Question

calculati : sin2x-cos²x

Answer 1

sin(2x)-cos^2(x)=2*sin(x)*cos(x)-cos(x)*cos(x)=cos(x)*[2sin(x)-cos(x)].
Daca am avem in vedere ecuatia;
sin(2x)-cos^2(x)=0- atunci rezolvarea acesteia va fi urmatoarea;
1) cos(x)=0, deci x1=π/2(rad)=90° si x2=3π/2(rad)=270°;
2) 2*sin(x)-cos(x)=0 =>
2*sin(x)=cos(x) |:cos(x);
deci 2*tg(x)=1, deci tg(x)=1/2.
Conform calculelor mele; x3=26,57° iar x4=206,57°.

Answer 2

Calculati : sin2x-cos²x

[tex]\it sin2x - cos^2x = sin2x-(1-sin^2x) = sin2x-1+sin^2x = \\ \\ =sin^2x+sin2x-1[/tex]

$\it sin2x - cos^2x = 2sinxcosx-cos^2x = cosx(2sinx-cosx)$