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4cos(2π/7) cos(π/7)-1=2cos (2π/7) Prove this

4cos(2π/7) cos(π/7)-1=2cos (2π/7)


Prove this

Grade:Upto college level

4 Answers

Sher Mohammad IIT Delhi
askIITians Faculty 174 Points
10 years ago
4cos(2p/7) cos(p/7)-1=1.24
2cos (2p/7)=1.24


sher mohammad
B.Tech, IIT Delhi.


Vineeth
35 Points
10 years ago
{2[2sin(P/7)Cos(P/7)]Cos(2P/7)}/Sin(P/7)-1 (By dividing numerator and denominator with Sin(P/7) =[2sin(2P/7)Cos(2P/7)]/Sin(P/7)-1 =Sin(4P/7) -1 Sin(P/7) =Sin(P-4P/7) -1 Sin(P/7) =Sin(3P/7) -1 Sin(p/7) =Sin(3P/7)-Sin(P/7) Sin(P/7) =2Cos[(3P+P)/2*7]Sin[(3P-P)/2*7] {SinC-SinD=2Cos[(C+D)/2]Sin[(C-D)/2]} Sin(P/7) =2Cos(2P/7)Sin(P/7) (by cancelling Sin(P/7) in Numerator and denominator) Sin(P/7) =2Cos(2P/7)
Vineeth
35 Points
10 years ago
{2[2sin(P/7)Cos(P/7)]Cos(2P/7)}/Sin(P/7)-1 (By dividing numerator and denominator with Sin(P/7) =[2sin(2P/7)Cos(2P/7)]/Sin(P/7)-1 =Sin(4P/7) -1 Sin(P/7) =Sin(P-4P/7) -1 Sin(P/7) =Sin(3P/7) -1 Sin(p/7) =Sin(3P/7)-Sin(P/7) Sin(P/7) =2Cos[(3P+P)/2*7]Sin[(3P-P)/2*7] Sin(P/7) =2Cos(2P/7)Sin(P/7) (by cancelling Sin(P/7) in Numerator and denominator) Sin(P/7) =2Cos(2P/7)
Vineeth
35 Points
10 years ago
{2[2sin(?/7)Cos(?/7)]Cos(2?/7)}/Sin(?/7)-1 (By dividing numerator and denominator with Sin(?/7) =[2sin(2?/7)Cos(2?/7)]/Sin(?/7)-1 =Sin(4?/7) -1 Sin(?/7) =Sin(?-4?/7) -1 Sin(?/7) =Sin(3?/7) -1 Sin(p/7) =Sin(3?/7)-Sin(?/7) Sin(?/7) =2Cos[(3?+?)/2*7]Sin[(3?-?)/2*7] Sin(?/7) =2Cos(2?/7)Sin(?/7) (by cancelling Sin(?/7) in Numerator and denominator) Sin(?/7) =2Cos(2?/7)

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