The array-response of an antenna can be defined as:
$$A(Z) = \sum_{-N}^{N} w_n Z^n$$
where $Z = \exp(-i \omega \Delta t) = \exp(-ik\Delta x \sin \alpha)$
According to my textbook, if we let $w_n = 1$ for all $n$, the expression above can be written as:
$$A(Z) = \frac{\sin [(2N+1)k \Delta x \sin \alpha /2]}{(2N+1)\sin[k \Delta x \sin \alpha /2]}$$
where we have divded with $(2N+1)$ so that $A(1) = 1$
I tried to solve this algebraically to see how we get this expresison, but can't seem to figure it out. If we let $w_n = 1$ for all $n$, then we have:
$$A(Z) = \sum_{-N}^{N} Z^n$$
which is a geometric series which can be expressed as:
$$A(Z) = \frac{Z^{-N} - Z^{N+1}}{1 - Z}$$
However, I can't seem to get further than this. I tried plugging in $z = \exp(-ik\Delta x \sin \alpha)$ but only get a very complicated expression that I am unable to simplify. If anyone can give me any tips as to how to proceed with this derivation once I have obtained $A(Z) = \frac{Z^{-N} - Z^{N+1}}{1 - Z}$, then I would be very grateful!
If we have $w_n=1$, $\forall n$, you get
$$A(Z)=\sum_{n=-N}^N Z^n=z^{-N}\sum_{n=0}^{2N}Z^n=z^{-N}\frac{1-z^{2N+1}}{1-z}=\\ =\frac{z^{-(2N+1)/2}-z^{(2N+1)/2}}{z^{-1/2}-z^{1/2}}$$
If you subsitute $Z=\exp(-ik\Delta x\sin \alpha)$ and divide by $2N+1$ you finally get the expression given in your question.
(Of course you need to know that $2i\sin x = e^{ix}-e^{-ix}$).