Show that :
tan [(n+1)a/2] = (sin a + sin 2a + ... + sin na)/(cos a + cos 2a + ... + cos na)

     

(A d2 solution)

Consider the expression  (sin a + sin 2a + ... + sin na)

By adding pairwise the 1st and the last, the 2nd  ad the penultimate we get ( n+1)/2 sums  like ( sin ka + sin (n+1-k)a) which equals

.5(n+1)*2 sin((n+1)/2)*cos((2ka-n-1)a/2     #!

The (cos a + cos 2a + ... + cos na)  equals, by the same token to

.5(n+1)*2 cos((n+1)/2)*cos((2ka-n-1)a/2    #2

Dividing   #!  by #2  we get  tan [(n+1)a/2]

qed

Rem : It is easy to reason out that both formulas are valid both for even and odd number of terms.

Edited on August 24, 2014, 1:31 pm

Posted by Ady TZIDON on 2014-08-24 11:46:28