How to transform equation $(1)$ to equation $(2)$?
$$u(r,t)= -\frac{Ak}{r} \left[ \frac{1}{kr} \cos k(ct-r)-\sin k(ct-r) \right] \tag{1}$$
$$u(r,t)=\frac{A}{r} \sqrt{\frac{1}{r^2}+k^2} \sin[k(ct-r)-\theta] \tag{2}$$ where $\tan \theta=\dfrac{1}{kr}$
Original images: equation $(1)$ and equation $(2)$
We can write $$-\frac{Ak}{r}[\frac{1}{kr}\cos k(ct-r) -\sin k(ct-r)]$$ $$\Rightarrow -\frac{A}{r^2}[-\sin k(ct-r)kr +\cos k(ct-r)]$$ $$\Rightarrow -\frac{A\sqrt{k^2r^2+1}}{r^2}[-\sin k(ct-r)\frac{kr}{\sqrt{k^2r^2+1}} + \cos k(ct-r)\frac{1}{\sqrt{k^2r^2+1}}]$$
We know that in a triangle with $\tan \theta = \frac{1}{kr}$, $\sin \theta =\frac{1}{\sqrt{k^2r^2+1}} , \cos \theta = \frac{kr}{\sqrt{k^2r^2+1}}$. Hence, our equation simplifies to, $$\Rightarrow -\frac{A\sqrt{k^2r^2+1}}{r^2}[-\sin k(ct-r)\cos \theta +\cos k(ct-r)\sin \theta]$$ which upon rearrangement gives us, $$\frac{A}{r}\sqrt{\frac{1}{r^2}+k^2}\sin [k(ct-r)-\theta]$$ Hope it helps.