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Question-56282

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Question Number 56282 by Tawa1 last updated on 13/Mar/19
Commented by mr W last updated on 13/Mar/19
sin^2 1°+sin^2 2°+...+sin^2 44°+sin^2 45°+sin^2 46°+...+sin^2 88°+sin^2 89°+sin^2 90° %sin^2 1°+sin^2 2°+...+sin^2 44°+sin^2 45°+cos^2 44°+...+cos^2 2°+cos^2 1°+sin^2 90° =(sin^2 1°+cos^2 1)°+(sin^2 2°+cos^2 2°)+...+(sin^2 44°+cos^2 44°)+sin^2 45°+sin^2 90° =(sin^2 1°+cos^2 1°)+(sin^2 2°+cos^2 2°)+...+(sin^2 44°+cos^2 44°)+sin^2 45°+sin^2 90° =(1)+(1)+...+(1)+sin^2 45°+sin^2 90° =44+(1/2)+1 =45.5
$$\mathrm{sin}^{\mathrm{2}} \:\mathrm{1}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}°+…+\mathrm{sin}^{\mathrm{2}} \:\mathrm{44}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{45}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{46}°+…+\mathrm{sin}^{\mathrm{2}} \:\mathrm{88}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{89}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{90}° \\ $$$$\%\mathrm{sin}^{\mathrm{2}} \:\mathrm{1}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}°+…+\mathrm{sin}^{\mathrm{2}} \:\mathrm{44}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{45}°+\mathrm{cos}^{\mathrm{2}} \:\mathrm{44}°+…+\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}°+\mathrm{cos}^{\mathrm{2}} \:\mathrm{1}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{90}° \\ $$$$=\left(\mathrm{sin}^{\mathrm{2}} \:\mathrm{1}°+\mathrm{cos}^{\mathrm{2}} \:\mathrm{1}\right)°+\left(\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}°+\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}°\right)+…+\left(\mathrm{sin}^{\mathrm{2}} \:\mathrm{44}°+\mathrm{cos}^{\mathrm{2}} \:\mathrm{44}°\right)+\mathrm{sin}^{\mathrm{2}} \:\mathrm{45}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{90}° \\ $$$$=\left(\mathrm{sin}^{\mathrm{2}} \:\mathrm{1}°+\mathrm{cos}^{\mathrm{2}} \:\mathrm{1}°\right)+\left(\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}°+\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}°\right)+…+\left(\mathrm{sin}^{\mathrm{2}} \:\mathrm{44}°+\mathrm{cos}^{\mathrm{2}} \:\mathrm{44}°\right)+\mathrm{sin}^{\mathrm{2}} \:\mathrm{45}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{90}° \\ $$$$=\left(\mathrm{1}\right)+\left(\mathrm{1}\right)+…+\left(\mathrm{1}\right)+\mathrm{sin}^{\mathrm{2}} \:\mathrm{45}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{90}° \\ $$$$=\mathrm{44}+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1} \\ $$$$=\mathrm{45}.\mathrm{5} \\ $$
Commented by Tawa1 last updated on 13/Mar/19
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by Tawa1 last updated on 13/Mar/19
I want to also understand how he solve using complex numbers. please.
$$\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{also}\:\mathrm{understand}\:\mathrm{how}\:\mathrm{he}\:\mathrm{solve}\:\mathrm{using}\:\:\mathrm{complex}\:\mathrm{numbers}.\:\mathrm{please}. \\ $$
Commented by Tawa1 last updated on 13/Mar/19
And how he got this step in his solution: (((1 − e^(i(n + 1)(π/(90))) )/(1 − e^((iπ)/(90)) )))
$$\mathrm{And}\:\mathrm{how}\:\mathrm{he}\:\mathrm{got}\:\mathrm{this}\:\mathrm{step}\:\mathrm{in}\:\mathrm{his}\:\mathrm{solution}:\:\:\:\:\:\left(\frac{\mathrm{1}\:−\:\mathrm{e}^{\mathrm{i}\left(\mathrm{n}\:+\:\mathrm{1}\right)\frac{\pi}{\mathrm{90}}} }{\mathrm{1}\:−\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{90}}} }\right) \\ $$
Commented by malwaan last updated on 14/Mar/19
very easy thank you so much
$${very}\:{easy} \\ $$$${thank}\:{you}\:{so}\:{much} \\ $$

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