I need to write $3\sin(x) + 4\cos(x)$ in the form of $r\sin(x-a)$.
Expanding $r\sin(x-a): r\sin(x)\cos(a)-r\sin(a)\cos(x)$
Comparing the two forms (The original equation and the expanded form): $3=r\cos(a)$ and $4=-r\sin(a)$.
Getting $r$:
$\begin{align*} 3^2 + 4^2 &= r^2 \cos(a)^2 + r^2 \sin(a)^2\\ 25 &= r^2 (\cos(a)^2 + \sin(a)^2)\\ 25 &= r^2\\ r &= -5, 5 \end{align*}$
Getting $a$:
$\begin{align*} \frac{-r\sin(a)}{ r\cos(a)} &= \frac{3}{4}\\ \tan(a) &= -\frac{3}{4}\\ a &= \arctan(-3/4)\\ a &= -36.87, 143.13\\ \end{align*}$
The questions:- There are two values for both $r$ and $a$, how should I choose the values to be in the final form?