Răspuns:
[tex]cos(a - b) = \frac{-3}{8}[/tex]
Explicație pas cu pas:
Datele problemei:
Fie a, b ∈R astfel incat sin(a) + sin(b) = 1 si cos(a) + cos(b) = 1/2 .
Ce se cere?
Sa se afle cos(a-b).
Formule utilizate:
Formula fundamentala a trigonometriei:
[tex]sin^2a + cos^2a = 1[/tex]
Formula pentru cos(a - b):
[tex]cos(a - b) = cos(a)cos(b) + sin(a)sin(b)[/tex]
Folosim relatiile date in enunt:
sin(a) + sin(b) = 1 [tex]| ( \:) ^2[/tex] => [tex]sin^2a+sin^2b + 2sin(a)sin(b) = 1\\[/tex] (relatia 1)
cos(a) + cos(b) = 1/2 [tex]| (\:)^2[/tex] => [tex]cos^2a+cos^2b + 2cos(a)cos(b) = \frac{1}{4}[/tex] (relatia 2)
Adunam relatiile 1 si 2 si obtinem:
[tex]sin^2a+sin^2b + 2sin(a)sin(b) + cos^2a+cos^2b + 2cos(a)cos(b) = 1 + \frac{1}{4} \\\\Grupam \: convenabil\: termenii \: si \: rescriem[/tex]
[tex]sin^2a+cos^2a+sin^2b + cos^2b + 2sin(a)sin(b)+2cos(a)cos(b) = \frac{5}{4} \\\\ Cum \:\: sin^2a+cos^2a = 1 \: \: si\:\:sin^2b + cos^2b = 1, \: avem\: ca:\\ \\ 1 +1 + 2sin(a)sin(b)+2cos(a)cos(b) = \frac{5}{4} \\ \\ 2sin(a)sin(b)+2cos(a)cos(b) = \frac{5}{4} - 2= \frac{-3}{4} \\\\\\ Noi\: vrem \:sa\: calculam :sin(a)sin(b)+cos(a)cos(b) \\ \:\: deci\: vom\: imparti \:prin\: 2\: relatia \:de\: mai\: sus.\\sin(a)sin(b)+cos(a)cos(b) = \frac{-3}{4 * 2} = \frac{-3}{8} \\ \\ \\ => cos(a - b) = \frac{-3}{8}[/tex]
