[latex]tg alpha = frac{sin alpha }{cos alpha } = frac{2 sqrt{5} }{5} - extgreater cos alpha = frac{5}{2 sqrt{5} } sin alpha \ sin^2 alpha +cos^2 alpha =1 \ sin^2 alpha + frac{25}{20} sin^2 alpha =1 \ frac{9}{4} sin^2 alpha =1 \ sin alpha = frac{2}{3} [/latex] sin a = - 2/3 odpada, bo a jest kątem ostrym a) [latex] frac{6}{sinalpha } - 8frac{cos^2 alpha }{sin^2 alpha } = frac{6}{sin alpha } - frac{8}{tg^2 alpha } = frac{6}{ frac{2}{3} } - frac{8}{ frac{20}{25} } =9-10=-1[/latex] b) [latex] frac{(1-sin^2 alpha)^2}{1-sin^2 alpha } =1-sin^2 alpha =1- frac{4}{9} = frac{5}{9} [/latex]
[latex]a) frac{6sin alpha -8cos^2alpha}{sin^2alpha} = frac{6sinalpha}{sin^2alpha}-frac{8cos^2alpha}{sin^2alpha } = 6frac{1}{sinalpha}-8ctg^2alpha = 6frac1{sinalpha}-8frac{1}{tg^2alpha}[/latex] Musimy wyznaczyć sinα. [latex]tg alpha =frac{sin alpha }{cos alpha }[/latex] [latex]sin^2 alpha + cos^2 alpha = 1\cos^2 alpha = 1-sin^2 alpha \cos alpha = sqrt{1-sin^2 alpha } vee cos alpha = -sqrt{1-sin^2 alpha [/latex] Ponieważ zaznaczono, że kąt alfa jest ostry, więc wszystkie funkcje tryg. tego kąta będzie dodatnie. [latex]tg alpha =frac{sinalpha}{sqrt{1-sin^2alpha}} = frac{2sqrt5}{5}\\5sinalpha = 2sqrt5*sqrt{1-sin^2alpha}} |^2\25sin^2alpha = 4*5*(1-sin^2alpha)\25sin^2alpha=20-20sin^2alpha\45sin^2alpha = 20 |:45\sin^2alpha = frac{20}{45} = frac49 |sqrt{}\sinalpha = frac23[/latex] [latex]6frac1{sinalpha}-8frac{1}{tg^2alpha} = 6*frac{1}{frac23}-8frac1{(frac{2sqrt5}{5})^2} = 6*frac32-8*(frac5{2sqrt5})^2 = 9-8*frac{25}{4*5} = \\=9-8*frac5{4} = 9-2*5 = -1[/latex] b) [latex]frac{sin^4 alpha -2sin^2alpha+1}{1-sin^2alpha} = frac{(sin^2alpha-1)^2}{-(sin^2alpha-1)} = -(sin^2alpha-1)[/latex] Już wyliczyliśmy, że: [latex]sin alpha = frac23[/latex] więc: [latex]-(sin^2 alpha -1) = -((frac23)^2-1) = -(frac49-1) = -(-frac59) = frac59[/latex]
