Question Number 135019 by faysal last updated on 09/Mar/21
Answered by Dwaipayan Shikari last updated on 09/Mar/21
$${sin}\mathrm{6}°{sin}\mathrm{42}°{sin}\mathrm{66}°{sin}\mathrm{78}° \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left({cos}\mathrm{60}°−{cos}\mathrm{72}°\right)\left({cos}\mathrm{36}°−{cos}\mathrm{120}°\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{4}}\right)\left(\frac{\sqrt{\mathrm{5}}+\mathrm{3}}{\mathrm{4}}\right)=\frac{\mathrm{1}}{\mathrm{16}}=\Lambda \\ $$$${cos}\mathrm{6}°{cos}\mathrm{42}°{cos}\mathrm{66}°{cos}\mathrm{78}° \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left({cos}\mathrm{60}°+{cos}\mathrm{72}°\right)\left({cos}\mathrm{120}°+{cos}\mathrm{36}°\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{16}}=\zeta \\ $$$$\frac{\Lambda}{\zeta}=\mathrm{1} \\ $$