Question

sa se stabileasca daca au loc egalitatile
a) sin (120°-x) = sin x + sin (60°-x)
b) cos x- cos (x+60°) + cos (x+120°)

Answer 1

Răspuns:

$a) sin(120-x)=sin(x)+sin(60-x)$

$=sin(120)*cos(-x)+sin(-x)*cos(120)=sin(x)+sin(60)*cos(-x)+sin(-x)*cos(60)$

$=\frac{\sqrt{3} }{2} *cos(x)-sin(x)*(-\frac{1}{2})=sin(x)+ \frac{\sqrt{3} }{2}*cos(x)-sin(x)*\frac{1}{2}$

$=\frac{\sqrt{3}*cosx+sinx }{2}=sinx+\frac{\sqrt{3}*cosx-sinx }{2}$ (Inmultim toata ecuatia cu 2)

$=\sqrt{3}*cosx+sinx=2sinx+\sqrt{3}*cosx-sinx\\=\sqrt{3} *cosx+sinx= \sqrt{3} *cosx+sinx=$

Egalitatea este adevarata.

$b) cos(x)-cos(x+60)+cos(x+120)=0$

$=cos(x)-(cos(x)*cos(60)-sin(x)*sin(60))+cos(x)*cos(120)-sin(x)*sin(120)$

$=cos(x)-cos(x)*\frac{1 }{2} +sin(x)*\frac{\sqrt{3} }{2} +cos(x)*(-\frac{1 }{2})-sin(x)*\frac{\sqrt{3} }{2}$

$=cos(x)+\frac{-cos(x)+sin(x)*\sqrt{3} }{2}+\frac{-cos(x)-sin(x)*\sqrt{3} }{2}$

$=cos(x)+(-\frac{2*cos(x)}{2})=cos(x)-cos(x)=0$

Egalitatea este adevarata.