Trigonometric Algebra

Our equation, $16\prod _{ k=1 }^{ 4 }{ \cos { { 2 }^{ k-1 } } }$ , is $16\cos { 1 } \cos { 2 } \cos { 4 } \cos { 8 }$ .

Let this be $S$ . Multiply both side by $\sin { 1 }$ . So that it would be $\sin { 1 } \cdot 16\cos { 1 } \cos { 2 } \cos { 4 } \cos { 8 } =S\sin { 1 }$ .

But we know that, $2\sin { C } \cos { C } =\sin { 2C }$ from basic trigonometric identity. Which means that $2\sin { 1 } \cos { 1 } =\sin { 2 }$ .

So, $\sin { 1 } \cdot 16\cos { 1 } \cos { 2 } \cos { 4 } \cos { 8 } =8\sin { 2 } \cos { 2 } \cos { 4 } \cos { 8 }$ .

But then again, $2\sin { 2 } \cos { 2 } =\sin { 4 }$ . Therefore

$8\sin { 2 } \cos { 2 } \cos { 4 } \cos { 8 } =4\sin { 4 } \cos { 4 } \cos { 8 } =2\sin { 8 } \cos { 8 } =\sin { 16 }$ .

So we conclude that $S\sin { 1 } =\sin { 16 }$ . Divide both sides by $\sin { 1 }$ , we get $S=\frac { \sin { 16 } }{ \sin { 1 } }$ .

But $\csc { C } =\frac { 1 }{ \sin { C } }$ . Therefore, $S=\sin { 16 } \csc { 1 }$ . From this, we evaluate A to be 16 and B to be 1. Answer is 16.