How do you simplify #sin((2pi)/7)+sin((4pi)/7)+sin((8pi)/7)# ?
2 Answers
Explanation:
=
=
=
=
=
=
=
=
Explanation:
Let:
#alpha = cos ((2pi)/7) + i sin ((2pi)/7)#
Then:
#alpha^7 = cos (2pi) + i sin (2pi) = 1#
So:
#0 = alpha^7-1 = (alpha-1)(alpha^6+alpha^5+alpha^4+alpha^3+alpha^2+alpha+1)#
We find:
#(alpha+alpha^2+alpha^4)^2#
#= (alpha)^2+(alpha^2)^2+(alpha^4)^2+2(alphaalpha^2+alpha^2alpha^4+alpha^4alpha)#
#= alpha^2+alpha^4+alpha^8+2(alpha^3+alpha^6+alpha^5)#
#= alpha+alpha^2+alpha^4+2(alpha+alpha^2+alpha^3+alpha^4+alpha^5+alpha^6-alpha-alpha^2-alpha^4)#
#= alpha+alpha^2+alpha^4+2(-1-alpha-alpha^2-alpha^4)#
#= -(alpha+alpha^2+alpha^4)-2#
So
#t^2+t+2 = 0#
which we can solve by completing the square:
#0 = t^2+t+2#
#color(white)(0) = (t+1/2)^2+7/4#
#color(white)(0) = (t+1/2)^2-(sqrt(7)/2i)^2#
#color(white)(0) = ((t+1/2)-sqrt(7)/2i)((t+1/2)+sqrt(7)/2i)#
#color(white)(0) = (t+1/2-sqrt(7)/2i)(t+1/2+sqrt(7)/2i)#
So:
#alpha+alpha^2+alpha^4 = t = -1/2+-sqrt(7)/2i#
That is:
#cos((2pi)/7)+cos((4pi)/7)+cos((8pi)/7)+i(sin((2pi)/7)+sin((4pi)/7)+sin((8pi)/7)) = -1/2+-sqrt(7)/2i#
Equating imaginary parts:
#sin((2pi)/7)+sin((4pi)/7)+sin((8pi)/7) = +-sqrt(7)/2#
Which sign is correct?
Note that:
#sin((2pi)/7) > 0#
#sin((4pi)/7) > 0#
#sin(-(2pi)/7) < sin((-pi)/7) = sin((8pi)/7)#
Hence:
#sin((2pi)/7)+sin((4pi)/7)+sin((8pi)/7) > 0#
So:
#sin((2pi)/7)+sin((4pi)/7)+sin((8pi)/7) = sqrt(7)/2#
