How to transform $y = 2 \sin(x) - \cos(x)$ into the format $y = z \sin(x - b)$ How can I transform $y = 2 \sin(x) - \cos(x)$ into the format $y = z \sin(x - b)$ for some $z,b\in\Bbb R$? I have been playing with various trig identities and looking through various trigonometry references but cannot so much as find a starting point. Any hints or pushes in the right direction would be much appreciated. Helping a struggling high school study and find myself greatly out of practice.

$z \sin (x-b)=z \cos b \sin x-z \sin b \cos x$
so we have to solve
$z \cos b =2;\;z \sin b =1$
from the second we get $b=\arcsin\frac{1}{z}$. We plug this in the first
$z\cos\left(\arcsin\frac{1}{z}\right)=2\to z\sqrt{1-\frac{1}{z^2}}=2 \to z=\sqrt{5}$
$b=\arcsin\frac{1}{\sqrt{5}}$
Therefore
$$y = 2 \sin x - \cos x \longrightarrow y=\sqrt{5} \sin\left(x-\arcsin\frac{1}{\sqrt{5}}\right)$$
