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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Prove that
Using nth root of unity

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Expert2022-01-19Added 699 answers

Step 1 P=k=1n1sin(kπn) =(2i)1nk=1n1(eikπneikπn) =(2i)1nein(n1)2πnk=1n1(e2ikπn1) =(2)1nk=1n1(ξk1) =21nk=1n1(1ξk) where ξ=e2iπn Now note that xn1=(x1)k=0n1 and xn1=k=0n1(xξk) Cancelling (x1) we have k=1n1(xξk)=k=0n1xk Substituting x=1 we have k=1n1(1ξk)=n P=n21n Step 2 In order to note that xn1=k=0n1(xξk), note that 1, ξ, , ξn1 are roots of xn1. Therefore by polynomial reminder theorem we have xn1=Q(x)k=0n1(xξk) 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 |1e2ikπn==2sin(kπn), k=1, , n1 to establish immediately that P=k=1n1sin(kπn)=21nk=1n1|1e2ikπn|, and =21n|k=1n1(1e2ikπn)| continue by applying the foregoing logic to the product to obtain P=n21n


Expert2022-01-19Added 597 answers

Step 1 Consider zn=1, each root is ξk=cos2kπn+isin2kπn=ei2kπn k=0, 1, 2, , n1 So, we have zn1=k=0n1(zξk) (z1)(zn1++z2+z+1)=(zξ0) k=1n1(zξk) (z1)(zn1++z2+z+1)=(z1) k=1n1(zξk) zn1++z2+z+1=k=1n1(zξk) By substituting z=1, n=k=1n1(1ξk) Next, take the modulus on both sides, |n|=n=|k=1n1(1ξk)|=k=1n1|(1ξk)| 1ξk=1(cos2kπn+isin2kπn)=2sin kπn(sinkπnicoskπn) |1ξk|=2sinkπn So, n=2n1k=1n1sinkπn k=1n1sinkπn=n2n1


Skilled2022-01-24Added 488 answers

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

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