Question

Дано Tgα=2.
Найти значение выражения: $\frac{sin^{4} \alpha-cos^{4}\alpha }{sin^{6} \alpha-cos^{6}\alpha }$

Answer 1

$\frac{sin^4(x)-cos^4(x)}{sin^6(x)-cos^6(x)} = \frac{(sin^2(x)+cos^2(x))(sin^2(x) - cos^2(x))}{(sin^2(x) - cos^2(x))(sin^4(x) + sin^2(x)cos^2(x) +cos^4(x))}$

$= \frac{1}{((sin^2(x) + cos^2(x))^2 -sin^2(x)cos^2(x))} = \frac{1}{1-0.25sin^2(2x)}=\frac{4}{4-sin^2(2x)}$

$tg(x)=2\\tg^2(x) = 4\\1+tg^2(x) = \frac{1}{cos^2(x)}\\cos^2(x) =\frac{1}{5}\\cos(x) =\frac{1}{\sqrt{5}}\\sin(x) = \sqrt{1-cos^2(x)} =\frac{2}{\sqrt{5}}\\sin(2x) = 2sin(x)cos(x) = 2*\frac{2}{\sqrt{5}}*\frac{1}{\sqrt{5}} = \frac{4}{5}$

$\frac{4}{4-sin^2(2x)} = \frac{4}{4-\frac{16}{25} }=\frac{4}{3.36} =\frac{400}{336} = \frac{25}{21}\\ Answer:\frac{25}{21}$