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Question

Find m if $$\sin { A } \sin { \left( { 60 }^{ o }-A \right)  } \sin { \left( { 60 }^{ o }+A \right)  } =\cfrac { 1 }{ m} \sin { 3A } \quad $$


Solution

$$LHS=\sin { A } \sin { (60-A) } \sin { (60+A) } \\ =\sin { A } [\sin { ^{ 2 }60 } -\sin { ^{ 2 }A } ]\\ =\sin { A } [\cfrac { 3 }{ 4 } -\sin { ^{ 2 }A } ]\\ =\cfrac { \sin { A }  }{ 4 } [3-4\sin { ^{ 2 }A } ]\\ =\cfrac { 3\sin { A } -4\sin { ^{ 3 }A }  }{ 4 } \\ =\cfrac { \sin { 3A }  }{ 4 } \\ =RHS$$

Mathematics

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