Compute ​

~ 45.5     is what I find  

Compute \(\sum \limits_{x=0^\circ}^{90^\circ} \cos^2(x)\)

\(\begin{array}{|rcll|} \hline &&\mathbf{ \sum \limits_{x=0^\circ}^{90^\circ} \left(\cos^2(x) \right) } \\ &=& \sum \limits_{x=0^\circ}^{44^\circ} \left(\cos^2(x)\right)+\cos^2(45^\circ)+\sum \limits_{x=46^\circ}^{90^\circ} \left(\cos^2(x)\right) \\ &=& \sum \limits_{x=0^\circ}^{44^\circ} \left(\cos^2(x)\right)+\sum \limits_{x=46^\circ}^{90^\circ} \left(\cos^2(x)\right)+\cos^2(45^\circ) \\ &=& \sum \limits_{x=0^\circ}^{44^\circ} \left(\cos^2(x)\right)+\sum \limits_{x=0^\circ}^{44^\circ} \left(\cos^2(90^\circ-x)\right)+\cos^2(45^\circ) \\ &=& \sum \limits_{x=0^\circ}^{44^\circ} \left(\cos^2(x)\right)+\sum \limits_{x=0^\circ}^{44^\circ} \left(\sin^2(x)\right)+\cos^2(45^\circ) \\ &=& \sum \limits_{x=0^\circ}^{44^\circ} \left(\cos^2(x)+\sin^2(x)\right)+\cos^2(45^\circ) \\ &=& \sum \limits_{x=0^\circ}^{44^\circ} \left(1\right)+\cos^2(45^\circ) \\ &=& 45+\cos^2(45^\circ) \quad | \quad \cos(45^\circ) = \dfrac{\sqrt{2}}{2} \\ &=& 45+ \left(\dfrac{\sqrt{2}}{2}\right)^2 \\ &=& 45+ \dfrac{2}{4} \\ &=& 45+ \dfrac{1}{2} \\ &=& \mathbf{45.5} \\ \hline \end{array}\)

\(\phantom{=} \displaystyle\sum_{x = 0^{\circ}}^{90^{\circ}} \cos^2 x\\ = \cos^2 0^{\circ} + \cos^2 1^{\circ} + \cdots + \cos^2 90^{\circ}\\ = 1 + (\cos^2 1^{\circ} + \cos^2 2^{\circ} + \cos^2 3^{\circ} + \cdots + \cos^2 89^{\circ}) + 0\\ = 1 + (\cos^2 1^{\circ} + \cos^2 89^{\circ}) + (\cos^2 2^{\circ} + \cos^2 88^{\circ}) + \cdots + (\cos^2 44^{\circ} + \cos^2 46^{\circ}) + \cos^2 45^{\circ}\\ \boxed{\text{Note that }\cos^2 n^{\circ} + \cos^2 (90 - n)^{\circ} = 1}\\ = 1 + 1 + 1 + 1 + \cdots + 1 + \dfrac{1}{2}\\ \boxed{\text{There are }\color{red}44\color{black}\text{ 1's}}\\ = 44 \dfrac{1}{2}\\ = \boxed{\dfrac{89}{2}}\)

Sorry that's wrong. It should like this: cos^2 0+cos^2 1+cos^2 2 ... cos^2 90 (in degrees). heureka was correct