Răspuns:
Formule folosite:
[tex]cos(x)*cos(y)=\frac{cos(x+y)+cos(x-y)}{2}[/tex]
[tex]2cosx*cosy=cos(x+y)+cos(x-y)[/tex]
[tex]cos^2(x+y)+cos^2(x-y)-cos(2x)*cos(2y)=?[/tex]
Ne vom ocupa cu cos^2(x+y)+cos^2(x-y) mai intai:
[tex]cos^2(x+y)+cos^2(x-y)=(cos(x+y)+cos(x-y))^2-2cos(x+y)*cos(x-y)[/tex]
[tex]=(cos(x+y)+cos(x-y))^2-2*\frac{cos(x+y+x-y)+cos(x+y-x+y)}{2}[/tex]
[tex]=(cos(x+y)+cos(x-y))^2-cos(2x)-cos(2y)[/tex]
[tex]=(cos(x+y)+cos(x-y))^2-(cos^2x-sin^2x+cos^2y-sin^2y)[/tex]
[tex]=(cos(x+y)+cos(x-y))^2-(cos^2x-(1-cos^2x)+cos^2y-(1-cos^2y))[/tex]
[tex]=(cos(x+y)+cos(x-y))^2-(2cos^2x+2cos^2y-2)[/tex]
[tex]=(cos(x+y)+cos(x-y))^2-2cos^2x-2cos^2y+2[/tex]
Acum de -cos(2x)*cos(2y)
[tex]-(cos^2x-sin^2x)*(cos^2y-sin^2y)[/tex]
[tex]=-(cos^2x-(1-cos^2x))*(cos^2y-(1-cos^2y))[/tex]
[tex]=-(2cos^2x-1)*(2cos^2y-1)[/tex]
[tex]=-(2cos^2x*2cos^2y-2cos^2x-2cos^2y+1)[/tex]
[tex]=-((cos(x+y)+cos(x-y))^2-2cos^2x-2cos^2y+1)[/tex]
[tex]=-((cos(x+y)+cos(x-y))^2+2cos^2x+2cos^2y-1[/tex]
Inlocuim:
[tex](cos(x+y)+cos(x-y))^2-2cos^2x-2cos^2y+2-(cos(x+y)+cos(x-y))^2+2cos^2x+2cos^2y-1=[/tex]
[tex]=2-1=1[/tex] "Expresia nu depinde de x si de y"
