Finding the Role of R in Trigonometric Equations: An Algebraic Exploration

[solve]

Homework Statement

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

Homework Equations

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 solve!

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

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]

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

you left something out!

 

[Mentallic]

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?

You have it backwards! If

\sin(x)\cos(t)+\cos(x)\sin(t)\equiv a\cdot\cos(x)+b\cdot\sin(x)

then a=\sin(t) and b=\cos(t).

 

[solve]

you left something out!

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

 

[solve]

You have it backwards! If

\sin(x)\cos(t)+\cos(x)\sin(t)\equiv a\cdot\cos(x)+b\cdot\sin(x)

then a=\sin(t) and b=\cos(t).

Thanks.

 

[tiny-tim]

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

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

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

 

[solve]

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

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 sin^2(t)+ cos^2(t)= 1. 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 R= \sqrt{A^2+ B^2} 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)​