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sin-9-sin-21-sin-39-2-golden-ratio-m-n-

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Question Number 153553 by mnjuly1970 last updated on 08/Sep/21
sin(9) + sin(21)+sin(39)=^? (ϕ/( (√2))) ϕ:= golden ratio m.n
$$ \\ $$$$\:{sin}\left(\mathrm{9}\right)\:+\:{sin}\left(\mathrm{21}\right)+{sin}\left(\mathrm{39}\right)\overset{?} {=}\frac{\varphi}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\varphi:=\:{golden}\:{ratio} \\ $$$$\:{m}.{n} \\ $$
Answered by bramlexs22 last updated on 08/Sep/21
sin (9°)+sin (21°)+sin (39°)= 2sin 15° cos 6°+sin (45°−6°)= 2sin (45°−30°)cos 6°+((√2)/2) cos 6°−((√2)/2) sin 6°= 2cos 6°(((√6)/4)−((√2)/4))+((√2)/2)cos 6°−((√2)/2)sin 6° = ((√6)/2) cos 6°−((√2)/2)sin 6° = (√((6+2)/4)) cos (6°−330°) =(√2) cos 324°=(√2) cos (270°+54°) =(√2) sin 54° = (√2) cos 36° =(√2) (1−2sin^2 18°) =(√2)(1−2((((√5)−1)/4))^2 )=(√2) (1−(((6−2(√5))/8))) =(√2) (((2+2(√5))/8))=(√2)((((√5)+1)/4))=((√2)/2)((((√5)+1)/2)) =(1/( (√2)))ϕ=(ϕ/( (√2)))
$$\:\mathrm{sin}\:\left(\mathrm{9}°\right)+\mathrm{sin}\:\left(\mathrm{21}°\right)+\mathrm{sin}\:\left(\mathrm{39}°\right)= \\ $$$$\:\mathrm{2sin}\:\mathrm{15}°\:\mathrm{cos}\:\mathrm{6}°+\mathrm{sin}\:\left(\mathrm{45}°−\mathrm{6}°\right)= \\ $$$$\:\mathrm{2sin}\:\left(\mathrm{45}°−\mathrm{30}°\right)\mathrm{cos}\:\mathrm{6}°+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\mathrm{cos}\:\mathrm{6}°−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\mathrm{sin}\:\mathrm{6}°= \\ $$$$\:\mathrm{2cos}\:\mathrm{6}°\left(\frac{\sqrt{\mathrm{6}}}{\mathrm{4}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\right)+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{cos}\:\mathrm{6}°−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{sin}\:\mathrm{6}°\:= \\ $$$$\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\:\mathrm{cos}\:\mathrm{6}°−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{sin}\:\mathrm{6}°\:=\:\sqrt{\frac{\mathrm{6}+\mathrm{2}}{\mathrm{4}}}\:\mathrm{cos}\:\left(\mathrm{6}°−\mathrm{330}°\right) \\ $$$$=\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{324}°=\sqrt{\mathrm{2}}\:\mathrm{cos}\:\left(\mathrm{270}°+\mathrm{54}°\right) \\ $$$$=\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{54}°\:=\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{36}° \\ $$$$=\sqrt{\mathrm{2}}\:\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{18}°\right) \\ $$$$=\sqrt{\mathrm{2}}\left(\mathrm{1}−\mathrm{2}\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} \right)=\sqrt{\mathrm{2}}\:\left(\mathrm{1}−\left(\frac{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{8}}\right)\right) \\ $$$$=\sqrt{\mathrm{2}}\:\left(\frac{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{8}}\right)=\sqrt{\mathrm{2}}\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\varphi=\frac{\varphi}{\:\sqrt{\mathrm{2}}}\:\: \\ $$
Commented by Tawa11 last updated on 08/Sep/21
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$

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