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Question Number 146404 by iloveisrael last updated on 13/Jul/21
 trigonometry
$$\:\mathrm{trigonometry} \\ $$
Commented by iloveisrael last updated on 13/Jul/21
Answered by gsk2684 last updated on 13/Jul/21
f(x)=((1−cos^2 x+cos^( 4) x)/(1−sin^2 x+sin^4 x))  =((1−cos^2 x(1−cos^2 x))/(1−sin^2 x(1−sin^2 x)))  =((1−cos^2 xsin^2 x)/(1−sin^2 xcos^2 x))=1  f(x)=1 ∀ x
$${f}\left({x}\right)=\frac{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{cos}^{\:\mathrm{4}} {x}}{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{sin}\:^{\mathrm{4}} {x}} \\ $$$$=\frac{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} {x}\right)}{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} {x}\right)} \\ $$$$=\frac{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}\mathrm{cos}\:^{\mathrm{2}} {x}}=\mathrm{1} \\ $$$${f}\left({x}\right)=\mathrm{1}\:\forall\:{x} \\ $$

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