If $\sin(\alpha)+\cos(\alpha)=a,$ denote $|\sin(\alpha)-\cos(\alpha)|$ in terms of $a$ I attempted to solve it using the following identities: $1.\ a^2-b^2.\ 2.\ (a\pm b)^2.\ 3. a^3\pm b^3.\ 4.\ (a\pm b)^3.$
Since none of my efforts led me to the correct answer (which is $\sqrt{2-a^2}$), I found it better not to write my lengthy work towards solving it. 
I mostly got the expression $2\sin(\alpha)\cos(\alpha).$ So, any helpful hints, comments or answers are welcome!

$$\begin{align}
\sin b+\cos b&=a\\
\sin^2 b+2\sin b\cos b+\cos^2 b&=a^2\\
-2(\sin^2b+\cos^2b)&\quad -2\\
-\sin^2b+2\sin b\cos b-\cos^2 b&=a^2-2\\
-(\sin b-\cos b)^2&=a^2-2\\
(\sin b-\cos b)^2&=2-a^2\\
|\sin b-\cos b|&=\sqrt{2-a^2}
\end{align}$$
