sin120-cos30=? Ma wyjść sin

sin120⁰-cos30⁰=sin(180⁰-60⁰)-cos(90⁰-60⁰)=sin60⁰-sin60⁰=sin0⁰

[latex]sin120^{o}-cos60^{o} = sin(180^{o}-60^{o})-cos(90^{o}-60^{o}) =\\=sin60^{o}-sin60^o} = sin0^{o[/latex]