Deriving the expansion of $\sin (\alpha - \beta)$ using $\sin x = \sqrt{1-\cos^2 x}$

Question

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 — James, Jun 05 '20 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. — Ethan Bolker, Jun 05 '20 at 18:59
So, basically, to combine both the comments, this approach will only work if $0 < \alpha < \dfrac{\pi}{2}$ and $0 < \beta < \dfrac{\pi}{2}$ and to make it applicable for all angles, I should try a different approach. Thanks! — Rajdeep Sindhu, Jun 05 '20 at 19:02

score 2 · Accepted Answer · answered Jun 05 '20 at 20:37

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. – Rajdeep Sindhu Jun 05 '20 at 20:51 — Rajdeep Sindhu, Jun 05 '20 at 20:51

score 0 · Answer 2 · answered Jun 05 '20 at 19:38

0

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) ?

answered Jun 05 '20 at 19:38

Narasimham

40,495

Shouldn't I simplify it too? – Rajdeep Sindhu Jun 05 '20 at 19:45
You certainly can simplify. But why bring in new complications along with the ' simplifications ? – Narasimham Jun 05 '20 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. – Rajdeep Sindhu Jun 05 '20 at 20:08
Your simplification inside the radical sign is fine. No need to simplify $\pm$ sign outside the radical. – Narasimham Jun 05 '20 at 20:15

Deriving the expansion of $\sin (\alpha - \beta)$ using $\sin x = \sqrt{1-\cos^2 x}$

2 Answers2