Y(max) + y(min) - ?y=(2sin3x + 2cos3x)^2

Question

Y(max) + y(min) - ?
y=(2sin3x + 2cos3x)^2

Answer 1

Y=(2sin3x+2cos3x)^2
y=(2*√2*(√2/2 *sin3x+√2/2* cos3x)^2=8*sin^2(3x+π/4)
      0 <=sin^2 (3x+π/4)<=1<br>         0<=y<=8<br>ymax+ymin=0+8=8
Ответ:8