Question

cos^2 din 120+sin^2 din 60

Answer 1

$\cos120^0 =\cos(180^0 -60^0)= - \cos 60^0 =\dfrac{1}{2} \\\;\\ \sin 60^0 = \dfrac{\sqrt3}{2}$

Expresia din enunt devine:


$\left(-\dfrac{1}{2}\right)^2+ \left( \dfrac{\sqrt3}{2}\right)^2 =\dfrac{1}{4}+\dfrac{3}{4} =1.$

Answer 2

   
[tex]\texttt{Folosim formula: } \boxed{sin^2x + cos^2x =1} \\ \\ cos^2 120^o + sin^2 60^o = \\ =cos^2 120^o + sin^2 (180^o-60^o) = \\ =cos^2 120^o + sin^2 (120^o) = \\ =sin^2 120^o + cos^2 (120^o) = \boxed{1}[/tex]