#sum_(k=1)^5 cos(k(pi/11)) = # ?

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Answer 1 · 2017-10-23T03:13:25

#LHS=cos^2 (pi /11) + cos^2((2 pi )/11 )+ cos^2 ((3 pi)/ 11) + cos^2(( 4 pi)/ 11) + cos^2(( 5 pi )/11) )#

#=1/2(2cos^2 (pi /11) + 2cos^2((2 pi )/11 )+ 2cos^2 ((3 pi)/ 11) + 2cos^2(( 4 pi)/ 11) + 2cos^2(( 5 pi )/11) )#

#=1/2(5+cos ((2pi) /11) + cos((4 pi )/11 )+ cos((6pi)/ 11) + cos(( 8pi)/ 11) + cos(( 10pi )/11) )# #color(red)(["using "2cos^2theta=(1+cos2theta)])#

#=1/2(5+cos ((6pi) /11) + cos((4 pi )/11 )+cos((2pi)/ 11) + cos(( 8pi)/ 11) +cos((pi-pi /11) ))#

#=1/2(5 + 2cos((5 pi )/11 )cos((pi)/ 11) + 2cos(( 5pi )/11) cos((3pi)/11)-cos(pi/11))#

#=1/2(5 + 2cos((5 pi )/11 )(cos((pi)/ 11) + cos((3pi)/11))-cos(pi/11))#

#=1/2(5 + 2cos((5 pi )/11 )*2cos((2pi)/ 11) cos((pi)/11)-cos(pi/11))#

#=1/2(5 + cos((5 pi )/11 )/sin(pi/11)*2cos((2pi)/ 11) *2sin(pi/11)cos((pi)/11)-cos(pi/11))#

#=1/2(5 + cos((5 pi )/11 )/sin(pi/11)*2cos((2pi)/ 11) sin((2pi)/11)-cos(pi/11))#

#=1/2(5 + (2cos((5 pi )/11 )sin((4pi)/11))/(2sin(pi/11))-cos(pi/11))#

#=1/2(5 + (sin((9 pi )/11 )-sin((pi)/11))/(2sin(pi/11))-cos(pi/11))#

#=1/2(5 + sin(pi-(2 pi )/11 )/(2sin(pi/11))-sin((pi)/11)/(2sin(pi/11))-cos(pi/11))#

#=1/2(5 + sin((2 pi )/11 )/(2sin(pi/11))-1/2-cos(pi/11))#

#=1/2(5 + (2sin(pi /11 )cos(pi/11))/(2sin(pi/11))-1/2-cos(pi/11))#

#=1/2(5 + cos(pi/11)-1/2-cos(pi/11))#

#=1/2(5-1/2)#

#=1/2*9/2=9/4=RHS#

Answer 2 · 2017-10-23T11:38:59

See below.

Using the identity

#(e^(ix) +e^(-ix))/2= cosx# then

#cos^2x = 1/4(e^(2ix)+2+e^(-2ix))= 1/2+1/4(e^(2ix)+e^(-2ix))# then

#S= sum_(k=1)^5 cos(k(pi/11)) = 5/2+1/4(sum_(k=1)^5e^(2i k (pi/11)) + sum_(k=1)^5 e^(-2ik (pi/11)))=#

#=5/2+1/4((e^(2i6(pi/11))-1)/(e^(2i(pi/11))-1)+(e^(-2i6(pi/11))-1)/(e^(-2i(pi/11))-1)-2)#

but

#(e^(2i6(pi/11))-1)/(e^(2i(pi/11))-1)+(e^(-2i6(pi/11))-1)/(e^(-2i(pi/11))-1)=1# so

#S = 5/2-1/4 = 9/4#