# Trignometry identities

There is issue I have with this question, specifically part b.

a) Show that $(\sin A - \cos A)^2 = 1-\sin(2A)$

b) Hence, find the exact value of $\sin 15-\cos 15$.

I used what I worked out in part a.

$(\sin 15-\cos 15)^2 = 1-\sin 30 = 1 - \frac 1 2 = \frac 1 2$

and I thought the answer for the question would be

$\pm \sqrt{\frac 1 2}$

however the answers only shows one solution:

$-\sqrt{\frac 1 2}$

and when I check $\sin 15-\cos 15$ on the calculator, that is the correct answer. I'm wondering why doing it algebraically doesn't work.

Notice that $0<\theta<45^{\circ}$ implies $\;\sin \theta<\frac{\sqrt{2}}{2}<\cos \theta$, $\;\;\;$then $\;\;\;\sin 15^{\circ}-\cos15^{\circ}<0$.
$$\sin15^\circ-\cos15^\circ=\sqrt2\sin(15^\circ-45^\circ)=-\sqrt2\sin30^\circ<0$$
• @WanyuTang, Do you know $\sin(A-B)$ formula & $$\sin45^\circ=\cos45^\circ=\dfrac1{\sqrt2}$$ – lab bhattacharjee Oct 9 '15 at 3:58
You get $|ans|=1/\sqrt{2}$. This does not imply that $ans$ necessarily has both positive and negative value. It can be both or either. Here $ans$ is not a variable its an unknown (at first) quantity.