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Note that $\cos \theta = \sin \left (\frac {\pi}{2} - \theta\right)$

$\therefore 2\cos^{2}(1) + 2\cos^{2} (2) + ... + 2\cos ^{2} (89) + 2\cos^{2} (89) + 2\cos^{2} (90)$

$= 2[(\cos^{2}(1) + \cos^{2}(89)) + (\cos^{2}(2) + \cos^{2}(88)) + ... + (\cos^{2}(43) + \cos^{2}(47)) + (\cos^{2}(44) + \cos^{2} (46))] + 2cos^{2} (45)$

$= 2[(\cos^{2}(1) + \sin^{2}(1)) + (\cos^{2}(2) + \sin^{2}(2)) + ... + (\cos^{2}(43) + \sin^{2}(43)) + (\cos^{2}(44) + \sin^{2} (44))] + 2\left(\frac {1}{\sqrt{2}}\right)^{2}$

$= 2(44) + 1 = \boxed{89}$

Solution:

That equation can be simplified to:

$2cos^{2}1 + 2cos^{2}89 + 2cos^{2}2 + 2cos^{2}88 + ... + 2cos^{2}44 + 2cos^{2}46 + 2cos^{2}45$

-> $2cos^{2}1 + 2cos^{2}89 + 2cos^{2}2 + 2cos^{2}88 + ... + 2cos^{2}44 + 2cos^{2}46 + 2cos^{2}45$

Since $sin^{2}\theta = cos^{2}(90-\theta)$ , then, the equation above is simply equal to:

-> $(2cos^{2}(1) + 2sin^{2}(1)) + (2sin^{2}(2) + 2sin^{2}(2)) + ... + (2cos^{2}(44) + 2sin^{2}(44)) + (2cos^{2}(45))$

Also, $sin^{2}\theta + cos^{2}\theta = 1$ , then,

-> $2*89/2$

-> $\boxed{89}$