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

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