Need this for a presentation please help

As suggested by Kav10, you need to use Cosine law for triangles ABC and ACE, and that that AC=CE. 

Using Cosine law on the trianles ABC we have
\[AC^2=AB^2+BC^2-2AB \cdot BC \cos B=4^2+4^2-2\cdot 4\cdot 4 \cos B.\]
Hence
\[AC^2=32(1-\cos B).             (1)\]

Using Cosine law on the trianles ACE we have
\[AE^2=AC^2+CE^2-2AC \cdot CE\cos (ACE),\]
since $AC=CE$ we get 
\[1^2=AC^2+AC^2-2AC^2 \cos (ACE)=2AC^2(1-\cos (ACE)),\]
or
\[\frac{1}{2AC^2}=(1-\cos (ACE)).             (2)\]
Multiplying (1) and (2) we get 
\[\frac{1}{2}=32(1-\cos B)(1-\cos (ACE)).\]
Hence 
\[(1-\cos B)(1-\cos (ACE))=\frac{1}{64}\]