How to show that Acosx+Bsinx=Ccos(x+a) ?

[trelek2]

How to show that Acosx+Bsinx=Ccos(x+a) ??

I'm really bad with trig identities, how do we show the following:
Acosx+Bsinx=Ccos(x+a)

 

[daudaudaudau]

Try to work out what cos(x)+sin(x) is. I'll give you a hint. sin(x)=cos(x-Pi/2), so you have cos(x)+sin(x)=cos(x)+cos(x-Pi/2). Now use the formula 2*cos(a)cos(b)=cos(a+b)+cos(a-b).

 

[trelek2]

I still get crap, I don't know how to get to the answer they want. What do I do with the A and B in front of both of the cos terms?

 

[qntty]

What are you trying to do? Are you trying to find (A,B,C) such that A \cos{x}+B \sin{x}=C \cos{(x+a)}?

 

[trelek2]

No, this has to be true for any arbitrary constants A,B,C,a (a is the phase angle). We use it all the time. I just don't know the trig identities and I don't know how to show that Acosx+Bsinx=Ccos(x+a) is true.

 

[Integral]

No, this has to be true for any arbitrary constants A,B,C,a (a is the phase angle). We use it all the time. I just don't know the trig identities and I don't know how to show that Acosx+Bsinx=Ccos(x+a) is true.

This cannot be true for abitrary A, B, and C.
Prove by example.
A=1, B=1, C= 0

There has to be a relationship between A,B and C.

 

[Georgepowell]

Expand the RHS using the cos(A+B) identity.

cos(A+B)=cos(A)cos(B)-sin(A)sin(B) or

Ccos(x+a)=Ccos(x)cos(a)-Csin(x)sin(a)

If Acosx+Bsinx = Ccos(x)cos(a)-Csin(x)sin(a) then

Ccos(a)=A and
B=-Csin(a)

Then you can solve those simultaneous equations.