acos(x)+bsin(x) = Rsin(x+t)

solve

1. The problem statement, all variables and given/known data

acos(x)+bsin(x)=Rsin(x+t)

2. Relevant equations
3. The attempt at a solution

Is there any way to show how R is "placed" in acos(x)+bsin(x)=Rsin(x+t) algebraically?
I mean I could, probably, do acos(x)+bsin(x)=sin(t)cos(x)+ cos(t)sinx(x), but still somehow need R in it. Does R give the equation more balance? ;)

Well, we also have x=Rcost and y=Rsint in addition to double angle identities, but I still can't seem to find satisfying algebraic justification for R's existence in f(x)= Rsin(x+t).

Please, help me figure it out.

Thanks.

tiny-tim

hi solve!

just expand sin(x+t), and equate coefficients

solve

Hi,tiny-tim.

See, sin(x+t)=sin(x)cos(t)+ cos(x)sin(t)= acos(x)+ bsin(x)

Ok, a=cos(t) and b=sin(t), but how or why does R end up there?

tiny-tim

you left something out!

Mentallic

You have it backwards! If

[tex]\sin(x)\cos(t)+\cos(x)\sin(t)\equiv a\cdot\cos(x)+b\cdot\sin(x)[/tex]

then [itex]a=\sin(t)[/itex] and [itex]b=\cos(t)[/itex].

solve

Does that happen to be R? If yes, I'd like to know why and how R got there.

solve

Thanks.

tiny-tim

Rsin(x+t) = Rsin(x)cos(t) + Rcos(x)sin(t) = acos(x)+ bsin(x)

solve

Ok.

acos(x)+bsin(x)=Rsin(x+t)

Why is there R in the above equation? Yes, you can expand it, but that doesn't explain what kind of reasoning keeps R in f(x)=Rsin(x+t)

HallsofIvy

Of course, sin(x+ t)= sin(x)cos(t)+ cos(x)sin(t) so that a= sin(t) and b= cos(t). But note that sin(t) and cos(t) are between -1 and 1 and that [itex]sin^2(t)+ cos^2(t)= 1[/itex]. In order to be able to write A sin(x)+ B cos(x), for any A and B, as "R sin(x+ t)" we must "normalize" the coefficients: letting [itex]R= \sqrt{A^2+ B^2}[/itex] so that Asin(x)+ Bcos(x)= R((A/R)sin(x)+ (B/R)cos(x)) and then let cos(t)= A/R and sin(t)= B/R.

tiny-tim

Rsin(x+t)

expand sin(x+t) …

Rsin(x+t)