Question Number 148754 by mathdanisur last updated on 30/Jul/21
Commented by liberty last updated on 31/Jul/21
$$\mathrm{ans}:\:\mathrm{850} \\ $$
Answered by liberty last updated on 31/Jul/21
$$\:\frac{\mathrm{sin}\:^{\mathrm{6}} \:\mathrm{36}°}{\mathrm{cos}\:^{\mathrm{6}} \:\mathrm{36}°}\:+\:\frac{\mathrm{sin}\:^{\mathrm{6}} \:\mathrm{72}°}{\mathrm{cos}\:^{\mathrm{6}} \:\mathrm{72}°}\:= \\ $$$$\frac{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\left(\mathrm{2cos}\:\mathrm{72}°\mathrm{sin}\:\mathrm{36}°\right)^{\mathrm{6}} +\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\left(\mathrm{2sin}\:\mathrm{72}°\mathrm{cos}\:\mathrm{36}°\right)^{\mathrm{2}} }{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\left(\mathrm{2cos}\:\mathrm{72}°\mathrm{cos}\:\mathrm{36}°\right)^{\mathrm{6}} } \\ $$$$=\frac{\left(\mathrm{sin}\:\mathrm{108}°−\mathrm{sin}\:\mathrm{36}°\right)^{\mathrm{6}} +\left(\mathrm{sin}\:\mathrm{108}°+\mathrm{sin}\:\mathrm{36}°\right)^{\mathrm{6}} }{\left(\mathrm{sin}\:\mathrm{18}°\left(\mathrm{1}−\mathrm{2sin}\:\mathrm{18}°\right)\right)^{\mathrm{6}} } \\ $$$$=\frac{\mathrm{cos}\:^{\mathrm{6}} \:\mathrm{18}°\left(\mathrm{1}−\mathrm{2sin}\:\mathrm{18}°\right)^{\mathrm{6}} +\mathrm{cos}\:^{\mathrm{6}} \:\mathrm{18}°\left(\mathrm{1}+\mathrm{2sin}\:\mathrm{18}°\right)^{\mathrm{6}} }{\mathrm{sin}\:^{\mathrm{6}} \:\mathrm{18}°\left(\mathrm{1}−\mathrm{2sin}\:\mathrm{18}°\right)^{\mathrm{6}} } \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{finish}\:\mathrm{it} \\ $$
Commented by liberty last updated on 31/Jul/21
$$\mathrm{cos}\:\mathrm{18}°=\frac{\sqrt{\mathrm{2}\sqrt{\mathrm{5}}+\mathrm{10}}}{\mathrm{4}} \\ $$$$\mathrm{sin}\:\mathrm{18}°=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\: \\ $$
Commented by mathdanisur last updated on 31/Jul/21
$${Thank}\:{you}\:{Ser} \\ $$