Instead of doing your problem I'll do one exactly in every detail
like your problem.  I do this to prevent students from ignoring
what we do on here and look only at the answer in attempt to fool
a teacher into thinking they did the problem themselves.  So instead
of f(t) = 8sint+6cost, I will do the problem for f(t) = 15sint+8cost.  

y= Asin(Bt+ø) 

let y = f(t) and use the double angle sine formula:

f(t) = Asin(Bt+ø)

f(t) = Asin(Bt)cos(ø)+ Acos(Bt)sin(ø)

We compare that to

f(t) = 15sin(t) + 8cos(t)

Its obvious that we can let B=1. So we have:

f(t) = Asin(t+ø)

f(t) = Asin(t)cos(ø) + Acos(t)sin(ø)

f(t) = 15sin(t) + 8cos(t)

Equating corresponding terms:

 15sin(t) = Asin(t)cos(ø)         8cos(t) = Acos(t)sin(ø)
       15 = Acos(ø)                     8 = Asin(ø)
 
Divide equals by equals and get equals:

     

     

     

     

Now we draw a right triangle that has an angle with tangent of 8/15.

We do that by putting 8 on the opposite side and 15 on the adjacent
side, so that , and we use the 
Pythagorean theorem to find the hypotenuse








From that triangle, cos(ø) = 15/17 

15 = Acos(ø)


 



So the answer is found by substituting in:

f(t) = Asin(t+ø)

f(t) = 17sin(t+28.07248694°)

Now do yours EXACTLY like this one.

Edwin