Got stuck in sine summation

I got this doubt while solving a problem-

\sum_{r=2}^{n}{}\left|A_{1} A_{r}\right| = 2\sum_{r=2}^{n}{}sin(r-1)\frac{\Pi }{n}

I got the answer as 2cot\frac{\Pi }{2n}

The answer given is 4cosec\frac{\Pi }{n}

Please help

3 Answers

1
Euclid ·

@swordfish i also get same ans as you 2cotÎ 2n

and 4cosecΠn = 2cotΠ2ncos2Π2n which can never become equal to Nr unless any other given condition like n→∞ or anything such!!!

1
venkateshan ·

dats ok but wht abt the LHS.... we 've solved the RHS and gt the ans as 2cot pi/2n..... i cud nt understand abt LHS

1
swordfish ·

@Venkatesha- I told that I got stuck with it while solving a problem-----so don't worry about how LHS came...But still I will post the full question for you-

Let A1 , A2......An be the vertices of a regular polygon of n sides inscribed in a circle of radius unity and centre at the origin. Let A1 correspond to the complex number 1.

Q) \sum_{r=2}^{n}{}\left|A_{1}A_{r} \right| =

a)2cosecÎ /n b) 4cosecÎ /n c) 2sinÎ /n d) 4sinÎ /n

The answer given is 4cosecÎ /n

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