zad 1 tgα=sinα/cosα {sin²α+cos²α=1 {sinα/cosα=√2 --- {sin²α+cos²α=1 {sinα=√2cosα --- {(√2cosα)²+cos²α=1 {sinα=√2cosα --- I. {2cos²α+cos²α=1 II.{sinα=√2cosα ---------------------- [zajmuję się równaniem I] [latex]3cos^{2}alpha-1=0\ \ 3(os^{2}alpha-frac{1}{3})=0\ \ 3(cosalpha-frac{sqrt{3}}{3})(cosalpha+frac{sqrt{3}}{3})=0\ \ cosalpha=frac{sqrt{3}}{3} vee cosalpha=-frac{sqrt{3}}{3}[/latex] ---------------------- {cos₁α=√3/3 v cos₂α=-√3/3 {sin₁α=√6/3 v sin₂α=-√6/3 Wartość wyrażenia dla cos₁α i sin₁α: [latex]frac{cosalpha+sinalpha}{sinalpha-cosalpha}= frac{frac{sqrt{3}+sqrt{6}}{3}}{frac{sqrt{6}-sqrt{3}}{3}}=\ \ =frac{sqrt{3}+sqrt{6}}{3}*frac{3}{sqrt{6}-sqrt{3}} =frac{(sqrt{3}+sqrt{6})(sqrt{6}+sqrt{3})}{sqrt{6}^{2}-sqrt{3}^{2}}=\ \ =frac{3+2sqrt{18}+6}{6-3}=frac{9+2sqrt{9*2}}{3}=frac{9+6sqrt{2}}{3}=3+2sqrt{2}[/latex] Wartość wyrażenia dla cos₂α i sin₂α: [latex]frac{cosalpha+sinalpha}{sinalpha-cosalpha}= frac{frac{-sqrt{3}+-sqrt{6}}{3}}{frac{-sqrt{6}+sqrt{3}}{3}}=\ \ =-frac{sqrt{3}+sqrt{6}}{3}*frac{3}{sqrt{3}-sqrt{6}} =-frac{(sqrt{3}+sqrt{6})(sqrt{6}+sqrt{3})}{sqrt{3}^{2}-sqrt{6}^{2}}=\ \ =-frac{3+2sqrt{18}+6}{3-6}=-frac{9+2sqrt{9*2}}{-3}=frac{9+6sqrt{2}}{3}=3+2sqrt{2}[/latex] [Wartość wyrażenia dla cos₁α i sin₁α=Wartość wyrażenia dla cos₂α i sin₂α] ========================================================== [latex]sinalpha-cosalpha=frac{1}{4} |^{2}\ \ (sinalpha-cosalpha)^{2}=(frac{1}{4})^{2}\ \ sin^{2}alpha-2sinalpha*cosalpha+cos^{2}alpha=frac{1}{16}\ \ 1-2sinalpha*cosalpha=frac{1}{16}\ \ -2sinalpha*cosalpha=frac{1}{16}-1\ \ -2sinalpha*cosalpha=-frac{15}{16} |*(-frac{1}{2})\ \ sinalpha*cosalpha=frac{15}{32}[/latex]
