1. sinα = √3/4 tgα = ? Korzystamy z jedynki trygonometrycznej: sin²α+cos²α = 1 cos²α = 1-sin²α = 1-(√3/4)² = 16/16 -3/16 = 13/16 cosα = √13/4 tgα = sinα/cosα = √3/4 : √13/4 = √3/4 * 4/√13 = √3/√13 * √13/√13 = √39/13 2. sinα = √2/5 tgα = ? cos²α = 1-sin²α = 1-(√2/5)² = 25/25 - 2/25 = 23/25 cosα = √23/5 tgα = sinα/cosα = √2/5 : √23/5 = √2/√23 * √23/√23 = √46/23
[latex]1.\ sinalpha = frac{sqrt3}{4}\ sin^2alpha+cos^2alpha = 1\ cos^2alpha = 1-sin^2alpha \ cos^2alpha =1-(frac{sqrt3}{4})^2\ cos^2alpha= 1-frac{3}{16}\ cos^2alpha=frac{13}{16}\ cosalpha=frac{sqrt{13}}{4}\ tgalpha= sinalpha/cosalpha \ tgalpha=frac{sqrt3}{4} : frac{sqrt{13}}{4} \ tgalpha= frac{sqrt3}{4} * frac{4}{sqrt13} \ tgalpha=frac{3*sqrt{13}}{13}=frac{sqrt39}{13} [/latex] [latex]2.\ sinalpha = frac{sqrt2}{5}\ sin^2alpha+cos^2alpha=1\ cos^2alpha=1-sin^2alpha\ cos^2alpha= 1-(frac{sqrt2}{5})^2\ cos^2alpha = 1 - frac{2}{25}\ cos^2alpha=frac{23}{25}\ cosalpha=frac{sqrt{23}}{5}\ tgalpha= frac{sinalpha}{cosalpha}\ tgalpha=frac{sqrt2}{5}: frac{sqrt{23}}{5}\ tgalpha=frac{sqrt2}{5}* frac{5}{sqrt{23}}\ tgalpha=frac{sqrt2*sqrt{23}}{23}=frac{sqrt{46}}{23}[/latex] licze na naj
