Răspuns:
Ex.3) Verifica daca egalitatea este adevarata
[tex](tg^4x-6*tg^2x+1)*cos^4x+2*sin^2(2x)=1[/tex]
[tex](\frac{sin^4x}{cos^4x}-6*\frac{sin^2x}{cos^2x}+1)*cos^4x+2sin^2(2x)=1\\\\[/tex] (Desfacem paranteza)
[tex]\frac{sin^4x}{cos^4x} *cos^4x-6*\frac{sin^2x}{cos^2x}*cos^4x+cos^4x+2sin^2(2x)=1[/tex]
[tex]sin^4x-6sin^2x*cos^2x+cos^4x+2sin^2(2x)=1[/tex]
[tex]sin^4x+cos^4x-6sin^2x*cos^2x+2sin^2(2x)=1[/tex]
[tex]1-6sin^2x*cos^4x+2sin^2(2x)=1[/tex]
[tex]-6sin^2x*cos^2x+2sin^2(2x)=0[/tex]
Voi face separat [tex]2sin^2(2x)[/tex]
[tex]2sin^2(2x)=2(sin2x)(sin2x)=2(2sinx*cosx)(2sinx*cosx) =8sin^2x*cos^2x[/tex]
Inlocuim in ecuatie:
[tex]-6sin^2x*cos^2x+8sin^2x*cos^2x=0[/tex]
[tex]2sin^2x*cos^2x=0[/tex] => egalitatea este falsa
Ex.4) Aduceti la o forma mai simpla expresia
[tex]E=\frac{sin(45+x)-cos(45+x)}{sin(45+x)+cos(45+x)}[/tex]
Voi face separat sin(45+x)-cos(45+x) si sin(45+x)+cos(45+x) pentru a fi mai clar
[tex]sin(45+x)-cos(45+x)=sin(45)*cos(x)+sin(x)*cos(45)-(cos(45)*cos(x)-sin(45)*sin(x))[/tex]
[tex]=\frac{\sqrt{2} }{2}*cos(x)+sin(x)*\frac{\sqrt{2}}{2}-\frac{\sqrt{2} }{2} *cos(x)+\frac{\sqrt{2} }{2}*sin(x)=2*sin(x)*\frac{\sqrt{2} }{2}[/tex]
[tex]=sin(x)*\sqrt{2}[/tex]
Acum sin(45+x)+cos(45+x):
[tex]sin(45+x)+cos(45+x)=sin(45)*cos(x)+sin(x)*cos(45)+cos(45)*cos(x)-sin(45)*sin(x)[/tex]
[tex]=\frac{\sqrt{2} }{2}*cos(x)+sin(x)*\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}*cos(x)-\frac{\sqrt{2} }{2}*sin(x)=2*cos(x)*\frac{\sqrt{2} }{2}=cos(x)*\sqrt{2}[/tex]
Inlocuim:
[tex]\frac{sin(x)*\sqrt{2} }{cos(x)*\sqrt{2} }=\frac{sin(x)}{cos(x)}=tg(x)[/tex]
