Question

How can find the value of sin 2pi/7 + sin 4pi/7 +sin 8pi/7?

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Answer 1

Sin(8π/7) + Sin(4π/7) + Sin(2π/7)

= 2.Sin(6π/7).Cos(2π/7) + Sin(2π/7) [As Sin(A) + Sin(B) = 2.Sin((A+B)/2).Cos((A+B)/2)]

=2.Sin(6π/7).Cos(2π/7) + 2.Sin(π/7).Cos(π/7)

=2.Sin(π - 6π/7).Cos(2π/7) + 2.Sin(π/7).Cos(π/7) [As Sin(π - A) = Sin(A)]

=2.Sin(π/7).Cos(2π/7) + 2.Sin(π/7).Cos(π/7)

= 2.Sin(π/7).[Cos(2π/7) + Cos(π/7)]

= 2.Sin(π/7).[2.Cos(3π/14). Cos(π/14)]

Answer 2

${\huge{\red{\mathfrak{answer}}}}$

B+Ai=e2iπ/7+e4iπ/7+e8iπ/7

B−Ai=e−2iπ/7+e−4iπ/7+e−8iπ/7

So

B2+A2=(B+Ai)(B−Ai)=3+2cos2π/7+2cos4π/7+2cos6π/7

But cos(2kπ/7)=cos(−2kπ/7)=cos(2(7−k)π/7).

So this means that B2+A2=3+∑

6

k=1

cos(2kπ/7).

But that sum ∑

6

k=0

cos(2kπ/7)=0, so you get that:

B2+A2=3+(−1)=2

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