Saturday, March 29, 2014

Prove that cos(2pi/7)+cos(4pi/7)+cos(6pi/7)=-1/2

We'll multiply both sides by sin (
/7)


sin( /7)cos(2 /7)+sin( /7)cos(4
/7)+sin( /7)cos(6 /7) = -sin( /7)/2


We'll
transform the product in sum:


(1/2)[sin( /7-2 /7) +
sin( /7+2 /7)] + (1/2)[sin( /7-4 /7) + sin( /7+4 /7)] +
(1/2)[sin( /7-6 /7) + sin( /7+6 /7)]= -sin(
/7)/2


We'll divide by (1/2) both
sides:


-sin( /7) + sin(3 /7) - sin(3 /7) +
sin(5 /7) - sin(5 /7) +  sin(7 /7) = -sin(
/7)


We'll eliminate like
terms:


-sin( /7) + sin( )= -sin(
/7)


But sin( ) =
0


-sin( /7) = -sin(
/7)


Since we've get the same results both
sides, the given identity is verified.

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