Consider the signal where $\delta > 0$ and small,
$$y(x,t)=A\sin{(kx - \omega (k) t)} + A\sin{((k + \delta)x - \omega (k + \delta) t)},$$
show that this can be rewritten as
$$y(x,t)=2A \cos{\left ( \frac{\delta}{2} (x - \omega ' (k)t \right )} \sin{(kx - \omega (k) t)} + O(\delta).$$
So far I have used a Taylor expansion to obtain $\omega (k + \delta)$ in terms of $k$ and $\delta$ like so,
$$\omega (k + \delta) = \omega (k) + \delta \omega ' (k) + O(\delta^2),$$
do I need to then use $\sin u + \sin v = 2\sin(\frac{u+v}{2})\cos(\frac{u-v}{2})$ or do I need to substitute my Taylor expansion in and then use the trigonometric identity?
$$A\sin{((k + \delta)x + \omega (k + \delta) t)} = A\sin[(kx- \omega t) - \delta(kx-\omega t)]$$ Using $\sin(a+b) = \sin a \cos b + \cos a \sin b$, $$A\sin{((k + \delta)x - \omega (k + \delta) t)} = A\sin(kx- \omega t)\cos(\delta[kx- \omega t]) + A\cos(kx- \omega t)\sin(\delta[kx- \omega t])$$ which when replaced in your main equation $$y(x,t)=A\sin{(kx - \omega (k) t)} + A\sin{((k + \delta)x - \omega (k + \delta) t)}$$ will give $$y(x,t)=A\Big(1+\cos(\delta[kx- \omega t]) \Big)\sin{(kx - \omega (k) t)} + A\cos(kx- \omega t)\sin(\delta[kx- \omega t])$$ Using $2\cos^2 x = 1 + \cos 2x$ we get $$y(x,t)=2A\Big(\cos^2(\frac{\delta}{2}[kx- \omega t]) \Big)\sin{(kx - \omega (k) t)} + A\cos(kx- \omega t)\sin(\delta[kx- \omega t])$$ The term $A\cos(kx- \omega t)\sin(\delta[kx- \omega t])$ is $O(\delta)$ because $\sin x \sim x$ for small $x$, whereas the only difference between my exact first term and your approximated second term is the squaring of cosine. But with different frequencies.