Find the sum: Cos^2 (0) + Cos^2 (2) + Cos^2 (4) + Cos^2 (356) + Cos^2 (358) + Cos^2 (360)
Thanks in advance.
1. I assume that the argument of the trigonometric functions is in degree (?).
2. You can simplify the given term a little bit:
Since
$\displaystyle \cos^2(0^\circ) = \cos^2(360^\circ) = 1$
$\displaystyle \cos^2(2^\circ) = \cos^2(358^\circ)$
$\displaystyle \cos^2(4^\circ) = \cos^2(356^\circ)$
you'll get:
$\displaystyle \begin{array}{l}
\cos^2 (0^\circ) + \cos^2 (2^\circ) + \cos^2 (4^\circ) + \cos^2 (356^\circ) + \cos^2 (358^\circ) + \cos^2 (360^\circ)=\\
2 + 2\cos^2(2^\circ)+2\cos^2(4^\circ)\end{array}$
Since $\displaystyle 2\cos^2(2^\circ) = \cos(4^\circ)+1$ the term simplifies to:
$\displaystyle 2+1+ \cos(4^\circ)+2\cos^2(4^\circ) = \boxed{3+ \cos(4^\circ)(1+2\cos(4^\circ))}$