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
