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Question Number 148754 by mathdanisur last updated on 30/Jul/21

Commented by liberty last updated on 31/Jul/21

ans: 850

$$\mathrm{ans}:\:\mathrm{850} \\ $$

Answered by liberty last updated on 31/Jul/21

((sin^6 36°)/(cos^6 36°)) + ((sin^6 72°)/(cos^6 72°)) = (((1/2^6 )(2cos 72°sin 36°)^6 +(1/2^6 )(2sin 72°cos 36°)^2 )/((1/2^6 )(2cos 72°cos 36°)^6 )) =(((sin 108°−sin 36°)^6 +(sin 108°+sin 36°)^6 )/((sin 18°(1−2sin 18°))^6 )) =((cos^6 18°(1−2sin 18°)^6 +cos^6 18°(1+2sin 18°)^6 )/(sin^6 18°(1−2sin 18°)^6 )) you can finish it

$$\:\frac{\mathrm{sin}\:^{\mathrm{6}} \:\mathrm{36}°}{\mathrm{cos}\:^{\mathrm{6}} \:\mathrm{36}°}\:+\:\frac{\mathrm{sin}\:^{\mathrm{6}} \:\mathrm{72}°}{\mathrm{cos}\:^{\mathrm{6}} \:\mathrm{72}°}\:= \\ $$$$\frac{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\left(\mathrm{2cos}\:\mathrm{72}°\mathrm{sin}\:\mathrm{36}°\right)^{\mathrm{6}} +\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\left(\mathrm{2sin}\:\mathrm{72}°\mathrm{cos}\:\mathrm{36}°\right)^{\mathrm{2}} }{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\left(\mathrm{2cos}\:\mathrm{72}°\mathrm{cos}\:\mathrm{36}°\right)^{\mathrm{6}} } \\ $$$$=\frac{\left(\mathrm{sin}\:\mathrm{108}°−\mathrm{sin}\:\mathrm{36}°\right)^{\mathrm{6}} +\left(\mathrm{sin}\:\mathrm{108}°+\mathrm{sin}\:\mathrm{36}°\right)^{\mathrm{6}} }{\left(\mathrm{sin}\:\mathrm{18}°\left(\mathrm{1}−\mathrm{2sin}\:\mathrm{18}°\right)\right)^{\mathrm{6}} } \\ $$$$=\frac{\mathrm{cos}\:^{\mathrm{6}} \:\mathrm{18}°\left(\mathrm{1}−\mathrm{2sin}\:\mathrm{18}°\right)^{\mathrm{6}} +\mathrm{cos}\:^{\mathrm{6}} \:\mathrm{18}°\left(\mathrm{1}+\mathrm{2sin}\:\mathrm{18}°\right)^{\mathrm{6}} }{\mathrm{sin}\:^{\mathrm{6}} \:\mathrm{18}°\left(\mathrm{1}−\mathrm{2sin}\:\mathrm{18}°\right)^{\mathrm{6}} } \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{finish}\:\mathrm{it} \\ $$

Commented by liberty last updated on 31/Jul/21

cos 18°=((√(2(√5)+10))/4) sin 18°=(((√5)−1)/4)

$$\mathrm{cos}\:\mathrm{18}°=\frac{\sqrt{\mathrm{2}\sqrt{\mathrm{5}}+\mathrm{10}}}{\mathrm{4}} \\ $$$$\mathrm{sin}\:\mathrm{18}°=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\: \\ $$

Commented by mathdanisur last updated on 31/Jul/21

Thank you Ser

$${Thank}\:{you}\:{Ser} \\ $$

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