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## Main Question or Discussion Point

Given that

n = Π[2 – 2cos(kπ/ n)] ........ (where Π is the product sign , from k = 1 to n-1 )

as

cos2@ = 1 – 2(sin@)^2

then

2 – 2cos(kπ/ n) = 4[sin(kπ/ 2n)]^2 , for k = 1 , 2 , 3 , … n-1

then

n = Π[4[sin(kπ/ 2n)]^2] = [4^(n-1)] Π[sin(kπ/ 2n)]^2

but the book then said

Π[sin(kπ/ n)]^2 = n / [4^(n-1)]

why ?

why is not the sin(kπ/ 2n) but sin(kπ/ n) ?

n = Π[2 – 2cos(kπ/ n)] ........ (where Π is the product sign , from k = 1 to n-1 )

as

cos2@ = 1 – 2(sin@)^2

then

2 – 2cos(kπ/ n) = 4[sin(kπ/ 2n)]^2 , for k = 1 , 2 , 3 , … n-1

then

n = Π[4[sin(kπ/ 2n)]^2] = [4^(n-1)] Π[sin(kπ/ 2n)]^2

but the book then said

Π[sin(kπ/ n)]^2 = n / [4^(n-1)]

why ?

why is not the sin(kπ/ 2n) but sin(kπ/ n) ?