How does one correctly write an equation of a wave? Below is the problem statement:
A transverse wave travels along the line with a speed $20 \mathrm{m/s}$. Two points which are located on the distance $12\mathrm{meters}$ and $15\mathrm{meters}$ from the source of oscillations are oscillating harmonically with the amplitudes $0.1\mathrm{meters}$, and the phase difference is $135^\circ$. Write the wave equation.
Now these are the only given information (there is no picture or anything to the problem), and problem also asked about other stuff like finding the wavelength, etc. which aren't the issue here. The issue is how to write the equation in the first place. First of all, I know the equation for the displacement of this line is going to look something like this:
$$x(t)=\text{Amplitude}\cdot \text{trigonometric function}(wt+\phi)$$
But my question is:


*

*Does it matter which trigonometric function I use for this? If so, why does it matter, and if not, why it doesn't matter?

*I've looked up about the phase difference ($\phi$), and in many places it stated that, the value of phase difference in the equation represents how much the wave is offset from the point of reference. So, in this example, can I take one of the points of oscillation (For example, the one that is $12\mathrm{meters}$ from the source) as the point of reference and write its equation with condition $\phi = 0$, and for the other one write the equation with condition $\phi = 135^\circ$? 
 A: To answer your first question, $cos$ and $sin$ are the same function but shifted by $\frac{\pi}{2}$ (the cosine is shifted to the left respect to sine)
$$cos(x-\frac{\pi}{2}) = sin(x)$$
Before answering your second question, remember a wave function is a function of time and position. Not position in function of time. It looks like this: 
$$ \Psi (x,t) = A*cos(\omega t - k x + \phi) = A * cos(\Phi) $$
where $\Psi$ is the displacement from the equilibrium (where a point $x$ would be if there was no wave) and $\Phi$ is the phase ($\phi$ is a global phase).
So, the phase diference betewen the 2 points would be $\Delta\Phi  = (\omega t - k*12m + \phi) - (\omega t - k*15m + \phi) = k*3m$ 
The phase diference is what matters. $\phi$ is canceled as all the points have the same global phase. 
Global phase only shifts the origin (of time or space). As you said, you could take as the origin of space the point that is at 12 meters from the source, it wont matter because the phase difference keeps being $k*3m$.
Observation: $\omega t$, $k x$ and $\phi$ are in radians (not in degrees as you said).
