# Homework Help: Asinx + bcosx

1. Oct 8, 2014

### physior

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. Oct 8, 2014

### HallsofIvy

That comes from the "sum formula" that says that $$sin(x+ w)= sin(x)cos(w)+ cos(x)sin(w)$$. 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 $a^2+ b^2= cos^2(w)+ sin^2(w)= 1$ which is of course, not always true! But if we define $R= \sqrt{a^2+ b^2}$ we can "multiply and divide" by R: $asin(x)+ bcos(x)= R((a/R)sin(x)+ (b/R)cos(x))$ and now $(a/R)^2+ (b/R)^2= \frac{a^2+ b^2}{R^2}= \frac{a^2+ b^2}{a^2+ b^2}= 1$. So we can say $cos(w)= a/R$ and $sin(w)= b/R$ so that $w= arccos(a/R)= arcsin(b/R)$. Or, since $cos(w)= a/R$ and $sin(w)= b/R$, $tan(w)= sin(w)/cos(w)= (b/R)/(a/R)= b/a.$.

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