1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Asinx + bcosx

  1. Oct 8, 2014 #1
    hello!

    I would like to know how to solve asinx + bcosx = c, with a, b, c being any real numbers (constants)

    First, are there any limitations for the above to be valid?

    Second, I was introduced to a solution but I cannot fully understand the procedure.

    Let's say we have asinx + bcosx = c
    We can solve this by using:
    R= root of a^2 plus b^2
    and
    asinx + bcosx = Rsin(x+w)
    and
    tan(w)=b over a

    the first question is, do the above are valid for a, b being either positive or negative?

    Second, when we find the "w" using the calculator (by inversing its tan), how do we find which exactly value of w we must use?

    Third, when we inverse the sin(x+w), by the calculator, how do we find which exactly values of x+w we must use?

    Thanks!
     
  2. jcsd
  3. Oct 8, 2014 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    That comes from the "sum formula" that says that [tex]sin(x+ w)= sin(x)cos(w)+ cos(x)sin(w)[/tex]. If we are given asin(x)+ bcos(x), comparing to the previous formula, we would like to have a= cos(w) and b= sin(w). But that is not always possible, because that would require that [itex]a^2+ b^2= cos^2(w)+ sin^2(w)= 1[/itex] which is of course, not always true! But if we define [itex]R= \sqrt{a^2+ b^2}[/itex] we can "multiply and divide" by R: [itex]asin(x)+ bcos(x)= R((a/R)sin(x)+ (b/R)cos(x))[/itex] and now [itex](a/R)^2+ (b/R)^2= \frac{a^2+ b^2}{R^2}= \frac{a^2+ b^2}{a^2+ b^2}= 1[/itex]. So we can say [itex]cos(w)= a/R[/itex] and [itex]sin(w)= b/R[/itex] so that [itex]w= arccos(a/R)= arcsin(b/R)[/itex]. Or, since [itex]cos(w)= a/R[/itex] and [itex]sin(w)= b/R[/itex], [itex]tan(w)= sin(w)/cos(w)= (b/R)/(a/R)= b/a.[/itex].

    You ask: "when we find the "w" using the calculator (by inversing its tan), how do we find which exactly value of w we must use?"
    Any if the values can be used. Use whichever fits your needs. Typically, that is the smallest positive value just because it is simplest.
     
    Last edited: Oct 10, 2014
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Asinx + bcosx
Loading...