Strange trigonometric proof. I was trying to find out how to prove 
$$ \sin(A-\arcsin(0.3 \ \sin \ A)) \ \cdot \ \sin(A+\arcsin(0.3 \ \sin \ A)) \ = \ 0.91 \ \sin^2 \ A \ \ . $$
When I put this equation into my calculator both sides appear to be exactly the same, but I have no idea how to prove it.

$$\sin { \left( A-\arcsin { \left( 0.3\sin { \left( A \right)  }  \right)  }  \right)  } \cdot \sin { \left( A+\arcsin { \left( 0.3\sin { \left( A \right)  }  \right)  }  \right) =0.91\sin ^{ 2 }{ \left( A \right)  }  }$$
Solution
 :
$$\left( \sin { A\cos { \left( \arcsin { \left( 0.3\sin { \left( A \right)  }  \right)  }  \right)  } -\cos { A } \sin { \left( \arcsin { \left( 0.3\sin { \left( A \right)  }  \right)  }  \right)  }  }  \right) \ast \left( \sin { A\cos { \left( \arcsin { \left( 0.3\sin { \left( A \right)  }  \right)  }  \right)  } +\cos { A } \sin { \left( \arcsin { \left( 0.3\sin { \left( A \right)  }  \right)  }  \right)  }  }  \right) =0.91\sin ^{ 2 }{ A } \\ \left( \sin { A\sqrt { 1-0.09\sin ^{ 2 }{ A }  } -0.3\cos { A } \sin { A }  }  \right) \ast \left( \sin { A\sqrt { 1-0.09\sin ^{ 2 }{ A }  } +0.3\cos { A } \sin { A }  }  \right) =\sin ^{ 2 }{ A\left( 1-0.09\sin ^{ 2 }{ A }  \right) -0.09\cos ^{ 2 }{ A } \sin ^{ 2 }{ A } = } \\ =\sin ^{ 2 }{ A } \left( 1-0.09\sin ^{ 2 }{ A } -0.09\cos ^{ 2 }{ A }  \right) =\sin ^{ 2 }{ A } \left( 1-0.09\left( \sin ^{ 2 }{ A } +\cos ^{ 2 }{ A }  \right)  \right) =\sin ^{ 2 }{ A\left( 1-0.09 \right) =091\sin ^{ 2 }{ A }  } $$
