Question

Решите, пожалуйста
Картинка прилагается

Answer 1

$sin alpha -cos alpha =frac{1}{2}; ; to \\(sin alpha -cos alpha )^2=frac{1}{4}\\sin^2 alpha +cos^2 alpha -2sin alpha cos alpha =frac{1}{4}\\1-sin2 alpha =frac{1}{4}; ; to ; ; sin2 alpha =1-frac{1}{4}=frac{3}{4}\\(sin^2 alpha +cos^2 alpha )^2=1^2=1\\sin^4 alpha +cos^4 alpha +2sin^2 alpha cos^2 alpha =1\\sin^4 alpha +cos^4 alpha +2(frac{1}{2}sin2 alpha )^2=1\\sin^4 alpha +cos^4 alpha =1-2cdot frac{1}{4}cdot sin^22 alpha$

$sin^4 alpha +cos^4 alpha =1-frac{1}{2}cdot frac{9}{16}= frac{23}{32}$

Answer 2

sin a - cos a = 1/2
(sin a - cos a)^2 = sin^2 a - 2sin a*cos a + cos^2 a = (1/2)^2 = 1/4
2sin a*cos a = sin 2a = 1 - 1/4 = 3/4
1^2 = (sin^2 a + cos^2 a)^2 = sin^4 a + cos^4 a + 2sin^2 a*cos^2 a
1 = sin^4 a + cos^4 a + (2sin a*cos a)^2/2 = sin^4 a + cos^4 a + 1/2*sin^2(2a)
sin^4 a + cos^4 a = 1 - 1/2*sin^2(2a) = 1 - 1/2*9/16 = 1 - 9/32 = 23/32