Question

ma poate ajuta cineva va rog cat de repede? multumesc :)​

Answer 1

$\displaystyle x, y\in\left(0, \frac{\pi}2\right)\Rightarrow \sin x>0,\sin y>0, \cos x>0, \cos y>0\\\sin^2x+\cos^2x=1\Rightarrow\cos^2x=1-\sin^2x=\\=1-\frac{169}{196}=\frac{27}{196}\Rightarrow|\cos x|=\sqrt\frac{27}{196}=\frac{3\sqrt3}{14}\\\text{Cum }\cos x>0\Rightarrow \cos x=\frac{3\sqrt3}{14}\\\\\sin^2y+\cos^2y=1\Rightarrow\sin^2y=1-\cos^2y=\\=1-\frac{48}{49}=\frac{1}{49}\Rightarrow|\sin y|=\frac{1}{7}\\\\\text{Cum } \sin y>0\Rightarrow \sin y =\frac{1}{7}$

$\displaystyle\sin(x-y)=\sin x\cos y-\cos x\sin y=\frac{13}{14}\cdot\frac{4\sqrt3}{7}-\frac{3\sqrt3}{14}\cdot\frac{1}{7}=\\=\frac{49\sqrt3}{2\cdot7^2}=\frac{\sqrt3}{2}$