Matematyka
france612
2017-06-23 06:44:04
Wyznacz sin alfa,cos alfa,tg alfa gdy alfa należy(90*,180*) oraz: a)sin alfa=3/5 b)cos alfa=-1/3 c) tg alfa= -4/3
Odpowiedź
wojtek0000
2017-06-23 11:26:44

[latex]alphain(90^{circ},180^{circ})Rightarrow sinalpha>0landcosalpha<0land analpha<0[/latex] [latex]sinalpha=cfrac{3}{5}[/latex] [latex]sin^2alpha=cfrac{9}{25}[/latex] [latex]cos^2alpha=1-sin^2alpha=cfrac{16}{25}[/latex] [latex]cosalpha=-cfrac{4}{5}[/latex] [latex] analpha=cfrac{sinalpha}{cosalpha}=-cfrac{3}{4}[/latex] [latex]cosalpha=-cfrac{1}{3}[/latex] [latex]cos^2alpha=cfrac{1}{9}[/latex] [latex]sin^2alpha=1-cos^2alpha=cfrac{8}{9}[/latex] [latex]sinalpha=cfrac{sqrt{8}}{3}=cfrac{2sqrt{2}}{3}[/latex] [latex] analpha=cfrac{sinalpha}{cosalpha}=-2sqrt{2}[/latex] [latex] analpha=-cfrac{4}{3}[/latex] [latex]cfrac{sinalpha}{cosalpha}=-cfrac{4}{3}[/latex] [latex]cfrac{sin^2alpha}{cos^2alpha}=cfrac{16}{9}[/latex] [latex]sin^2alpha=cfrac{16}{9}cos^2alpha[/latex] [latex]sin^2alpha+cos^2alpha=1[/latex] [latex]cfrac{16}{9}cos^2alpha+cos^2alpha=1[/latex] [latex]cfrac{25}{9}cos^2alpha=1[/latex] [latex]cos^2alpha=cfrac{9}{25}[/latex] [latex]cosalpha=-cfrac{3}{5}[/latex] [latex]sin^2alpha=cfrac{16}{9}cdotcfrac{9}{25}=cfrac{16}{25}[/latex] [latex]sinalpha=cfrac{4}{5}[/latex]

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