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Question Number 151959 by john_santu last updated on 24/Aug/21

Answered by iloveisrael last updated on 24/Aug/21

(sin^2 x)^3 +(cos^2 x)^3 = (1/4) ⇒(1)(sin^4 x+cos^4 x−(sin xcos x)^2 )=(1/4) ⇒(sin^2 x+cos^2 x)^2 −3(sin xcos x)^2 =(1/4) ⇒(sin xcos x)^2 =((1−(1/4))/3)=(1/4) ⇔(1/(sin^6 x))+(1/(cos^6 x)) = ((1/4)/((sin^2 x cos^2 x)^3 )) =(1/(((1/4))^2 )) = 16.

$$\left(\mathrm{sin}\:^{\mathrm{2}} {x}\right)^{\mathrm{3}} +\left(\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{3}} =\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\left(\mathrm{1}\right)\left(\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}−\left(\mathrm{sin}\:{x}\mathrm{cos}\:{x}\right)^{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\left(\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{3}\left(\mathrm{sin}\:{x}\mathrm{cos}\:{x}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\left(\mathrm{sin}\:{x}\mathrm{cos}\:{x}\right)^{\mathrm{2}} =\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Leftrightarrow\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{6}} {x}}+\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{6}} {x}}\:=\:\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\left(\mathrm{sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} }\:=\:\mathrm{16}. \\ $$

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