Question

E =cos²(x+y) + cos²(x-y) -cos(2x)*cos(2y)

Arătați că nu depinde de x și de y ​

Answer 1

Răspuns:

Formule folosite:

$cos(x)*cos(y)=\frac{cos(x+y)+cos(x-y)}{2}$

$2cosx*cosy=cos(x+y)+cos(x-y)$

$cos^2(x+y)+cos^2(x-y)-cos(2x)*cos(2y)=?$

Ne vom ocupa cu cos^2(x+y)+cos^2(x-y) mai intai:

$cos^2(x+y)+cos^2(x-y)=(cos(x+y)+cos(x-y))^2-2cos(x+y)*cos(x-y)$

$=(cos(x+y)+cos(x-y))^2-2*\frac{cos(x+y+x-y)+cos(x+y-x+y)}{2}$

$=(cos(x+y)+cos(x-y))^2-cos(2x)-cos(2y)$

$=(cos(x+y)+cos(x-y))^2-(cos^2x-sin^2x+cos^2y-sin^2y)$

$=(cos(x+y)+cos(x-y))^2-(cos^2x-(1-cos^2x)+cos^2y-(1-cos^2y))$

$=(cos(x+y)+cos(x-y))^2-(2cos^2x+2cos^2y-2)$

$=(cos(x+y)+cos(x-y))^2-2cos^2x-2cos^2y+2$

Acum de -cos(2x)*cos(2y)

$-(cos^2x-sin^2x)*(cos^2y-sin^2y)$

$=-(cos^2x-(1-cos^2x))*(cos^2y-(1-cos^2y))$

$=-(2cos^2x-1)*(2cos^2y-1)$

$=-(2cos^2x*2cos^2y-2cos^2x-2cos^2y+1)$

$=-((cos(x+y)+cos(x-y))^2-2cos^2x-2cos^2y+1)$

$=-((cos(x+y)+cos(x-y))^2+2cos^2x+2cos^2y-1$

Inlocuim:

$(cos(x+y)+cos(x-y))^2-2cos^2x-2cos^2y+2-(cos(x+y)+cos(x-y))^2+2cos^2x+2cos^2y-1=$

$=2-1=1$ "Expresia nu depinde de x si de y"