Question

Sa se arate ca sin 40 grade*sin 140 grade=cos^2 130 grade

Answer 1

[tex]sin140^0=sin(180^0-40^0)=sin40^0\\ sin40^0*sin140^0=sin^240^0(1)\\ cos130^0=cos(90^0+40^0)=cos90^0*cos40^0-sin90^0*sin40^0=\\ =-sin40^0\\ cos^2130^0=(-sin40^0)^2=sin^240^0(2)\\ Din~(1)+(2)=\ \textgreater \ sin40^0*sin140^0=cos^2130^0[/tex]