Question

ex 3 și 4 rezolvate urgent pls ​

Answer 1

Răspuns:

Ex.3) Verifica daca egalitatea este adevarata

$(tg^4x-6*tg^2x+1)*cos^4x+2*sin^2(2x)=1$

$(\frac{sin^4x}{cos^4x}-6*\frac{sin^2x}{cos^2x}+1)*cos^4x+2sin^2(2x)=1$ (Desfacem paranteza)

$\frac{sin^4x}{cos^4x} *cos^4x-6*\frac{sin^2x}{cos^2x}*cos^4x+cos^4x+2sin^2(2x)=1$

$sin^4x-6sin^2x*cos^2x+cos^4x+2sin^2(2x)=1$

$sin^4x+cos^4x-6sin^2x*cos^2x+2sin^2(2x)=1$

$1-6sin^2x*cos^4x+2sin^2(2x)=1$

$-6sin^2x*cos^2x+2sin^2(2x)=0$

Voi face separat $2sin^2(2x)$

$2sin^2(2x)=2(sin2x)(sin2x)=2(2sinx*cosx)(2sinx*cosx) =8sin^2x*cos^2x$

Inlocuim in ecuatie:

$-6sin^2x*cos^2x+8sin^2x*cos^2x=0$

$2sin^2x*cos^2x=0$ => egalitatea este falsa

Ex.4) Aduceti la o forma mai simpla expresia

$E=\frac{sin(45+x)-cos(45+x)}{sin(45+x)+cos(45+x)}$

Voi face separat sin(45+x)-cos(45+x) si sin(45+x)+cos(45+x) pentru a fi mai clar

$sin(45+x)-cos(45+x)=sin(45)*cos(x)+sin(x)*cos(45)-(cos(45)*cos(x)-sin(45)*sin(x))$

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

$=sin(x)*\sqrt{2}$

Acum sin(45+x)+cos(45+x):

$sin(45+x)+cos(45+x)=sin(45)*cos(x)+sin(x)*cos(45)+cos(45)*cos(x)-sin(45)*sin(x)$

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

Inlocuim:

$\frac{sin(x)*\sqrt{2} }{cos(x)*\sqrt{2} }=\frac{sin(x)}{cos(x)}=tg(x)$