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Question Number 209670 by mokys last updated on 18/Jul/24

$$findthesumof{sin}^2\left(1\right)+...+{sin}^2\left(90\right)$$

Answered by Ar Brandon last updated on 18/Jul/24

$$S=\underset{\theta =0}{\sum ^{90}}{\sin }^2\theta$$$$=\frac{1}{2}+\underset{\theta =0}{\sum ^{44}}({\sin }^2\theta +{\cos }^2\theta )$$$$=\frac{1}{2}+\underset{\theta =0}{\sum ^{44°}}\left(1\right)=\frac{1}{2}+45=\frac{91}{2}$$

Commented by mokys last updated on 18/Jul/24

$$thankyousirbutcanyougivemehow$$$$$$$$\underset{\theta =0}{\sum ^{90}}{sin}^2\theta =\frac{1}{2}+\underset{\theta =0}{\sum ^{{44}^{°}}}({sin}^2\theta +{cos}^2\theta )$$

Commented by Ar Brandon last updated on 18/Jul/24

$$\underset{\theta =0}{\sum ^{90}}{\sin }^2\theta$$$$={\sin }^2\left(1\right)+\cdot \cdot \cdot +{\sin }^2\left(45°\right)+\cdot \cdot \cdot +{\sin }^2\left(90°\right)$$$$={\sin }^2\left(1\right)+\cdot \cdot \cdot +{\sin }^2\left(44°\right)+{\left(\frac{1}{\sqrt{2}}\right)}^2+{\sin }^2\left(46°\right)+\cdot \cdot \cdot +{\sin }^2\left(90°\right)$$$$={\sin }^2\left(1\right)+\cdot \cdot \cdot +{\sin }^2\left(44°\right)+\frac{1}{2}+{\cos }^2\left(44°\right)+\cdot \cdot \cdot +{\cos }^2\left(0°\right)$$$$=\underset{\theta =0°}{\sum ^{44°}}{\sin }^2\theta +\frac{1}{2}+\underset{\theta =0°}{\sum ^{44°}}{\cos }^2\theta =\frac{1}{2}+\underset{\theta =0°}{\sum ^{44°}}({\sin }^2\theta +{\cos }^2\theta )$$

Answered by Sutrisno last updated on 18/Jul/24

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

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