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.
 A: \begin{eqnarray}
y&=&a\sin x+b\cos x\\
\frac{y}{\sqrt{a^2+b^2}}&=&\frac{a}{\sqrt{a^2+b^2}}\sin x+\frac{b}{\sqrt{a^2+b^2}}\cos x
\end{eqnarray}
Let $\cos\phi=\frac{a}{\sqrt{a^2+b^2}}$ and $\sin\phi=-\frac{b}{\sqrt{a^2+b^2}}$ then
\begin{eqnarray}
\frac{y}{\sqrt{a^2+b^2}}&=&\sin x\cos\phi-\cos x\sin\phi\\
&=&\sin{(x-\phi)}\\
y&=&(\sqrt{a^2+b^2})\sin{(x-\phi)}
\end{eqnarray}
A: Hint: $\;z \sin(x-b) = z\big(\sin(x)\cos(b)-\sin(b)\cos(x)\big)\,$. Identifying the coefficients of $\sin(x)\,$, $\cos(x)\,$, try to determine $z$ and $b$ such that $z \cos(b)=2$ and $z \sin(b)=1$.
A: $$y=2\sin x-\cos x$$
To make it in $$sin(A-B)=\sin A \cos B-\sin B\cos A$$
$$c=\sqrt{2^2+(-1)^2}$$
$$c=\sqrt5$$
$$y=\sqrt5 (\frac{2}{\sqrt5}\sin x-\frac{1}{\sqrt5}\cos x)$$
Taking new angle such that 
$$\cos \phi=\frac{2}{\sqrt 5}$$
$$\sin \phi =\frac{1}{\sqrt 5}$$
$$\phi=0.463$$
The angle is in the first quadrant since sin and cos are both positive.
$$y=\sqrt 5\sin(x-0.463)$$
A: write it in the form $$\sqrt{5}(\frac{2}{\sqrt{5}}\sin(x)-\frac{1}{\sqrt{5}}\cos(x))$$
where $$\cos(\phi)=\frac{1}{\sqrt{5}}$$
$$\sin(\phi)=-\frac{1}{\sqrt{5}}$$
A: $z \sin (x-b)=z \cos b \sin x-z \sin b \cos x$
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
$$y = 2 \sin x - \cos x \longrightarrow y=\sqrt{5} \sin\left(x-\arcsin\frac{1}{\sqrt{5}}\right)$$
