2012/032) If a=2π/2013, what is cos2a+cos4a+cos6a+..........+cos 2012a…

First we need to show that

Cos a + cos 2a + cos 3a + … + cos na = 0 ..1

When a = 2π/n

For a proof I propose as below

2 cos ka sin a/2 = sin (k+1/2) a – sin(k-1/2) a

If we take it a sum from k = 1 to n we get telescopic sum

2 ( cos a + … + cos na) sin a/2 = sin (na +1/2) a – sin ½ a = sin (1/2 a) – sin ½ a = 0

So cos a + cos 2a + cos 3a + …. + cos n a = 0 …1

Using this we can prove the desired result

As we know that n = 2013 and an = 2pi

Cos a = cos 2012 a
Cos 3a = cos 2010 a
So for each odd on left there is an even on the right and exception is cos 2013a = 1

Cos a + cos 3a + cos 5a + …+ cos 2011 a = cos 2a + cos 4a + cos 6a + … + cos 2012 a ..2

From (1) group even terms together and odd ones also)

Cos 2a + cos 4a + cos 6a + … + cos 2012 a + ( cos a + cos 3a + cos 5a + … cos 2011 a ) + 1 = 0

Putting from (2) we get
2(cos 2a + cos 4a + … cos 2012a) + 1 = 0
Or (cos 2a + cos 4a + … cos 2012a) = - 1/2