Câu hỏi 11

\({A^2} = A_1^2 + A_2^2 + 2.{A_1}.{A_2}.\cos \left( {\Delta \varphi } \right) = A_1^2 + 36 + 12{A_1}.\cos ({120^0}) \Rightarrow {A^2} = A_1^2 + 36 - 6{A_1} = {({A_1} - 3)^2} + 27\)

\(A \ge \sqrt {27}  \Rightarrow {A_{\min }} = \sqrt {27}  \Leftrightarrow {A_1} = 3cm\)

Khi đó :\(\tan \varphi  = {{{A_1}\sin {\varphi _1} + {A_2}.\sin {\varphi _2}} \over {{A_2}.\cos {\varphi _1} + {A_2}.\cos {\varphi _2}}} = {{3.\sin {\pi  \over 6} + 6.\sin {{ - \pi } \over 2}} \over {3.\cos {\pi  \over 6} + 6.\cos {{ - \pi } \over 2}}} =  - \sqrt 3  \Rightarrow \varphi  =  - {\pi  \over 3}\)