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Question Number 165314 by mathlove last updated on 29/Jan/22

$$\sin {}^{2}1+\sin {}^{2}2+\sin {}^{2}3+.....+{sin}^{2}90=?$$

Commented by cortano1 last updated on 29/Jan/22

$$45.5$$

Commented by shunmisaki007 last updated on 29/Jan/22

$$$$$${\sin }^2\left(1°\right)+{\sin }^2\left(2°\right)+{\sin }^2\left(3°\right)+...+{\sin }^2\left(90°\right)$$$$=({\sin }^2\left(1°\right)+{\sin }^2\left(89°\right))+({\sin }^2\left(2°\right)+{\sin }^2\left(88°\right))+({\sin }^2\left(3°\right)+{\sin }^2\left(87°\right))+...+({\sin }^2\left(44°\right)+{\sin }^2\left(46°\right))+{\sin }^2\left(45°\right)+{\sin }^2\left(90°\right)$$$$=({\sin }^2\left(1°\right)+{\cos }^2\left(1°\right))+({\sin }^2\left(2°\right)+{\cos }^2\left(2°\right))+({\sin }^2\left(3°\right)+{\cos }^2\left(3°\right))+...+({\sin }^2\left(44°\right)+{\cos }^2\left(44°\right))+{\left(\frac{\sqrt{2}}{2}\right)}^2+1$$$$=44+\frac{1}{2}+1$$$$=45\frac{1}{2}\mathrm{or}45.5...★$$

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