Consider $f(x)=a\sin(x)+b\cos(x)$
where $a,b$ are some real constants.
Putting $f(x)=R\sin(\alpha+x)$, I got $$f(x)=\pm\sqrt{a^2+b^2}\sin\left(\arctan\left(\frac{b}{a}\right)+x \right) \tag{1}$$ According to my Desmos graphs: $$f(x)=+\sqrt{a^2+b^2}\sin\left(\arctan\left(\frac{b}{a}\right)+x \right) \tag{2}$$ for $a>0$
and $$f(x)=-\sqrt{a^2+b^2}\sin\left(\arctan\left(\frac{b}{a}\right)+x \right) \tag{3}$$ for $a<0$.
So the sign $\pm$ taken is independent of the value of $b$.
I tried to prove this by discussing the possible values taken by each of $a$ and $b$: e.g. when $a,b>0$ $$0<\arctan\left(\frac{b}{a}\right)<\frac{\pi}{2} \tag{4}$$ But I struggled to proceed in discussing $$\sin\left(\arctan\left(\frac{b}{a}\right)+x \right) $$ because $\sin()$ can take any values regardless of what $a,b$ are.
Is there a way to determine which sign should be taken in the RHS of $(1)$?
Denote $$ \alpha = \arctan \frac{b}{a}. $$ Then evidently $$ \tan \alpha = \frac{b}{a}. $$ Let's start with identity $$ \sin^2 \alpha + \cos^2 \alpha = 1 $$ and divide each term by $\cos^2 \alpha$ to get $$ \tan^2 \alpha + 1 = \frac{1}{\cos^2 \alpha} \implies \cos^2 \alpha = \frac{1}{1+\tan^2 \alpha} = \frac{a^2}{a^2 + b^2}. $$ This gives us also $$ \sin^2 \alpha = \frac{b^2}{a^2 + b^2}. $$
Now (looking at $\alpha$) you can decide when one should take positive/negative value for $\sin$ and $\cos$.
Suppose, for example $$ 0 \le \alpha < \frac{\pi}{2}. $$ This means that both sine and cosine will be nonnegative and we have $$ \sin \alpha = \frac{|b|}{\sqrt{a^2 + b^2}}, \; \cos \alpha = \frac{|a|}{\sqrt{a^2 + b^2}}. $$ This will give us $$ \sin(\alpha+x ) = \sin \alpha \cos x + \cos \alpha \sin x = \frac{|a|}{\sqrt{a^2 + b^2}} \sin x + \frac{|b|}{\sqrt{a^2 + b^2}} \cos x. $$
Now let's consider the case $$ 0 > \alpha > -\frac{\pi}{2}. $$ This means that sine will be negative and cosine will be positive. So $$ \sin \alpha = -\frac{|b|}{\sqrt{a^2 + b^2}}, \; \cos \alpha = \frac{|a|}{\sqrt{a^2 + b^2}}. $$ $$ \sin(\alpha+x ) = \sin \alpha \cos x + \cos \alpha \sin x = \frac{|a|}{\sqrt{a^2 + b^2}} \sin x - \frac{|b|}{\sqrt{a^2 + b^2}} \cos x. $$