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Stiind ca sinx+cosy=1/3 si siny+cosx=4/3,calculati sin(x+y)

Răspuns :

Ridicam la patrat ambele expresii
[tex](\sin{x}+\cos{y})^{2}=(\frac{1}{3})^{2}\Rightarrow \sin^{2}{x}+\cos^{2}{y}+2\sin{x}\cos{y}=\frac{1}{9}[/tex]
[tex](\sin{y}+\cos{x})^{2}=(\frac{4}{3})^{2}\Rightarrow \sin^{2}{y}+\cos^{2}{x}+2\sin{y}\cos{x}=\frac{16}{9}[/tex]
Stim ca formula de baza a trigonometriei este
[tex]\sin^{2}{x}+\cos^{2}{x}=1[/tex]
Si mai stim ca
[tex]\sin{(x+y)}=\sin{x}\cos{y}+\sin{y}\cos{x}[/tex]
Atunci adunam cele doua formule la patrat
[tex]\sin^{2}{x}+\cos^{2}{x}+\cos^{2}{y}+\sin^{2}{y}+2(\sin{x}\cos{y}+\sin{y}\cos{x})=1+1+2\sin{(x+y)}=\frac{1}{9}+\frac{16}{9}=\frac{17}{9}\Rightarrow 2\sin{(x+y)}=\frac{17}{9}-2=\frac{17-18}{9}=-\frac{1}{9}\Rightarrow \sin{(x+y)}=-\frac{1}{18}[/tex]