Sprawdź tożsamość: (sin α + 2cos α )² + (2sin α - cos α)² =4

w załączniku rozwiązanie

[latex](sinalpha+2cosalpha)^{2}+(2sinalpha - cosalpha)^2} = 4\\L = sin^{2}alpha + 4sinalphacosalpha+4sinalphacosalpha+4sin^{2}alpha - 4sinalphacosalpha+cosalpha=\\=sin^{2}alpha+cos^{2}alpha+4sin^{2}alpha+4cos^{2}alpha =\\=5(sin^{2}alpha+cos^{2}alpha) =5\\P = 4\\L e P[/latex] (sin²α + cos²α = 1)

Sprawdź tożsamość(sin[latex] alpha [/latex]+2cos[latex] alpha [/latex])²+(2sin[latex] alpha [/latex]-cos[latex] alpha [/latex]²=4

Sprawdź tożsamość(sin[latex] alpha [/latex]+2cos[latex] alpha [/latex])²+(2sin[latex] alpha [/latex]-cos[latex] alpha [/latex]²=4...