# $A \cos (x) + B \sin(x)$ with $A,B \in \mathbb{R}$ but not necessarily positive

In this question, it is shown that

$$A \sin(x) + B \cos (x) = R \sin ( x + \theta )$$

for $$A,B$$ real and positive, with

$$R = \sqrt{A^2 + B^2}$$

$$\theta = \arctan \left( \frac{B}{A} \right)$$

This is performed solving the system

$$\left\{ \begin{array}{c} R \cos (\theta) = A\\ R \sin (\theta) = B \end{array} \right.$$

for $$R$$ and $$\theta$$. What if, instead, $$A$$ and/or $$B$$ are real, but not positive?

Some trivial test on WolframAlpha shows that the $$\arctan \left( \frac{B}{A} \right)$$ term is computed the same way, considering the sign of $$A$$ and $$B$$.

$$R$$ is instead negative if $$A$$ or both $$A$$ and $$B$$ are negative. It is obtained by squaring and summing the two equations in the system and I can't figure out how a sign can be considered in this operation.

Let $$A = R \cos \theta$$ and $$B = R \sin \theta$$

So, $$y = A \sin x + B \cos x = R \cos \theta \sin x+ R \sin \theta \cos x = R \sin(x + \theta)$$

Also, $$R^2 = A^2 + B^2$$ or $$R = \sqrt{A^2 + B^2}$$

Let

$$\alpha = \tan^{-1} \left| \frac{B}{A} \right|$$

so that $$\alpha \in [0,\pi/2]$$

Let us assume $$R$$ to be always positive (conventional).

So we now have 4 cases.

CASE 1

$$A \ge 0$$ and $$B \ge 0$$

In this case, $$R \cos \theta \ge 0$$ and $$R \sin \theta \ge 0$$

Being $$R>0$$, this implies $$\sin \theta \ge 0$$ and $$\cos \theta \ge 0$$

So, $$\theta \in [0 ,\pi/2]$$

and

$$\alpha = \theta$$

CASE2

$$A \le 0$$ and $$B \ge 0$$

In this case, $$R \cos \theta \le 0$$ and $$R \sin \theta \ge 0$$

As $$R>0$$, $$\sin \theta \ge 0$$ and $$\cos \theta \le 0$$

So, $$\theta \in [ \pi/2, \pi]$$

$$\alpha = \pi - \theta$$

or $$\theta = \pi - \alpha$$

CASE3

$$A \le 0$$ and $$B \le 0$$

In this case, $$R \cos \theta \le 0$$ and $$R \sin \theta \le 0$$

As $$R>0$$, $$\sin \theta \le 0$$ and $$\cos \theta \le0$$

So, $$\theta \in [ \pi, 3\pi/2]$$ or $$\theta \in [ -\pi/2, -\pi]$$.

$$\alpha = \theta - \pi$$

or $$\theta = \alpha + \pi$$

CASE4

$$A \ge 0$$ and $$B \le 0$$

In this case, $$R \cos \theta \ge 0$$ and $$R \sin \theta \le 0$$

As $$R>0$$, $$\sin \theta \le 0$$ and $$\cos \theta \ge 0$$

So, $$\theta \in [3\pi/2, 2\pi]$$ or $$\theta \in [-\pi/2,0]$$.

$$\alpha = - \theta$$

or

$$\theta = - \alpha$$

So, at last $$y = R \sin( x + \theta)$$

Find the signs of $$A$$ and $$B$$, then the principal value $$\alpha$$. Then according to the quadrant, find $$\theta$$ in terms of $$\alpha$$ and substitute in the above expression.

Write your term in the form $$\sqrt{A^2+B^2}\left (\frac{A\cos(x)}{\sqrt{A^2+B^2}}+\frac{B\sin(x)}{\sqrt{A^2+B^2}}\right)$$

• Before or after writing the system? Sorry, I can't still figure it out. Also you swapped $A$ and $B$, I guess it is a typo. – BowPark May 10 '19 at 10:09

$$\newcommand{arccot}{\operatorname{arccot}}$$A method is, for all $$A,B\in\Bbb R$$ such that $$A\ne 0\lor B\ne 0$$, to consider the following manipulation $$\sqrt{A^2+B^2}\left(\frac A{\sqrt{A^2+B^2}}\sin x+\frac{B}{\sqrt{B^2+A^2}}\cos x\right)=\\=\sqrt{A^2+B^2}\sin\left(x+\theta\left(\frac A{\sqrt{A^2+B^2}},\frac {B}{\sqrt{A^2+B^2}}\right)\right)$$

where $$\theta(c,s)$$ is, provided that $$c^2+s^2=1$$, the one and only real number $$\theta\in (-\pi,\pi]$$ such that $$\cos \theta=c\land \sin\theta=s$$. There are a number of ways to write such number. Here are a few:

\begin{align}\theta(c,s)&=\begin{cases}\arccos c&\text{if }s\ge 0\\ -\arccos c&\text{if }s<0\end{cases}\\ \theta(c,s)&=\begin{cases}\arctan \frac sc&\text{if }c> 0\\ -\frac\pi2\operatorname{sgn}s&\text{if }c=0\\ \pi+\arctan \frac sc&\text{if }c<0\land s\ge 0\\ -\pi+\arctan\frac sc&\text{if }c<0\land s<0\end{cases}\\ \theta(c,s)&=\begin{cases}\arccot \frac cs&\text{if }s> 0\\ \pi\operatorname{sgn}c&\text{if }s=0\\ -\arccot c&\text{if }s<0\end{cases}\end{align}

Of course, a different choice of range for $$\theta(c,s)$$ (for instance $$[0,2\pi)$$) may yield different expressions.

• This is very thorough, but it doesn't show the way these expressions are obtained. Let's focus on the second one: $\theta(c,s)$ in terms of $\arctan$. How to obtain this? – BowPark May 10 '19 at 10:23
• Because $\cos\arctan x=\frac{1}{\sqrt{1+x^2}}$ and $\sin\arctan x=\frac{x}{\sqrt{1+x^2}}$. If you apply this to the four cases, you'll obtain $\cos\theta(c,s)=c$ and $\sin\theta(c,s)=s$. Then, you may notice that every so-defined $\theta(c,s)$ will actually be in $(-\pi,\pi]$. – Saucy O'Path May 10 '19 at 11:06

For $$R$$ the signs of $$A$$ and $$B$$ do not matter at all.

To get a correct angle for any $$(A,B) \ne (0,0)$$ you may have a look at the 2-argument arctangent function: $$θ=\rm{atan2}(B, A).$$

For each case in its definition you get the same signs as in your tests with WolframAlpha:

Case $$A > 0$$: $$R \sin\left(x + \rm{atan2}(B, A)\right) = R \sin\left(x + \arctan\left(\frac A B\right)\right)$$ Case $$A < 0$$: $$R \sin(x + \rm{atan2}(B, A)) = R \sin\left(x + \arctan\left(\frac A B\right) \pm \pi\right) = -R \sin\left(x + \arctan\left(\frac A B\right)\right)$$

Case $$A = 0$$ and $$B > 0$$: $$R \sin\left(x + \rm{atan2}(B, 0)\right) = R \sin\left(x + \frac \pi 2\right) = R \cos(x) = B \cos(x)$$ Case $$A = 0$$ and $$B < 0$$: $$R \sin\left(x + \rm{atan2}(B, 0)\right) = R \sin\left(x - \frac \pi 2\right) = -R \cos(x) = B \cos(x)$$

A simpler way of solving this, without restrictions or polar coordinates, is the following:

If $$A=0$$ there is nothing to prove. Otherwise: $$A \sin(x) + B \cos (x) =A \left( \sin(x) + \frac{B}{A} \cos (x) \right)$$

Now, let $$\tan(\theta)=\frac{B}{A}$$. Then $$A \sin(x) + B \cos (x) =A \left( \sin(x) + \tan(\theta)\cos (x) \right)=A \left( \sin(x) + \frac{\sin(\theta)}{\cos(\theta)}\cos (x) \right)\\=A \frac{ \sin(x)\cos(\theta) + \sin(\theta)\cos(x)}{\cos(\theta)}=\frac{A}{\cos(\theta)}\sin(x+\theta)$$

• A pretty smart and compact solution. Thanks! – BowPark May 10 '19 at 15:59

Take $$\theta=\arccos\frac AR$$. Then$$\sin^2\theta=1-\cos^2\theta=1-\frac{A^2}{R^2}=\left(\frac BR\right)^2$$amd therefore $$R\sin\theta=\pm\frac BR$$. If $$R\sin\theta=\frac BR$$, then that $$\theta$$ will do. Otherwise, take $$-\theta$$ instead.