We have to prove that (cospi/4+i*sinpi/4)^2008 is real.

cos (pi/4) = 1/sqrt 2

sin (pi/4) = 1/sqrt 2

(cos pi/4 + i*sin pi/4)^2008

=> [1/sqrt 2 + i*(1/sqrt 2)]^2008

=> [1/sqrt 2 + i*(1/sqrt 2)]^2^1004

=> [1/2 + i^2/2 + 2*(1/sqrt 2)(1/sqrt 2)]^1004

substitute i^2 = -1

=> [1/2 - 1/2 + 2*i*(1/sqrt 2)(1/sqrt 2)]^1004

=> [2*i*(1/2)]^1004

=> i^1004

=> i^2^502

=> (-1)^502

=> 1

**This proves that (cos pi/4 + i*sin pi/4)^2008 = 1 is real.**

We'll apply Moivre's rule and we'll get:

(cospi/4+i*sinpi/4)^2008 = (cos 2008pi/4+i*sin 2008pi/4)

We'll simplify and we'll get:

(cos 2008pi/4+i*sin 2008pi/4) = (cos 502pi+i*sin 502pi)

Since 502 is multiple of 2, we'll re-write 502pi as:

502pi = 251*2pi

(cos 2pi+i*sin 2pi) = 1 + i*0

(cos 502pi+i*sin 502pi) = 1

Since the imaginary part is 0, the given number is a real number:

**(cospi/4+i*sinpi/4)^2008 = 1**