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Question Number 4418 by Syaka last updated on 23/Jan/16
What the exact value of :  sin 10° (2cos 10° + 1)(2cos 10° − 1) = ....?
$${What}\:{the}\:{exact}\:{value}\:{of}\:: \\ $$$${sin}\:\mathrm{10}°\:\left(\mathrm{2}{cos}\:\mathrm{10}°\:+\:\mathrm{1}\right)\left(\mathrm{2}{cos}\:\mathrm{10}°\:−\:\mathrm{1}\right)\:=\:….? \\ $$$$ \\ $$
Answered by Yozzii last updated on 24/Jan/16
sin10°(2cos10°+1)(2cos10°−1)  =sin10°(4cos^2 10°−1)  =2(2sin10°cos10°)cos10°−sin10°  =2sin20°cos10°−sin10°  ={sin(20°+10°)+sin(20°−10°)}−sin10°  =sin30°+sin10°−sin10°  =0.5
$${sin}\mathrm{10}°\left(\mathrm{2}{cos}\mathrm{10}°+\mathrm{1}\right)\left(\mathrm{2}{cos}\mathrm{10}°−\mathrm{1}\right) \\ $$$$={sin}\mathrm{10}°\left(\mathrm{4}{cos}^{\mathrm{2}} \mathrm{10}°−\mathrm{1}\right) \\ $$$$=\mathrm{2}\left(\mathrm{2}{sin}\mathrm{10}°{cos}\mathrm{10}°\right){cos}\mathrm{10}°−{sin}\mathrm{10}° \\ $$$$=\mathrm{2}{sin}\mathrm{20}°{cos}\mathrm{10}°−{sin}\mathrm{10}° \\ $$$$=\left\{{sin}\left(\mathrm{20}°+\mathrm{10}°\right)+{sin}\left(\mathrm{20}°−\mathrm{10}°\right)\right\}−{sin}\mathrm{10}° \\ $$$$={sin}\mathrm{30}°+{sin}\mathrm{10}°−{sin}\mathrm{10}° \\ $$$$=\mathrm{0}.\mathrm{5} \\ $$

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