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

$$4\sin 36°\cos 72°\sin 108°?$$

Answered by bemath last updated on 02/Sep/20

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

Answered by nimnim last updated on 02/Sep/20

$$=4\sin 36.\cos (90-18).\sin (90+18)$$$$=4\sin 36.\sin 18\cos 18$$$$=2\sin 36\sin \left(2\times 18\right)$$$$={2\sin }^{2}36$$$$=2{\left[\frac{1}{4}\sqrt{10-2\sqrt{5}}\right]}^2$$$$=\frac{1}{8}(10-2\sqrt{5})$$$$=\frac{1}{4}(5-\sqrt{5})$$

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