Question-84117

Question Number 84117 by Power last updated on 09/Mar/20

Answered by TANMAY PANACEA last updated on 09/Mar/20

sin100+sin20+sin50 2sin60.cos40+cos40 cos40(1+(√3) ) 96×2×sin40.cos40+(1/2)(cos30−cos100) =96×2×sin40.cos40+((√3)/4)−((cos100)/2) =192sin40.cos40+((√3)/4)−(1/2)(2cos^2 50−1) =192sin40.cos40−sin^2 40+(((√3) +2)/4) wait

$${sin}\mathrm{100}+{sin}\mathrm{20}+{sin}\mathrm{50} \\ $$$$\mathrm{2}{sin}\mathrm{60}.{cos}\mathrm{40}+{cos}\mathrm{40} \\ $$$${cos}\mathrm{40}\left(\mathrm{1}+\sqrt{\mathrm{3}}\:\right) \\ $$$$\mathrm{96}×\mathrm{2}×{sin}\mathrm{40}.{cos}\mathrm{40}+\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\mathrm{30}−{cos}\mathrm{100}\right) \\ $$$$=\mathrm{96}×\mathrm{2}×{sin}\mathrm{40}.{cos}\mathrm{40}+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}−\frac{{cos}\mathrm{100}}{\mathrm{2}} \\ $$$$=\mathrm{192}{sin}\mathrm{40}.{cos}\mathrm{40}+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{cos}^{\mathrm{2}} \mathrm{50}−\mathrm{1}\right) \\ $$$$=\mathrm{192}{sin}\mathrm{40}.{cos}\mathrm{40}−{sin}^{\mathrm{2}} \mathrm{40}+\frac{\sqrt{\mathrm{3}}\:+\mathrm{2}}{\mathrm{4}} \\ $$$$ \\ $$$${wait} \\ $$$$ \\ $$

Answered by mind is power last updated on 09/Mar/20

i think Quation is ((96sin(80)sin(65)sin(35))/(sin(20)+sin(50)+sin(110))) sin(65)sin(35)=(1/2)(cos(65−35)−cos(65+35)) =(1/2)(cos(30)−cos(110)) sin(80).sin(65)sin(35)=((sin(80))/2)(cos(30)−cos(110)) =(1/2)(sin(80)cos(30)−sin(80)cos(110)) =(1/4)(sin(110)+sin(50)−sin(180)−sin(−20)) =(1/4)(sin(110)+sin(50)+sin(20)) ⇔ ((sin(65)sin(35)sin(80))/(sin(110)+sin(50)+sin(20)))=(1/4) ⇔((96sin(65)sin(35)sin(80))/(sin(110)+sin(20)+sin(50)))=((96)/4)=24

$${i}\:{think}\:{Quation}\:{is} \\ $$$$\frac{\mathrm{96}{sin}\left(\mathrm{80}\right){sin}\left(\mathrm{65}\right){sin}\left(\mathrm{35}\right)}{{sin}\left(\mathrm{20}\right)+{sin}\left(\mathrm{50}\right)+{sin}\left(\mathrm{110}\right)} \\ $$$${sin}\left(\mathrm{65}\right){sin}\left(\mathrm{35}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\left(\mathrm{65}−\mathrm{35}\right)−{cos}\left(\mathrm{65}+\mathrm{35}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\left(\mathrm{30}\right)−{cos}\left(\mathrm{110}\right)\right) \\ $$$${sin}\left(\mathrm{80}\right).{sin}\left(\mathrm{65}\right){sin}\left(\mathrm{35}\right)=\frac{{sin}\left(\mathrm{80}\right)}{\mathrm{2}}\left({cos}\left(\mathrm{30}\right)−{cos}\left(\mathrm{110}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({sin}\left(\mathrm{80}\right){cos}\left(\mathrm{30}\right)−{sin}\left(\mathrm{80}\right){cos}\left(\mathrm{110}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left({sin}\left(\mathrm{110}\right)+{sin}\left(\mathrm{50}\right)−{sin}\left(\mathrm{180}\right)−{sin}\left(−\mathrm{20}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left({sin}\left(\mathrm{110}\right)+{sin}\left(\mathrm{50}\right)+{sin}\left(\mathrm{20}\right)\right) \\ $$$$\Leftrightarrow \\ $$$$\frac{{sin}\left(\mathrm{65}\right){sin}\left(\mathrm{35}\right){sin}\left(\mathrm{80}\right)}{{sin}\left(\mathrm{110}\right)+{sin}\left(\mathrm{50}\right)+{sin}\left(\mathrm{20}\right)}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Leftrightarrow\frac{\mathrm{96}{sin}\left(\mathrm{65}\right){sin}\left(\mathrm{35}\right){sin}\left(\mathrm{80}\right)}{{sin}\left(\mathrm{110}\right)+{sin}\left(\mathrm{20}\right)+{sin}\left(\mathrm{50}\right)}=\frac{\mathrm{96}}{\mathrm{4}}=\mathrm{24} \\ $$$$ \\ $$$$ \\ $$

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