Question

Решить
1) номер 8 (3)
2) номер 9 (1)
3) номер 9 (2)

Answer 1

$y=10cos^2x-6sin xcos x+2sin^2x=8cos^2x-3sin2x+2=\ \ \ =8cdot dfrac{1+cos2x}{2} -3sin2x+2=4cos 2x-3sin2x+6=\ \ \ = 5cos(2x-arccos frac{4}{5} )+6$

$-1 leq cos(2x-arccos frac{4}{5} ) leq 1\ \ -5 leq 5cos(2x-arccos frac{4}{5} ) leq 5\ \ \ 1 leq 5cos(2x-arccos frac{4}{5} )+6 leq 11$

Область значений : $E(y)=[1;11]$

$9.(1),,, y=sin^4 x+cos^4x=sin^4x+cos^4xpm2sin^2xcos^2x=\ \ \ =(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-0.5sin^22x$

$0 leq sin^22x leq 1\ \ -0.5 leq -0.5sin^22x leq 0\ \ 0.5 leq 1-0.5sin^22x leq 1$

Область значений : $E(y)=[0.5;1]$

$9.(2),, y=sin^6x+cos^6x=\ \ =(sin^ 2x+cos^2x)^3-3sin^4xcos^2x-3sin^2xcos^4x=\ \ \ =1-3sin^2xcos^2x(sin^2x+cos^2x)=1-3sin^2xcos^2x=\ \ \ =1-0.75sin^22x$

$0 leq sin^22x leq 1\ \ -0.75 leq -0.75sin^22x leq 0\ \ 0.25 leq 1-0.75sin^22x leq 1$

Область значений : $E(y)=[0.25;1]$