Question

$\frac{ { \sin }^{2}2x - 4 \sin{}^{2}x }{ \sin^{2}2x +4 \sin^{2}x - 4 } = \tan^{4}x$
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Answer 1

Răspuns:

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

Explicație pas cu pas:

folosesti formulele:

$sin 2x=2sin x cos x$

$sin^{2}x+cos^{2}x=1$