<s>[INST] 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.
 [/INST] \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}
</s>