# 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!

• Assume $\alpha > \beta$ and both angles are less than $\pi/2$ then what you get inside the square root is positive Jun 5, 2020 at 18:59
• Unfortunately, you can't get rid of the absolute values. Your method does not work because you cannot always recover the value of $\sin x$ from its square. Jun 5, 2020 at 18:59
• @James But, isn't this identity applicable for all angles? Jun 5, 2020 at 19:00
• then you cannot get rid of the absolute value Jun 5, 2020 at 19:00
• @EthanBolker So, I try a different approach, right? Jun 5, 2020 at 19:00

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.

• Thanks, both for the answer and the correction. Jun 5, 2020 at 20:51

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) ?

• Shouldn't I simplify it too? Jun 5, 2020 at 19:45
• You certainly can simplify. But why bring in new complications along with the ' simplifications ? Jun 5, 2020 at 19:58
• Well, I certainly didn't intend for complications to be present, thought it'd be simple. It almost did look simple till the modulus got in and thus, I thought I'd seek help to make it simple from the complicated state it's currently in. Jun 5, 2020 at 20:08
• Your simplification inside the radical sign is fine. No need to simplify $\pm$ sign outside the radical. Jun 5, 2020 at 20:15