Đáp án:
\[\cos \left( {x + y} \right) = - 1\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
\sin x + \sin y = \sqrt 3 \\
\cos x - \cos y = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{\left( {\sin x + \sin y} \right)^2} = 3\\
{\left( {\cos x - \cos y} \right)^2} = 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\sin ^2}x + 2\sin x.\sin y + {\sin ^2}y = 3\\
{\cos ^2}x - 2\cos x.\cos y + {\cos ^2}y = 1
\end{array} \right.\\
\Rightarrow \left( {{{\sin }^2}x + 2\sin x.\sin y + {{\sin }^2}y} \right) + \left( {{{\cos }^2}x - 2\cos x.\cos y + {{\cos }^2}y} \right) = 3 + 1\\
\Leftrightarrow \left( {{{\sin }^2}x + {{\cos }^2}x} \right) + \left( {{{\sin }^2}y + {{\cos }^2}y} \right) + 2\left( {\sin x\sin y - \cos x\cos y} \right) = 4\\
\Leftrightarrow 1 + 1 + 2\left( {\sin x\sin y - \cos x.\cos y} \right) = 4\\
\Leftrightarrow \sin x\sin y - \cos x\cos y = 1\\
\cos \left( {x + y} \right) = \cos x.\cos y - \sin x.\sin y = - 1
\end{array}\)
Vậy \(\cos \left( {x + y} \right) = - 1\)
