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
