The use of phasors can help in such a case?
In the diagram the reference phasor $WX$ represents $A\sin(kx-\omega t)$ with amplitude $A$ at some position $x$ and phasor $WY$ represents $A\sin(kx-\omega t+\phi)$ at the same position with the same amplitude $A$ and leading $A\sin(kx-\omega t)$ by $\phi$.
The diagram shows immediately that the resultant phasor $WY$ leads the reference phasor by $\frac \phi 2$ and so is of the form $\sin(kx-\omega t+\frac \phi 2)$
The amplitude of the resultant phasor is the length of $WY$ which can be found by application of the cosine rule on triangle $WXY$
$$WY^2 = A^2 + A^2 - 2 \,A\,A\cos (\pi - \phi) = A^2 \,2(1-\cos^2 \phi)= 4A^2\cos^2 \left (\frac \phi 2 \right)$$
$$\Rightarrow WY = 2A\cos \left (\frac \phi 2 \right)$$
The resultant phasor has an amplitude of $2A\cos \left (\frac \phi 2 \right)$ so can be written as $$2A\cos \left (\frac \phi 2 \right)\sin(kx-\omega t+\frac \phi 2)$$
Response to a question asked by the OP.
The standing wave can be dealt with in this way.
The $A\sin(kx +\omega t)$ variation can be thought of as a $A\sin(\omega t + \phi)$ variation at a position $x$ where $\phi = kx$.
This can be drawn as a phasor which is $\phi$ in advance of the reference phasor $A\sin{\omega t}$ phasor.
The $A\sin(kx - \omega t)$ variation at the same position $x$ is a little trickier to deal with in the following way.
$A\sin(kx - \omega t) = - A \sin(\omega t - kx) = A \sin(\omega t - kx - \pi) = A \sin(\omega t - (\phi + \pi))$
This is a phasor which lags behind the reference phasor $A\sin \omega t$ by $\phi + \pi$
So as you move position $x$ the phase $\phi$ changes to give a different but constant amplitude for the resultant phasor.
When $x=0 \Rightarrow \phi =0$ the resultant is $0$ and when $x= \frac{\pi}{2k} \left (= \frac{\lambda}{4}\right )$ the resultant is $2A$.
The resultant phasor is $\frac \pi 2$ ahead of the reference $\sin\omega t$ phasor ie it is $\cos \omega t$ and has an amplitude of $2A\sin \phi \Rightarrow 2A\sin\phi\cos \omega t$