Umm.. don't you use auxiliary method?
So let's say we have 3sinx + 4cosx. We need to put it in the form Asin(x+a)?
Well, 3sinx + 4cosx = Asin(x+a)
Asin(x+a) = A[sinxcosa + sinacosx]
=Asinxcosa + Asinacosx
You match the coefficients. So 3sinx matches with Asinxcosa, and 4cosx matches with Asinacosx.
3 is the coefficient of sinx, and so is Acosa. So Acosa=3
4 is the coefficient of cosx, and so is Asina. So Asina=4
You cancel out the A's and divide sina/cosa to get tana. So tana=4/3. Therefore a=53 degrees.
So you have Asin(x+53)
To find A you just root the sum of the square coefficients of the first equation (3sinx + 4cosx) so root(3^2+4^2) = 5. So A=5.
Therefore 3sinx + 4cosx = 5sin(x+53).
Sorry if it didn't make sense, best way i could explain.