Question

Довисти  тотожность : 

1.

$frac{sin^{2}2alpha - 4sin^{2}alpha}{sin^{2}alpha+4sin^{2}alpha -4} = tg^{4}alpha$ 

2.

$2cos^{2}alpha cdot sin^{2}alpha + cos^{4}alpha + sin^{4}alpha = 1$

3.Найдите cos α 

$cos^{4}alpha - sin^{4}alpha =frac{1}{8}$

 

 

 

Answer 1

$2) 2cos^{2}alpha cdot sin^{2}alpha + cos^{4}alpha + sin^{4}alpha = \ = sin^{4}alpha + 2cos^{2}alpha cdot sin^{2}alpha + cos^{4}alpha=(sin^{2}alpha+cos^{2}alpha)^2=1^2=1$

 

$3) cos^{4}alpha - sin^{4}alpha = (cos^{2}alpha - sin^{2}alpha)(cos^{2}alpha + sin^{2}alpha) = \ = cos^{2}alpha - sin^{2}alpha = cos2alpha = 2*cos^{2}alpha-1 \ 2*cos^{2}alpha-1=frac{1}{8} \ \ 2*cos^{2}alpha=frac{9}{8} \ \ cos^{2}alpha=frac{9}{16} \ \ cosalpha=бfrac{3}{4}=б0,75$