E) sin 210 = sin(180+30) = sin 180 * cos 30 + sin 30 * cos 180 = 0 - [tex] \frac{1}{2} [/tex] = [tex] \frac{-1}{2} [/tex]
F) sin 105 = sin(45+60) = sin(45)*cos(60) + sin(60)*cos(45) = [tex] \frac{ \sqrt{2} }{2} * \frac{1}{2} + \frac{ \sqrt{3} }{2}* \frac{ \sqrt{2} }{2} = \frac{ \sqrt{2} +\sqrt{6}}{2} [/tex]
H) sin (– π/2)-sin 3π/2 = -1 - (-1) = -1+1=0
I) (din formula cos(180-x) = +sin(x)) => sin(x) + cos(x)
mai mult habar nu am :)) dar cred ca ar trebui sa se rezolve asemanator
