SHM question help pls asap

The cos() and sin() curves are related by a shift of pi/2, so if you set alpha appropriately, theyre the same, so either is fine.

Okay so i wont loose mark if i did Rsin(wt)?

It would depend on the initial conditions. The shift of alpha means you can vary location of the sinusoidal solution, independent of whether you write
Rcos(wt+alpha)
or
Rsin(wt+alpha)
If you said the solution was Rsin(wt) then that would assume the displacement at time 0 was 0, so the "point" would be passing through the equilibrium position with a positive velocity.

Wdum by the intial conditions?
In an exam how would i know whether to re-write it as a function of sin or a function of cos?

You can write either
Rcos(wt+alpha)
Rsin(wt+alpha)
Theyre equivalent (subject to alpha being set appropriately).

The original solution was
Acos(wt) + Bsin(wt)
Here you have two constants A and B which are determined by the value of the displacement and velocity at time 0 (usually). You have a second order differential equation, you need two initial conditions (x(0)=..., x'(0)=...) to get a particular solution where the values of A and B are determined. The above representations are equivalent, but the two constants to determine are R and alpha.

This must be covered in your textbook in this section?

I dont think its covered in my textbook
Yeah i just came across another example and here they use cos instead!
So how do i know whether to use cos or sin, sorry im still slightly confused.

How did they go from there to there in one step? How did they know not to use the cos function if that makes sense?

I can guarantee that just about all of this will be covered in your textbook.

Third time, theyre equivalent, you can use either. However
Rcos(wt+alpha)
is probably more common as the intiial coditions are usually that initially the velocity is 0 and the particle has a maximum displacement. This is what a basic cos() curve represents.

okay thanks, sorry was just making sure because i have an exam today.