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Question Number 63423 by minh2001 last updated on 04/Jul/19

Prove that ((sin^2 (36°))/(cos^2 (72°)))+sin^2 (72°)cos(36°)=((45+11(√5))/(16))

$${Prove}\:{that}\: \\ $$$$\frac{{sin}^{\mathrm{2}} \left(\mathrm{36}°\right)}{{cos}^{\mathrm{2}} \left(\mathrm{72}°\right)}+{sin}^{\mathrm{2}} \left(\mathrm{72}°\right){cos}\left(\mathrm{36}°\right)=\frac{\mathrm{45}+\mathrm{11}\sqrt{\mathrm{5}}}{\mathrm{16}} \\ $$

Commented by Tony Lin last updated on 04/Jul/19

sin(36°)=((√(10−2(√5)))/4) cos(72°)=(((√5)−1)/4) sin(72°)=((√(10+2(√5)))/4) cos(36°)=(((√5)+1)/4) ((sin^2 (36°))/(cos^2 (72°)))+sin^2 (72°)cos(36°) =((10−2(√5))/(((√5)−1)^2 ))+((10+2(√5))/(16))×(((√5)+1)/4) =((10−2(√5))/(6−2(√5)))+((20+12(√5))/(64)) =((5−(√5))/(3−(√5)))+((5+3(√5))/(16)) =(((5−(√5))(3+(√5)))/4)+((5+3(√5))/(16)) =((10+2(√5))/4)+((5+3(√5))/(16)) =((45+11(√5))/(16))

$${sin}\left(\mathrm{36}°\right)=\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}} \\ $$$${cos}\left(\mathrm{72}°\right)=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}} \\ $$$${sin}\left(\mathrm{72}°\right)=\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}} \\ $$$${cos}\left(\mathrm{36}°\right)=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}} \\ $$$$\frac{{sin}^{\mathrm{2}} \left(\mathrm{36}°\right)}{{cos}^{\mathrm{2}} \left(\mathrm{72}°\right)}+{sin}^{\mathrm{2}} \left(\mathrm{72}°\right){cos}\left(\mathrm{36}°\right) \\ $$$$=\frac{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{16}}×\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}}+\frac{\mathrm{20}+\mathrm{12}\sqrt{\mathrm{5}}}{\mathrm{64}} \\ $$$$=\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{3}−\sqrt{\mathrm{5}}}+\frac{\mathrm{5}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{16}} \\ $$$$=\frac{\left(\mathrm{5}−\sqrt{\mathrm{5}}\right)\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)}{\mathrm{4}}+\frac{\mathrm{5}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{16}} \\ $$$$=\frac{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}+\frac{\mathrm{5}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{16}} \\ $$$$=\frac{\mathrm{45}+\mathrm{11}\sqrt{\mathrm{5}}}{\mathrm{16}} \\ $$

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