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Question Number 111513 by john santu last updated on 04/Sep/20
((cos^2 x))^(1/(3 )) + ((sin^2 x))^(1/(3 )) = (2)^(1/(3 )) find cos^2 (2x) ?
$$\:\:\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:^{\mathrm{2}} {x}}\:+\:\sqrt[{\mathrm{3}\:}]{\mathrm{sin}\:^{\mathrm{2}} {x}}\:=\:\sqrt[{\mathrm{3}\:}]{\mathrm{2}} \\ $$$${find}\:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)\:? \\ $$
Answered by bemath last updated on 04/Sep/20
(((cos^2 x))^(1/(3 )) + ((sin^2 x))^(1/(3 )) )^3 = 2 (a+b)^3 = a^3 +b^3 +3ab(a+b) ⇒1+3 ((sin^2 x cos^2 x))^(1/(3 )) (((sin^2 x))^(1/(3 )) +((cos^2 x))^(1/(3 )) ) = 2 ⇒3.(2)^(1/(3 )) (((1/4)sin^2 (2x)))^(1/(3 )) = 1 ⇒ 54((1/4)sin^2 (2x))= 1 ⇒ sin^2 (2x) = (4/(54)) →cos^2 (2x) = ((54−4)/(54)) → cos^2 (2x) = ((50)/(54)) = ((25)/(27))
$$\:\left(\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\:+\:\sqrt[{\mathrm{3}\:}]{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\:\right)^{\mathrm{3}} \:=\:\mathrm{2} \\ $$$$\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{3}} \:=\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{3ab}\left(\mathrm{a}+\mathrm{b}\right) \\ $$$$\Rightarrow\mathrm{1}+\mathrm{3}\:\sqrt[{\mathrm{3}\:}]{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\:\left(\sqrt[{\mathrm{3}\:}]{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\:+\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\right)\:=\:\mathrm{2} \\ $$$$\Rightarrow\mathrm{3}.\sqrt[{\mathrm{3}\:}]{\mathrm{2}}\:\:\sqrt[{\mathrm{3}\:}]{\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{2x}\right)}\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{54}\left(\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{2x}\right)\right)=\:\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{2x}\right)\:=\:\frac{\mathrm{4}}{\mathrm{54}}\:\rightarrow\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{2x}\right)\:=\:\frac{\mathrm{54}−\mathrm{4}}{\mathrm{54}} \\ $$$$\rightarrow\:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{2x}\right)\:=\:\frac{\mathrm{50}}{\mathrm{54}}\:=\:\frac{\mathrm{25}}{\mathrm{27}} \\ $$
Answered by som(math1967) last updated on 04/Sep/20
cos^2 x+sin^2 x+3^3 (√(cos^2 xsin^2 x)){^3 (√2)}=2 27×2sin^2 xcos^2 x=1 (2sinxcosx)^2 =(2/(27)) sin^2 2x=(2/(27)) cos^2 2x=1−(2/(27))=((25)/(27))
$$\mathrm{cos}^{\mathrm{2}} \mathrm{x}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}+\mathrm{3}^{\mathrm{3}} \sqrt{\mathrm{cos}^{\mathrm{2}} \mathrm{xsin}^{\mathrm{2}} \mathrm{x}}\left\{^{\mathrm{3}} \sqrt{\mathrm{2}}\right\}=\mathrm{2} \\ $$$$\mathrm{27}×\mathrm{2sin}^{\mathrm{2}} \mathrm{xcos}^{\mathrm{2}} \mathrm{x}=\mathrm{1} \\ $$$$\left(\mathrm{2sinxcosx}\right)^{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{27}} \\ $$$$\mathrm{sin}^{\mathrm{2}} \mathrm{2x}=\frac{\mathrm{2}}{\mathrm{27}} \\ $$$$\mathrm{cos}^{\mathrm{2}} \mathrm{2x}=\mathrm{1}−\frac{\mathrm{2}}{\mathrm{27}}=\frac{\mathrm{25}}{\mathrm{27}} \\ $$

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