Question Number 8289 by rhm last updated on 07/Oct/16
$${what}\:{is}\:{value} \\ $$$${sin}\:\mathrm{36}° \\ $$$${plese}\:{give}\:{me}\:{answer} \\ $$
Commented by rhm last updated on 07/Oct/16
$${sir}\:{answer} \\ $$
Commented by Rasheed Soomro last updated on 07/Oct/16
$${sin}\:\mathrm{36}°=\frac{\sqrt{\mathrm{2}\sqrt{\mathrm{5}}\:−\mathrm{10}}}{\mathrm{4}} \\ $$
Answered by ridwan balatif last updated on 15/Oct/16
$$\mathrm{let}:\:\:\mathrm{sin18}^{\mathrm{o}} =\mathrm{sin}\theta \\ $$$$\mathrm{sin36}^{\mathrm{o}} =\mathrm{cos54}^{\mathrm{o}} \\ $$$$\mathrm{sin2}\theta\:\:=\mathrm{cos3}\theta \\ $$$$\mathrm{2sin}\theta\mathrm{cos}\theta=\mathrm{cos2}\theta\mathrm{cos}\theta−\mathrm{sin}\theta\mathrm{sin2}\theta \\ $$$$\mathrm{2sin}\theta\mathrm{cos}\theta=\mathrm{cos2}\theta\mathrm{cos}\theta−\mathrm{sin}\theta.\left(\mathrm{2sin}\theta\mathrm{cos}\theta\right) \\ $$$$\mathrm{2sin}\theta=\mathrm{cos2}\theta−\mathrm{2sin}^{\mathrm{2}} \theta \\ $$$$\mathrm{2sin}\theta=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \theta−\mathrm{2sin}^{\mathrm{2}} \theta \\ $$$$\mathrm{4sin}^{\mathrm{2}} \theta+\mathrm{2sin}\theta−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{sin}\theta_{\mathrm{1},\mathrm{2}} =\frac{−\mathrm{b}\pm\sqrt{\mathrm{D}}}{\mathrm{2a}} \\ $$$$\mathrm{sin}\theta_{\mathrm{1},\mathrm{2}} =\frac{−\mathrm{2}\pm\sqrt{\mathrm{20}}}{\mathrm{2}.\mathrm{4}} \\ $$$$\mathrm{sin}\theta_{\mathrm{1},\mathrm{2}} =\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\mathrm{we}\:\mathrm{take}\:\mathrm{positive},\:\mathrm{because}\:\mathrm{sin18}^{\mathrm{o}} \:\mathrm{has}\:\mathrm{positive}\:\mathrm{value} \\ $$$$\mathrm{sin}\theta=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\mathrm{so},\:\mathrm{sin18}^{\mathrm{o}} =\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\:\:\:\mathrm{and}\:\mathrm{cos18}^{\mathrm{o}} =\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}} \\ $$$$\mathrm{sin36}^{\mathrm{o}} =\mathrm{2sin18}^{\mathrm{o}} \mathrm{cos18}^{\mathrm{o}} \\ $$$$\mathrm{sin36}^{\mathrm{o}} =\left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{8}}\right)\left(\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}\right) \\ $$