Wiedząc, że jest kątem ostrym i tgalfa=3 oblicz wartość wyrażenia . Sin^3alfa/cos^2alfa

tgα=sinα/cosα=3 ⇒ sinα = 3cosα 1/3sinα=cosα sin²α +cos²α=1 sin²α+1/9sin²α = 1 10/9sin²α=1 sinα=3/√10=3√10/10 sin³α/cos²α = tg²α * sinα = 9*3√10/10 = 27√10/10

[latex]tg alpha =3\ frac{sin alpha }{cos alpha }=2\ sin alpha =3cos alpha \ \ sin^2 alpha +cos^2 alpha =1\ (3cos alpha )^2+cos^2 alpha =1\ 9cos^2 alpha +cos^2 alpha =1\ 10cos^2 alpha =1\ cos^2 alpha =frac{1}{10}\ cos alpha =frac{1}{sqrt{10}}\ \ sin alpha =3cos alpha =frac{3}{sqrt{10}}\ \ \ frac{sin^3 alpha }{cod^2 alpha }=frac{sin^2 alpha cdot sin alpha }{cos^2 alpha }=tg^2 alpha cdot sin alpha =3^2cdot frac{3}{sqrt{10}}=frac{27sqrt{10}}{10}[/latex]