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Question Number 157884 by HongKing last updated on 29/Oct/21

(1/(sin(10°))) - 4 sin(70°) = ?

$$\frac{\mathrm{1}}{\mathrm{sin}\left(\mathrm{10}°\right)}\:-\:\mathrm{4}\:\mathrm{sin}\left(\mathrm{70}°\right)\:=\:? \\ $$

Answered by tounghoungko last updated on 29/Oct/21

(1/(sin 10°))−4(((√3)/2) cos 10°+(1/2)sin 10°) =(1/(sin 10°))−2(√3) cos 10°−2sin 10° =((1−2(√3) sin 10° cos 10°−2sin^2 10°)/(sin 10°)) =((1−(√3) sin 20°−2((1/2)−(1/2)cos 20°))/(sin 10°)) =((cos 20°−(√3) sin 20°)/(sin 10°)) =((2((1/2)cos 20°−((√3)/2)sin 20°))/(sin 10°)) =((2cos 80°)/(sin 10°)) = 2 .

$$\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{10}°}−\mathrm{4}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{cos}\:\mathrm{10}°+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{10}°\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{10}°}−\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{10}°−\mathrm{2sin}\:\mathrm{10}° \\ $$$$=\frac{\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{10}°\:\mathrm{cos}\:\mathrm{10}°−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{10}°}{\mathrm{sin}\:\mathrm{10}°} \\ $$$$=\frac{\mathrm{1}−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{20}°−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{20}°\right)}{\mathrm{sin}\:\mathrm{10}°} \\ $$$$=\frac{\mathrm{cos}\:\mathrm{20}°−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{20}°}{\mathrm{sin}\:\mathrm{10}°} \\ $$$$=\frac{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{20}°−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\mathrm{20}°\right)}{\mathrm{sin}\:\mathrm{10}°} \\ $$$$=\frac{\mathrm{2cos}\:\mathrm{80}°}{\mathrm{sin}\:\mathrm{10}°}\:=\:\mathrm{2}\:. \\ $$

Commented by HongKing last updated on 29/Oct/21

alot thankyou sir

$$\mathrm{alot}\:\mathrm{thankyou}\:\mathrm{sir} \\ $$

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