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Asinx + bcosx

  1. Oct 8, 2014 #1

    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
    asinx + bcosx = Rsin(x+w)
    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?

  2. jcsd
  3. Oct 8, 2014 #2


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    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 by a moderator: Oct 10, 2014
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