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.
 A: $$\sin { \left( A-\arcsin { \left( 0.3\sin { \left( A \right)  }  \right)  }  \right)  } \cdot \sin { \left( A+\arcsin { \left( 0.3\sin { \left( A \right)  }  \right)  }  \right) =0.91\sin ^{ 2 }{ \left( A \right)  }  }$$
Solution
 :
$$\left( \sin { A\cos { \left( \arcsin { \left( 0.3\sin { \left( A \right)  }  \right)  }  \right)  } -\cos { A } \sin { \left( \arcsin { \left( 0.3\sin { \left( A \right)  }  \right)  }  \right)  }  }  \right) \ast \left( \sin { A\cos { \left( \arcsin { \left( 0.3\sin { \left( A \right)  }  \right)  }  \right)  } +\cos { A } \sin { \left( \arcsin { \left( 0.3\sin { \left( A \right)  }  \right)  }  \right)  }  }  \right) =0.91\sin ^{ 2 }{ A } \\ \left( \sin { A\sqrt { 1-0.09\sin ^{ 2 }{ A }  } -0.3\cos { A } \sin { A }  }  \right) \ast \left( \sin { A\sqrt { 1-0.09\sin ^{ 2 }{ A }  } +0.3\cos { A } \sin { A }  }  \right) =\sin ^{ 2 }{ A\left( 1-0.09\sin ^{ 2 }{ A }  \right) -0.09\cos ^{ 2 }{ A } \sin ^{ 2 }{ A } = } \\ =\sin ^{ 2 }{ A } \left( 1-0.09\sin ^{ 2 }{ A } -0.09\cos ^{ 2 }{ A }  \right) =\sin ^{ 2 }{ A } \left( 1-0.09\left( \sin ^{ 2 }{ A } +\cos ^{ 2 }{ A }  \right)  \right) =\sin ^{ 2 }{ A\left( 1-0.09 \right) =091\sin ^{ 2 }{ A }  } $$
A: [I thought of another way since I left my comment above.]
Applying the "product-to-sum" formula for two sine factors,
$$ \sin \ \alpha  \ \cdot \ \sin \ \beta \ = \ \frac{1}{2} \ [ \ \cos(\alpha \ - \ \beta) \ - \ \cos(\alpha \ + \ \beta) \ ] \ \ , $$
with $ \ \alpha \ \ \text{and} \ \ \beta \ $ being sums and differences of angles, we can write
$$ \sin (\theta \ + \ \phi)  \ \cdot \ \sin \ (\theta \ - \ \phi) \ = \ \frac{1}{2} \ [ \ \cos \ 2 \phi \ - \ \cos \ 2 \theta \ ] \ \ . $$
So the left-hand side of your equation becomes
$$ \sin(A-\arcsin(0.3 \ \sin \ A)) \ \cdot \ \sin(A+\arcsin(0.3 \ \sin \ A)) \ = \ \frac{1}{2} \ [ \ \cos  ( 2 \ \cdot \arcsin \ [0.3 \ \sin \ A]) \ - \ \cos \ 2 A \ ] \ \ . $$
We're looking ultimately for an expression which has a factor of $ \ \sin^2  A \ $ , so we'll use the "double-angle" formula for cosine, $ \ \cos \ 2 \theta \ = \ \cos^2 \ \theta \ - \ \sin^2 \ \theta \ = \ 1 \ - \ 2 \ \sin^2 \ \theta \ $ . The left-hand side becomes
$$ = \ \frac{1}{2} \ [ \   ( 1 \ - \ 2 \ \sin^2 ( \ \arcsin \ [0.3 \ \sin \ A] \ ) \ ) \ - \ ( 1  \ - \ 2 \ \sin^2  A ) \ ]  $$
$$ = \ \frac{1}{2} \ [ \   2 \ \sin^2  A \  - \ 2 \ \sin^2 ( \ \arcsin \ [0.3 \ \sin \ A] \ ) \ ] \ = \     \sin^2  A \  - \   \sin^2 ( \ \arcsin \ [0.3 \ \sin \ A]  \ ) \ \ . $$
Now we need to consider the angle in the second term.  [Since arcsine has as its domain $ \ -\frac{\pi}{2} \ \le \ \theta \ \le \ \frac{\pi}{2} \ $ , we would have to deal with negative as well as positive angles.  But we can restrict the discussion to positive angles, since the argument for negative angles is similar.]  Suppose angle $ \ A \ $ has a sine value of $ \frac{y}{1} \ $ ; then the angle in the second term has a sine value of $ \ 0.3 \ y \ $ .  That angle has measure $ \ \arcsin \ (0.3 \ y ) \ $ , but we don't care what that number is, since we are going to take the sine of that angle and square it.  [You don't indicate at level you're familiar with working with trigonometric functions; if you are used to the inverse functions, this is something you already know and I apologize for the explanantion.]
Hence, the expression reduces to
$$  \sin^2  A \  - \   ( \ 0.3 \ \sin \ A \ )^2  \ = \ \sin^2  A \  - \    0.09 \ \sin^2 \ A  \  = \ 0.91 \ \sin^2 \ A \  \ . $$
