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I've just been doing a problem that involved this equation: $$ \frac{1}{\sin\left(\frac{\theta}{2}\right)}\left( \sin\left(b\theta-\frac{\theta}{2}\right)-\sin\left(a\theta-\frac{\theta}{2}\right) \right) =0$$ and I'm wondering whether it's ok to ignore the $\frac{1}{\sin(\theta/2)}$ expression?

My justification for ignoring it is since $|\sin(\theta/2)|\leq 1$ we know that $$\frac{1}{\sin\left(\frac{\theta}{2}\right)}$$ will never approach zero and thus can't introduce a root into the original equation; so we don't have to consider it when looking for roots. Is this a correct line of reasoning or will I miss solutions?

As a second related question: If I didn't ignore it and used it to simplify terms, for example if you expanded out the trig functions to give: $$ \frac{1}{\sin\left(\frac{\theta}{2}\right)}\left( \sin(b\theta)\cos\left(\frac{\theta}{2}\right)-\cos(b\theta)\sin\left(\frac{\theta}{2}\right)-\sin(a\theta)\cos\left(\frac{\theta}{2}\right)+\cos(a\theta)\sin\left(\frac{\theta}{2}\right) \right) $$ and then proceeded to cancel the $\sin\left(\frac{\theta}{2}\right)$ terms (I guess if we cancel them we have to assume that $\frac{1}{\sin\left(\frac{\theta}{2}\right)}\neq0$) how would that change the roots of the original equation?

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  • $\begingroup$ Yes, $p/q=0$ if and only if $p=0$ and $q\ne0$. $\endgroup$ – Rahul Apr 22 '15 at 22:59
  • $\begingroup$ @Rahul Thanks for the response; yeah that makes sense, it just seemed odd to me, the idea of ignoring it. If $q=\sin(\theta/2)\neq 0$, then would any solutions that make $p=0$ but also make $q=0$ have to be ignored? $\endgroup$ – Jay Apr 22 '15 at 23:04
  • $\begingroup$ You need to be mindful of the indeterminate case. It is okay to ignore the $\frac{1}{\sin\bigg(\frac{\theta}{2}\bigg)}$ as long as $a\ne b$. $\endgroup$ – John Joy Apr 22 '15 at 23:09
  • $\begingroup$ @JohnJoy Thanks for the answer, how did you determine that it's ok only when $a\neq b$. Since I guess that would make the numerator $0$ but how would assuming $a\neq b$ imply that $\theta/2\neq \pi n$? (and thus ensuring it doesn't become indeterminate) Thanks for the help! $\endgroup$ – Jay Apr 22 '15 at 23:14
  • $\begingroup$ You're right. I guess what my point was is that in problems like these that it is important to look at when the numerator and denominator become zero, and to note whether or not they become zero together. $\endgroup$ – John Joy Apr 22 '15 at 23:22

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