Zad. 1 [latex]sin alpha + cos alpha = frac{5}{4} /^2 \ (sin alpha + cos alpha)^2 = (frac{5}{4})^2 \ sin^2 alpha + 2sin alpha cdot cos alpha + cos^2 alpha = frac{25}{16} \ 1+ 2sin alpha cdot cos alpha = frac{25}{16} \ 2sin alpha cdot cos alpha = frac{25}{16} - 1\ 2sin alpha cdot cos alpha = frac{9}{16}[/latex] [latex](sin alpha - cos alpha)^2 = sin^2 alpha - 2sin alpha cdot cos alpha + cos^2 alpha =\ = 1- 2sin alpha cdot cos alpha= 1-frac{9}{16} = frac{7}{16}[/latex] [latex](sin alpha - cos alpha)^2 = frac{7}{16}[/latex] Zad. 2 [latex]sin alpha + cos alpha = frac{1}{sqrt{2}} /^2 \ (sin alpha + cos alpha)^2 = (frac{1}{sqrt{2}})^2 \ sin^2 alpha + 2sin alpha cdot cos alpha + cos^2 alpha = frac{1}{2} \ 1+ 2sin alpha cdot cos alpha = frac{1}{2} \ 2sin alpha cdot cos alpha = frac{1}{2} - 1\ 2sin alpha cdot cos alpha = - frac{1}{2}[/latex] [latex]|sin alpha - cos alpha| = sqrt{(sin alpha - cos alpha)^2} =\ = sqrt{sin^2 alpha - 2sin alpha cdot cos alpha + cos^2 alpha} =\ = sqrt{1- 2sin alpha cdot cos alpha} = sqrt{1- (- frac{1}{2})} = \ = sqrt{1+ frac{1}{2}} = sqrt{1frac{1}{2}} = sqrt{frac{3}{2}} = frac{sqrt{3}}{sqrt{2}} = \ = frac{sqrt{3} cdot sqrt{2}}{sqrt{2} cdot sqrt{2}} = frac{sqrt{6}}{2} \\ |sin alpha - cos alpha|= frac{sqrt{6}}{2}[/latex]
1. Kąt alfa jest kątem ostrym oraz sin alfa + cos alfa = 5/4 oblicz: a) (sin alfa - cos alfa)² 2. Wiedząc, że sin alfa + cos alfa = 1/pierwiastek2 oblicz: a) |sin alfa - cos alfa|
Odpowiedź
Dodaj swoją odpowiedź