The-sum-of-the-series-sin-sin-n-4-n-2-sin-n-6-n-2-n-terms-is-equal-to-1-sin-n-2-n-2-cos-2n-2-n-3-tann-4-cotn-

Question Number 13654 by Tinkutara last updated on 22/May/17

The sum of the series

\sin θ + \sin (\frac{n - 4}{n - 2}) θ + \sin (\frac{n - 6}{n - 2}) θ + \dots n terms

is equal to

(1) \sin (\frac{n θ}{2 - n})

(2) \cos (\frac{2 n θ}{2 - n})

(3) \tan n θ

(4) \cot n θ

Answered by ajfour last updated on 22/May/17

T_{r} = \sin (\frac{n - 2 r}{n - 2}) θ

T_{n - 1} = \sin (\frac{n - 2 (n - 1)}{n - 2}) θ = \sin (\frac{2 - n}{n - 2}) θ

T_{n - 2} = \sin (\frac{n - 2 (n - 2)}{n - 2}) θ = \sin (\frac{4 - n}{n - 2}) θ

\dots . . \dots . . \dots . . \dots . . \dots \dots

S = \sin (\frac{n - 2}{n - 2}) θ + \sin (\frac{2 - n}{n - 2}) θ

+ \sin (\frac{n - 4}{n - 2}) θ + \sin (\frac{4 - n}{n - 2}) θ

+ \sin (\frac{n - 6}{n - 2}) θ + \sin (\frac{6 - n}{n - 2}) θ

+ \dots . + \dots \dots .

+ [+ \sin (\frac{- n}{n - 2}) θ]

e x c e p t i n g t h e l a s t t e r m, r e s t

a l l s u m t o z e r o .

S o s u m o f s e r i e s = T_{n} = \sin (\frac{n θ}{2 - n}) .

Commented by Tinkutara last updated on 22/May/17

But I have calculated it using the formula

S_{n} = \frac{\sin \frac{n β}{2}}{\sin \frac{β}{2}} \sin [α + (n - 1) \frac{β}{2}]

where α is the first angle and β is the

common difference of the angles which

are in AP i . e . to find the sum \sin α +

\sin (α + β) + \sin (α + 2 β) + \dots +

\sin [α + (n - 1) β] .

Using α = θ and \frac{β}{2} = \frac{θ}{2 - n}, my answer

comes out to be \sin (\frac{3 n θ}{2 - n}) . Where

is the mistake ?

Commented by ajfour last updated on 22/May/17

S_{n} = \frac{\sin (\frac{n θ}{2 - n})}{\sin (\frac{θ}{2 n})} \sin (\frac{2 θ - n θ + n θ - θ}{2 - n})

\Rightarrow S_{n} = \sin (\frac{n θ}{2 - n}) .

Commented by Tinkutara last updated on 22/May/17

Thanks Sir! I understood .