Question

Evaluate sin(π\n)+sin(3π\n)+sin(5π\n)+...to n terms

Answer 1

Let
S = sin π/n + sin 3π/n + sin 5π/n +....to n terms
= sin [(2.1 - 1)π/n] + sin [(2.2 - 1)π/n] + sin [(2.3 - 1)π/n] +.....+ sin [(2.r - 1)π/n] +......+ sin [(2.n - 1)π/n]
= t(1) + t(2) +..........+ t(n),
where t(r) = rth term of the sum = sin [(2r - 1)π/n]

Note that
(n - r + 1)th term of the sum, t(n - r + 1)
= sin [(2.(n - r + 1) - 1)π/n]
= sin [(2.n - 2r + 1)π/n]
= sin [(2.n - (2r - 1))π/n]
= sin [2π - (2r - 1)π/n]
= -sin [(2r - 1)π/n]..................[as sin (2π - θ) = -sin θ]
= -t(r)
=> t(r) + t(n - r + 1) = 0..................(1)

Now, two cases may arise.

Case I : n is even. n = 2m, say.
In this case, putting r = 1, 2, 3............, (m - 1), m in (1)
t(1) + t(2m) = 0
t(2) + t(2m - 1) = 0
t(3) + t(2m - 2) = 0
.
.
.
.

t(m - 1) + t(m + 2) = 0
t(m) + t(m + 1) = 0

Summing all these
t(1) + t(2) +.........+ t(m - 1) + t(m) + t(m + 1) + t(m + 2) +.........+ t(2m-1) + t(2m) = 0
i.e. t(1) + t(2) +..........+ t(n) = 0
i.e. S = 0

Case II : n is odd. n = 2m + 1, say.
In this case, putting r = 1, 2, 3............, (m-1), m in (1)
t(1) + t(2m + 1) = 0
t(2) + t(2m) = 0
t(3) + t(2m - 1) = 0
.
.
.
.

t(m - 1) + t(m + 3) = 0
t(m) + t(m + 2) = 0

Summing all these
t(1) + t(2) +.........+ t(m - 1) + t(m) + t(m + 2) + t(m + 3) +.........+ t(2m) + t(2m + 1) = 0
So,
S = t(1) + t(2) +..........+ t(n)
= t(1) + t(2) +.........+ t(m - 1) + t(m) + t(m+1) + t(m + 2) + t(m + 3) +.........+ t(2m) + t(2m + 1)
= { t(1) + t(2) +.........+ t(m - 1) + t(m) + t(m + 2) + t(m + 3) +.........+ t(2m)
+ t(2m + 1) } + t(m+1)
= 0 + t(m+1) = t(m+1)

Now, t(m+1)
= sin [(2(m+1) - 1)π/n]
= sin [(2m + 2 - 1)π/n]
= sin [(2m + 1)π/n]
= sin [nπ/n] = sin π = 0
Hence, S = 0

Thus, in any case the required sum is 0.