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.


  • $\begingroup$ Assume $\alpha > \beta $ and both angles are less than $\pi/2$ then what you get inside the square root is positive $\endgroup$
    – James
    Jun 5, 2020 at 18:59
  • $\begingroup$ 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. $\endgroup$ Jun 5, 2020 at 18:59
  • $\begingroup$ @James But, isn't this identity applicable for all angles? $\endgroup$ Jun 5, 2020 at 19:00
  • $\begingroup$ then you cannot get rid of the absolute value $\endgroup$
    – James
    Jun 5, 2020 at 19:00
  • $\begingroup$ @EthanBolker So, I try a different approach, right? $\endgroup$ Jun 5, 2020 at 19:00

2 Answers 2


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.

  • $\begingroup$ Thanks, both for the answer and the correction. $\endgroup$ 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) ?

  • $\begingroup$ Shouldn't I simplify it too? $\endgroup$ Jun 5, 2020 at 19:45
  • $\begingroup$ You certainly can simplify. But why bring in new complications along with the ' simplifications ? $\endgroup$
    – Narasimham
    Jun 5, 2020 at 19:58
  • $\begingroup$ 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. $\endgroup$ Jun 5, 2020 at 20:08
  • $\begingroup$ Your simplification inside the radical sign is fine. No need to simplify $\pm$ sign outside the radical. $\endgroup$
    – Narasimham
    Jun 5, 2020 at 20:15

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