3sin(a-b)=sinb thì tan(a-b)=sina/(3+cosa)? 01/09/2021 Bởi Ariana 3sin(a-b)=sinb thì tan(a-b)=sina/(3+cosa)?
$\begin{array}{l} 3\sin \left( {a – b} \right) = \sin b\\ \Leftrightarrow 3\sin \left( {a – b} \right) = \sin \left[ {a – \left( {a – b} \right)} \right]\\ \Leftrightarrow 3\sin \left( {a – b} \right) = \sin a\cos \left( {a – b} \right) – \sin (a – b)\cos a\\ \Leftrightarrow \sin \left( {a – b} \right)\left( {3 + \cos a} \right) = \sin a\cos \left( {a – b} \right)\\ \Leftrightarrow \dfrac{{\sin \left( {a – b} \right)}}{{\cos \left( {a – b} \right)}} = \dfrac{{\sin a}}{{3 + \cos a}} \Leftrightarrow \tan \left( {a – b} \right) = \dfrac{{\sin a}}{{3 + \cos a}} \end{array}$ Bình luận
$\begin{array}{l} 3\sin \left( {a – b} \right) = \sin b\\ \Leftrightarrow 3\sin \left( {a – b} \right) = \sin \left[ {a – \left( {a – b} \right)} \right]\\ \Leftrightarrow 3\sin \left( {a – b} \right) = \sin a\cos \left( {a – b} \right) – \sin (a – b)\cos a\\ \Leftrightarrow \sin \left( {a – b} \right)\left( {3 + \cos a} \right) = \sin a\cos \left( {a – b} \right)\\ \Leftrightarrow \dfrac{{\sin \left( {a – b} \right)}}{{\cos \left( {a – b} \right)}} = \dfrac{{\sin a}}{{3 + \cos a}} \Leftrightarrow \tan \left( {a – b} \right) = \dfrac{{\sin a}}{{3 + \cos a}} \end{array}$