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Question Number 156841 by mnjuly1970 last updated on 16/Oct/21

Answered by gsk2684 last updated on 16/Oct/21

let C=cos (π/(2n+1)).cos ((2π)/(2n+1)).cos ((3π)/(2n+1))...cos (((n−1)π)/(2n+1)).cos ((nπ)/(2n+1)) let S=sin (π/(2n+1)).sin ((2π)/(2n+1)).sin ((3π)/(2n+1))...sin (((n−1)π)/(2n+1)).sin((nπ)/(2n+1)) 2^n .S.C=(2sin (π/(2n+1))cos (π/(2n+1)))(2sin ((2π)/(2n+1))cos ((2π)/(2n+1)))(2sin ((3π)/(2n+1))cos ((3π)/(2n+1)))...(2sin (((n−1)π)/(2n+1))cos (((n−1)π)/(2n+1)))(2sin ((nπ)/(2n+1))cos ((nπ)/(2n+1))) =sin ((2π)/(2n+1)).sin ((4π)/(2n+1)).sin ((6π)/(2n+1))...sin ((2(n−1)π)/(2n+1)).sin((2nπ)/(2n+1)) =sin ((2π)/(2n+1)).sin ((4π)/(2n+1)).sin ((6π)/(2n+1))...sin (π−((2π)/(2n+1))).sin(π−(π/(2n+1))) =sin (π/(2n+1)).sin ((2π)/(2n+1)).sin ((3π)/(2n+1))...sin (((n−1)π)/(2n+1)).sin((nπ)/(2n+1)) 2^n CS=S C=(1/2^n )

letC=cosπ2n+1.cos2π2n+1.cos3π2n+1...cos(n−1)π2n+1.cosnπ2n+1letS=sinπ2n+1.sin2π2n+1.sin3π2n+1...sin(n−1)π2n+1.sinnπ2n+12n.S.C=(2sinπ2n+1cosπ2n+1)(2sin2π2n+1cos2π2n+1)(2sin3π2n+1cos3π2n+1)...(2sin(n−1)π2n+1cos(n−1)π2n+1)(2sinnπ2n+1cosnπ2n+1)=sin2π2n+1.sin4π2n+1.sin6π2n+1...sin2(n−1)π2n+1.sin2nπ2n+1=sin2π2n+1.sin4π2n+1.sin6π2n+1...sin(π−2π2n+1).sin(π−π2n+1)=sinπ2n+1.sin2π2n+1.sin3π2n+1...sin(n−1)π2n+1.sinnπ2n+12nCS=SC=12n

Commented by mnjuly1970 last updated on 16/Oct/21

very nice sir..grateful...

verynicesir..grateful...

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