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!
Best Answer
You start wrong, I'm afraid: you can say that $$ \lvert\sin(\alpha-\beta)\rvert=\sqrt{1-\cos^2(\alpha-\beta)} $$ where you cannot omit the absolute value in the left-hand side. At the end you get $$ \lvert\sin(\alpha-\beta)\rvert=\lvert\sin\alpha\cos\beta-\cos\alpha\sin\beta\rvert $$ (you have a sign wrong, but it's just a typo). Now you could do a very long case analysis to show that $\sin(\alpha-\beta)$ has the same sign as $\sin\alpha\cos\beta-\cos\alpha\sin\beta$ for all $\alpha,\beta$.
Not something that I'd try myself.
I'd add that using $\pm$ doesn't help, because you would need to assign the proper sign anyway by the same case analysis.