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Question Number 111114 by bobhans last updated on 02/Sep/20

4 sin 36° cos 72° sin 108° ?

$$\mathrm{4}\:\mathrm{sin}\:\mathrm{36}°\:\mathrm{cos}\:\mathrm{72}°\:\mathrm{sin}\:\mathrm{108}°\:?\: \\ $$

Answered by bemath last updated on 02/Sep/20

(√(bemath)) we want to compute the value of 4 sin 36° cos 72° sin 108° . Let p = 4sin 36° cos 72° sin 108° { ((sin 108° = sin (90°+18°) = cos 18°)),((2cos 72° cos 18° = cos 90° + cos 54° = cos 54°)) :} p= 2sin 36° .cos 54° ⇒cos 54° = sin 36° p= 2 sin^2 (36°) = 1−(1−2sin^2 (36°)) p=1−cos 72° = 1−sin 18° p = 1−((((√5)−1)/4))=((4−(√5)+1)/4) p= ((5−(√5))/4)

$$\:\:\sqrt{\mathrm{bemath}} \\ $$$$\:\mathrm{we}\:\mathrm{want}\:\mathrm{to}\:\mathrm{compute}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\mathrm{4}\:\mathrm{sin}\:\mathrm{36}°\:\mathrm{cos}\:\mathrm{72}°\:\mathrm{sin}\:\mathrm{108}°\:. \\ $$$$\mathrm{Let}\:\mathrm{p}\:=\:\mathrm{4sin}\:\mathrm{36}°\:\mathrm{cos}\:\mathrm{72}°\:\mathrm{sin}\:\mathrm{108}°\: \\ $$$$\begin{cases}{\mathrm{sin}\:\mathrm{108}°\:=\:\mathrm{sin}\:\left(\mathrm{90}°+\mathrm{18}°\right)\:=\:\mathrm{cos}\:\mathrm{18}°}\\{\mathrm{2cos}\:\mathrm{72}°\:\mathrm{cos}\:\mathrm{18}°\:=\:\mathrm{cos}\:\mathrm{90}°\:+\:\mathrm{cos}\:\mathrm{54}°\:=\:\mathrm{cos}\:\mathrm{54}°}\end{cases} \\ $$$$\mathrm{p}=\:\mathrm{2sin}\:\mathrm{36}°\:.\mathrm{cos}\:\mathrm{54}° \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{54}°\:=\:\mathrm{sin}\:\mathrm{36}° \\ $$$$\mathrm{p}=\:\mathrm{2}\:\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{36}°\right)\:=\:\mathrm{1}−\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \left(\mathrm{36}°\right)\right) \\ $$$$\mathrm{p}=\mathrm{1}−\mathrm{cos}\:\mathrm{72}°\:=\:\mathrm{1}−\mathrm{sin}\:\mathrm{18}° \\ $$$$\mathrm{p}\:=\:\mathrm{1}−\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)=\frac{\mathrm{4}−\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{p}=\:\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$

Answered by nimnim last updated on 02/Sep/20

=4sin36.cos(90−18).sin(90+18) =4sin36.sin18cos18 =2sin36sin(2×18) =2sin^2 36 =2[(1/4)(√(10−2(√5)))]^2 =(1/8)(10−2(√5)) =(1/4)(5−(√5))

$$=\mathrm{4sin36}.\mathrm{cos}\left(\mathrm{90}−\mathrm{18}\right).\mathrm{sin}\left(\mathrm{90}+\mathrm{18}\right) \\ $$$$=\mathrm{4sin36}.\mathrm{sin18cos18} \\ $$$$=\mathrm{2sin36sin}\left(\mathrm{2}×\mathrm{18}\right) \\ $$$$=\mathrm{2sin}^{\mathrm{2}} \mathrm{36} \\ $$$$=\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}\right]^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{5}−\sqrt{\mathrm{5}}\right) \\ $$

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