задача от изпит 2015

[Делян Василев]

80842283_627624711377536_5027443319446175744_n.jpg (60.59 KiB) Прегледано 580 пъти

[Добромир Глухаров]

$A=sin2x-sin^23x+cos(\pi/3-3x)sin(3x-\pi/6)$

$A=sin2x-sin^23x+(sin(\pi/3-\pi/6)+sin(6x-\pi/3-\pi/6))/2$

$A=sin2x-sin^23x+(1/2+sin(6x-\pi/2))/2$

$A=sin2x-sin^23x+1/4-cos(6x)/2$

$A=sin2x-sin^23x+1/4-(cos^2(3x)-sin^2(3x))/2$

$A=sin2x+1/4-(cos^2(3x)+sin^2(3x))/2=sin2x-1/4$

$\max A=1-1/4=3/4$