Let-0-lt-lt-45-find-the-value-of-sin-2-45-sin-2-45-

Question Number 144539 by ZiYangLee last updated on 26/Jun/21

Let 0°<θ<45°, find the value of sin^2 (45°+θ)+sin^2 (45°−θ)

$$\mathrm{Let}\:\mathrm{0}°<\theta<\mathrm{45}°,\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{sin}^{\mathrm{2}} \left(\mathrm{45}°+\theta\right)+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{45}°−\theta\right) \\ $$

Answered by mitica last updated on 26/Jun/21

sin^2 (45+θ)+sin^2 (45−θ)= sin^2 (45+θ)+cos^2 ((90−(45−θ))= sin^2 (45+θ)+cos^2 ((45+θ)=1

$${sin}^{\mathrm{2}} \left(\mathrm{45}+\theta\right)+{sin}^{\mathrm{2}} \left(\mathrm{45}−\theta\right)= \\ $$$${sin}^{\mathrm{2}} \left(\mathrm{45}+\theta\right)+{cos}^{\mathrm{2}} \left(\left(\mathrm{90}−\left(\mathrm{45}−\theta\right)\right)=\right. \\ $$$${sin}^{\mathrm{2}} \left(\mathrm{45}+\theta\right)+{cos}^{\mathrm{2}} \left(\left(\mathrm{45}+\theta\right)=\mathrm{1}\right. \\ $$$$ \\ $$$$ \\ $$

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Let-0-lt-lt-45-find-the-value-of-sin-2-45-sin-2-45-

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