### Instruction: $ABCDE$ is inscribed in a circle with $AB = BC = CD = DE = 4$ and $AE = 1.$  Compute $(1 - \cos \angle B)(1 - \cos \angle ACE).$.### Response: I notice that the problem involves cosines of angles, which makes me think of the law of cosines, a formula that relates the lengths of the sides and the angles of a triangle.
The law of cosines says that for any triangle $ABC$, $a^2 = b^2 + c^2 - 2bc \cos A$, where $a, b, c$ are the lengths of the sides opposite to the angles $A, B, C$, respectively.
I can use the law of cosines to find $\cos \angle B$ and $\cos \angle ACE$, since I know the lengths of the sides of the triangles $ABC$ and $ACE$.
For triangle $ABC$, I have $a = 4, b = 4, c = 4$, so I can plug these values into the law of cosines and get $4^2 = 4^2 + 4^2 - 2 \cdot 4 \cdot 4 \cos B$, which simplifies to $0 = -16 \cos B$.
This implies that $\cos B = 0$, and therefore $\angle B = 90^\circ$.
For triangle $ACE$, I have $a = 1, b = 4, c = 4$, so I can plug these values into the law of cosines and get $1^2 = 4^2 + 4^2 - 2 \cdot 4 \cdot 4 \cos ACE$, which simplifies to $1 = 32 - 32 \cos ACE$.
This implies that $\cos ACE = \frac{31}{32}$.
Now I have the values of $\cos \angle B$ and $\cos \angle ACE$, so I can plug them into the expression I need to compute.
I get $(1 - \cos \angle B)(1 - \cos \angle ACE) = (1 - 0)(1 - \frac{31}{32}) = \frac{1}{32}$.