<s>[INST] Deriving the expansion of $\sin (\alpha - \beta)$ using $\sin x = \sqrt{1-\cos^2 x}$ I was deriving the expansion of  the expansion of $\sin (\alpha - \beta)$ given that $\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$
Now, my textbook has done it in a different manner but I thought of doing it using the simple trigonometric identity $\sin^2 x + \cos^2 x = 1 \implies \sin x = \sqrt{1-\cos^2 x}$. I thought that it would be pretty easy (it probably is), until I got stuck in the final part which included the modulus function.
Here's how I did it :
$$\sin (\alpha - \beta) = \sqrt {1 - \cos^2 (\alpha - \beta)} = \sqrt{1-(\cos \alpha \cos \beta + \sin \alpha \sin \beta)^2}$$
By substituting $1$ as $\sin^2\alpha + \cos^2\alpha$ and expanding $(\cos \alpha \cos \beta + \sin \alpha \sin \beta)^2$, we get :
$$\therefore \sin (\alpha - \beta) = \sqrt{\sin^2\alpha + \cos^2\alpha - \cos^2\alpha\cos^2\beta-
\sin^2\alpha\sin^2\beta-2\sin\alpha\sin\beta\cos\alpha\cos\beta}$$
$$\therefore \sin(\alpha-\beta) = \sqrt{\sin^2\alpha (1-\sin^2\beta)+\cos^2\alpha(1-\cos^2\beta)-2\sin\alpha\sin\beta\cos\alpha\cos\beta}$$
$$\therefore \sin(\alpha - \beta) = \sqrt{\sin^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta-2\sin\alpha\sin\beta\cos\alpha\cos\beta}$$
$$\therefore \sin(\alpha - \beta) = \sqrt{(\sin\alpha\cos\beta - \cos\alpha\sin\beta)^2} = |\sin\alpha\cos\beta-\cos\alpha\sin\beta|$$
Now, how do I get rid of the modulus sign? I do know that I must decide whether the expression inside the modulus functions in positive or negative, but I can't seem to decide how.
Thanks!
 [/INST] What is your question?
At start when you accepted 
$$\sin x =\pm \sqrt{1-\cos^2 x}, \tag1$$
at the end why don't you accept
$$\sin (\alpha - \beta) = \pm \sqrt{1-(\cos \alpha \cos \beta + \sin \alpha \sin \beta)^2}\tag2$$
in that very same sense? What is extra in (2) ?
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