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Question-7125




Question Number 7125 by Tawakalitu. last updated on 11/Aug/16
Commented by Tawakalitu. last updated on 11/Aug/16
I have seen a solution to this question
Ihaveseenasolutiontothisquestion
Answered by sandy_suhendra last updated on 13/Aug/16
[cos (π/7)+cos((3π)/7)+cos((5π)/7)]×((2 sin(π/7))/(2 sin (π/7)))  =((2sin(π/7)cos(π/7)+2sin(π/7)cos((3π)/7)+2sin(π/7)cos((5π)/7))/(2 sin(π/7)))  2sinAcosB=sin(A+B) + sin(A−B)  sin (−A) = −sinA  =((sin((2π)/7)+sin 0+sin((4π)/7)−sin((2π)/7)+sin((6π)/7)−sin((4π)/7))/(2 sin (π/7)))  = ((sin ((6π)/7))/(2 sin (π/7))) = ((sin [π−(π/7)])/(2 sin (π/7))) = ((sin (π/7))/(2 sin (π/7))) = (1/2)
[cosπ7+cos3π7+cos5π7]×2sinπ72sinπ7=2sinπ7cosπ7+2sinπ7cos3π7+2sinπ7cos5π72sinπ72sinAcosB=sin(A+B)+sin(AB)sin(A)=sinA=sin2π7+sin0+sin4π7sin2π7+sin6π7sin4π72sinπ7=sin6π72sinπ7=sin[ππ7]2sinπ7=sinπ72sinπ7=12
Commented by Rasheed Soomro last updated on 15/Aug/16
G^(OO) D Approach!
GOODApproach!

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