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
1@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!!!
1dats 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@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