Question #b954a

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Answer 1 · 2017-03-30T04:51:49

#0.#

We rearrange such of the terms

#cosalpha and cosbeta#, that, #alpha+beta=pi,# as shown below :

#:." The Exp.="{cos(2pi/9)+cos(7pi/9)}+{cos(4pi/9)+cos(5pi/9)}+{cos(pi/3)+cos(2pi/3)}#

#"Now, "alpha+beta=pi rArr cosalpha+cosbeta=cosalpha+cos(pi-alpha)#

#=cosalpha-cosalpha=0.#

#"Hence, the Exp.="{0}+{0}+{0}=0.#

Enjoy Maths.#

N.B.:

The Answer to the unedited Exp. is

#0+0+(-1)/3-cos(pi/3)=-1/3-1/2=-5/6.#

Answer 2 · 2017-03-30T04:53:28

0

#cos 2pi/9 + cos 4 pi/9 + cos 5 pi/9 + cos 7 pi/9 + cos (pi/3) + cos 2 pi/3#

#=cos 40^@ + cos 80^@ + cos 100^@ + cos 140^@ + cos 60^@ + cos 120^@#

note:
#cos 100^@ = - cos 80^@, cos 140^@= - cos 40^@ and cos 120^@ = - cos 60^@#

#=cos 40^@ + cos 80^@ - cos 80^@ - cos 40^@ + cos 60^@ - cos 60^@ = 0#