Question

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

Answer 1

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