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Question-69034

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Question Number 69034 by ahmadshah last updated on 18/Sep/19
Answered by $@ty@m123 last updated on 18/Sep/19
4cos x−3sec x−2tan x 4cos x−(3/(cos x))−((2sin x)/(cos x)) =((4cos^2 x−3−2sin x)/(cos x)) =((4cos^3 x−3cos x−2sin xcos x)/(cos^2 x)) =((cos 3x−sin 2x)/(cos^2 x)) =((cos 54−sin 36)/(cos^2 18)) =((cos 54−cos 54)/(cos^2 18)) =0
$$\mathrm{4cos}\:{x}−\mathrm{3sec}\:{x}−\mathrm{2tan}\:{x} \\ $$$$\mathrm{4cos}\:{x}−\frac{\mathrm{3}}{\mathrm{cos}\:{x}}−\frac{\mathrm{2sin}\:{x}}{\mathrm{cos}\:{x}} \\ $$$$=\frac{\mathrm{4cos}\:^{\mathrm{2}} {x}−\mathrm{3}−\mathrm{2sin}\:{x}}{\mathrm{cos}\:{x}} \\ $$$$=\frac{\mathrm{4cos}\:^{\mathrm{3}} {x}−\mathrm{3cos}\:{x}−\mathrm{2sin}\:{x}\mathrm{cos}\:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}} \\ $$$$=\frac{\mathrm{cos}\:\mathrm{3}{x}−\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}} \\ $$$$=\frac{\mathrm{cos}\:\mathrm{54}−\mathrm{sin}\:\mathrm{36}}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{18}} \\ $$$$=\frac{\mathrm{cos}\:\mathrm{54}−\mathrm{cos}\:\mathrm{54}}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{18}} \\ $$$$=\mathrm{0} \\ $$
Answered by MJS last updated on 18/Sep/19
sin 18° =−(1/4)+((√5)/4) cos 18° =((√(10+2(√5)))/4) ⇒ sec 18° =((2(√2))/( (√(5+(√5))))); tan 18° =(((−1+(√5))(√2))/(2(√(5+(√5))))) 4((√(10+2(√5)))/4)−3((2(√2))/( (√(5+(√5)))))−2((((√5)−1)(√2))/(2(√(5+(√5)))))= =(√(10+2(√5)))−((6(√2)+(1−(√5))(√2))/( (√(5+(√5)))))= =(√(10+2(√5)))−(((5+(√5))(√2))/( (√(5+(√5)))))= =(√(10+2(√5)))−(√(5+(√5)))(√2)= =(√(10+2(√5)))−(√(10+2(√5)))=0
$$\mathrm{sin}\:\mathrm{18}°\:=−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\mathrm{cos}\:\mathrm{18}°\:=\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}} \\ $$$$\Rightarrow\:\mathrm{sec}\:\mathrm{18}°\:=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}};\:\mathrm{tan}\:\mathrm{18}°\:=\frac{\left(−\mathrm{1}+\sqrt{\mathrm{5}}\right)\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}} \\ $$$$\mathrm{4}\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}}−\mathrm{3}\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}}−\mathrm{2}\frac{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}}= \\ $$$$=\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}−\frac{\mathrm{6}\sqrt{\mathrm{2}}+\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}}= \\ $$$$=\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}−\frac{\left(\mathrm{5}+\sqrt{\mathrm{5}}\right)\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}}= \\ $$$$=\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}−\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}\sqrt{\mathrm{2}}= \\ $$$$=\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}−\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}=\mathrm{0} \\ $$

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