Question Number 144539 by ZiYangLee last updated on 26/Jun/21
$$\mathrm{Let}\:\mathrm{0}°<\theta<\mathrm{45}°,\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$ $$\mathrm{sin}^{\mathrm{2}} \left(\mathrm{45}°+\theta\right)+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{45}°−\theta\right) \\ $$
Answered by mitica last updated on 26/Jun/21
$${sin}^{\mathrm{2}} \left(\mathrm{45}+\theta\right)+{sin}^{\mathrm{2}} \left(\mathrm{45}−\theta\right)= \\ $$ $${sin}^{\mathrm{2}} \left(\mathrm{45}+\theta\right)+{cos}^{\mathrm{2}} \left(\left(\mathrm{90}−\left(\mathrm{45}−\theta\right)\right)=\right. \\ $$ $${sin}^{\mathrm{2}} \left(\mathrm{45}+\theta\right)+{cos}^{\mathrm{2}} \left(\left(\mathrm{45}+\theta\right)=\mathrm{1}\right. \\ $$ $$ \\ $$ $$ \\ $$