Prove that if #alpha=pi/18#, #cos1alpha+cos2alpha+...+cos18alpha=-1#?

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Answer 1 · 2018-01-13T07:44:44

Please see below.

Recall that #cos(pi-A)=-cosA#

hence, #cos((17pi)/18)=cos(pi-pi/18)=-cos(pi/18)#

and hence #cos((17pi)/18)+cos(pi/18)=0#

or #cosalpha+cos17alpha=0#

Similarly #cos2alpha+cos16alpha=0#

#cos3alpha+cos15alpha=0#

#cos4alpha+cos14alpha=0#
...
...
#cos8alpha+cos10alpha=0#

as #cos9alpha=cos(pi/2)=0# and #cos18alpha=cospi=-1#

Hence #cosalpha+cos2alpha+...+cos18alpha=-1#

But as #cos0alpha=cos0=1#, we have

#cos0alpha+cos1alpha+cos2alpha+...+cos18alpha=0#

and #cos1alpha+cos2alpha+...+cos18alpha=-1#