If $\sin(\alpha)+\cos(\alpha)=a,$ denote $|\sin(\alpha)-\cos(\alpha)|$ in terms of $a$

Question

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!

Answer 1

$$1+2\sin{\alpha}\cos{\alpha}=a^2,$$ which says $$2\sin{\alpha}\cos{\alpha}=a^2-1.$$ Id est, $$|\sin\alpha-\cos\alpha|=\sqrt{1-2\sin\alpha\cos\alpha}=\sqrt{1-(a^2-1)}=\sqrt{2-a^2}$$

Answer 2

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