(1−e2ikπn)| continue by applying the foregoing logic to the product to obtain P=n21−n

Vasquez · Accepted Answer

Step 1 Consider zn=1, each root is ξk=cos⁡2kπn+isin⁡2kπn=ei2kπn k=0, 1, 2, ⋯, n−1 So, we have zn−1=∏k=0n−1(z−ξk) ⇒(z−1)(zn−1+⋯+z2+z+1)=(z−ξ0) ∏k=1n−1(z−ξk) ⇒(z−1)(zn−1+⋯+z2+z+1)=(z−1) ∏k=1n−1(z−ξk) ⇒zn−1+⋯+z2+z+1=∏k=1n−1(z−ξk) By substituting z=1, ⇒n=∏k=1n−1(1−ξk) Next, take the modulus on both sides, |n|=n=|∏k=1n−1(1−ξk)|=∏k=1n−1|(1−ξk)| 1−ξk=1−(cos⁡2kπn+isin⁡2kπn)=2sin kπn(sin⁡kπn−icos⁡kπn) |1−ξk|=2sin⁡kπn So, n=2n−1∏k=1n−1sin⁡kπn ∏k=1n−1sin⁡kπn=n2n−1

alenahelenash · Accepted Answer

Step 1 Here is a more 1st principles pf. I use a hint in Marsdens

Prove that prod_{k=1}^{n-1}\sin\frac{k\pi}{n}=\frac{n}{2^{n-1}} Using n^{th} root of unity (e^{\frac{2ki\pi}{n}})^{n}=1

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