Transform General EQ Acos(wt)+Bsin(wt)

  • Thread starter Thread starter RJLiberator
  • Start date Start date
  • Tags Tags
    General Transform
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 18K views
RJLiberator
Gold Member
Messages
1,094
Reaction score
63

Homework Statement



Show that Ccos(wt+phi) = Acos(wt)+Bsin(wt)

Homework Equations


Trig identity that states cos(wt+phi) = cos(wt)cos(phi)-sin(wt)sin(phi)

The Attempt at a Solution



Ccos(wt+phi)=(Ccos(phi))cos(wt)+(-Csin(phi))sin(wt)
let A = Ccos(phi)
Let B = -Csin(phi)

Ccos(wt+phi) = Acos(wt)+Bsin(wt)
and done.

Is this as simple as I have shown? Or am I making a critical mistake in letting A = Ccos(phi) and B = -Csin(phi)?
Is there a more rigorous way of doing this that would be expected?

Since phi is a constant, C is a constant, I would think that this is a suitable way to prove that these two sides are equal, but I can't help but feel a bit weak here about this.
 
Physics news on Phys.org
RJLiberator said:

Homework Statement



Show that Ccos(wt+phi) = Acos(wt)+Bsin(wt)

Homework Equations


Trig identity that states cos(wt+phi) = cos(wt)cos(phi)-sin(wt)sin(phi)

The Attempt at a Solution



Ccos(wt+phi)=(Ccos(phi))cos(wt)+(-Csin(phi))sin(wt)
let A = Ccos(phi)
Let B = -Csin(phi)

Ccos(wt+phi) = Acos(wt)+Bsin(wt)
and done.

Is this as simple as I have shown? Or am I making a critical mistake in letting A = Ccos(phi) and B = -Csin(phi)?
Is there a more rigorous way of doing this that would be expected?

Since phi is a constant, C is a constant, I would think that this is a suitable way to prove that these two sides are equal, but I can't help but feel a bit weak here about this.
Yes, it is just that simple.
 
  • Like
Likes   Reactions: RJLiberator
Excelent. Thank you for the confirmation, then I know I am on the right track.

Now the question states to express C and phi as a function of A and B. In this case, I set the equations equal to each other
Ccos(wt+phi) = Acos(wt)+Bsin(wt) and isolate C and phi, I assume.
 
RJLiberator said:
Excellent. Thank you for the confirmation, then I know I am on the right track.

Now the question states to express C and phi as a function of A and B. In this case, I set the equations equal to each other
Ccos(wt+phi) = Acos(wt)+Bsin(wt) and isolate C and phi, I assume.

Going in this direction can be a bit trickier. The results you have in the OP should help with this.
 
  • Like
Likes   Reactions: RJLiberator