Answer
$\sqrt{2}\sin(x+\frac{7\pi}{4})$
Work Step by Step
Use the identity $A\sin x+B\cos x=k\sin(x+\phi)$, where $k=\sqrt{A^2+B^2}$ and $\phi$ satisfies $\cos \phi=\frac{A}{\sqrt{A^2+B^2}}$ and $\sin \phi=\frac{B}{\sqrt{A^2+B^2}}$.
In this case, $k=\sqrt{1^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}$.
$\phi$ satisfies $\cos \phi=\frac{1}{\sqrt{1^2+(-1)^2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$ and $\sin \phi=\frac{-1}{\sqrt{1^2+(-1)^2}}=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}$. The only values $\phi$ in $[0, 2\pi)$ where $\cos \phi=\frac{\sqrt{2}}{2}$ are $\frac{\pi}{4}$ and $\frac{7\pi}{4}$, and of these two values, only $\phi=\frac{7\pi}{4}$ also satisfies $\sin\phi=-\frac{\sqrt{2}}{2}$, so $\phi=\frac{7\pi}{4}$.
Therefore, $\sin x-\cos x=\sqrt{2}\sin(x+\frac{7\pi}{4})$.
