I am working with some textbook trigonometry. Its proofs always take the LHS and show that it is equal to the RHS, but I find cross multiplication easier and faster. Can we indeed do it my way? For example:
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ryang
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Aniket Kumar
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Yes, this is a correct way to prove it – Zima Jun 25 '23 at 08:40
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2@Zima: Yes, this is a correct way to prove it -- Cross-multiplication is not correct unless the two terms being multiplied are never equal to zero. See What are the rules for proving trigonometric identities? AND Methodology for Verifying Trigonometric Identities AND May we permit identities to be established by equivalent equations? – Dave L. Renfro Jun 25 '23 at 08:52
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@DaveL.Renfro: Be careful, if the terms are equal to $0$ the whole equation is non-sense, since it's dividing by $0$. So when one writes an equation like the one in OP, one is assuming the quantities in the denominators are not equal to $0$, otherwise the equation wouldn't be defined in the first place. Hence, cross multiplying is well defined. – Zima Jun 25 '23 at 08:59
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1@Zima no that’s not a correct way to present a proof. To prove a statement $P$ is true, you can’t use $P\implies Q$, and $Q$ in order to conclude $P$ itself is true. Here, the correct proof is a simple fix in reversing the order of presentation, but still, your remark is very misleading (actually, outright wrong) and dangerous. – peek-a-boo Jun 26 '23 at 00:27
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Does this answer your question? What are the rules for proving trigonometric identities? – ryang Jun 26 '23 at 02:28
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1You can put restrictions on $A$ as $\tan A$ and $\cot A$ have undefined values at $\pi k/2$ and $\pi k$, respectively, before you cross-multiply. – bjcolby15 Jun 26 '23 at 15:09
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@peek-a-boo I strongly disagree with you, and also you didn' motivate your statements, you just said what I wrote is wrong and misleading. I said that what daveLrenfro explained in his comment is not valid, since if you write the fractions to begin with, the denominators are not $0$. Why do you say this is not correct? – Zima Jun 26 '23 at 16:49
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To write a proof formally, you should start with things that are already known to be true, and use those to derive new true statements until you reach the thing you're actually trying to prove. Working backwards from the result is a good way to figure out how to prove it, but you need to be careful to ensure that the steps you've taken are reversible.
So in your example, what you've written is what I would consider "scratch work", then the proof itself could proceed by writing the steps in reverse order. Alternatively, I might take a slightly tangential path by writing:
$$\begin{eqnarray} \frac{1 - \tan A}{1 + \tan A} - \frac{\cot A - 1}{\cot A + 1} &=& \frac{(1 - \tan A)(\cot A + 1) - (\cot A - 1)(1 + \tan A)}{(1 + \tan A)(\cot A + 1)} \\ &=& \frac{(\cot A - 1 + 1 + \tan A) - (\cot A - 1 + 1 - \tan A)}{(1 + \tan A)(\cot A + 1)} \\ &=& 0 \\ \implies \frac{1 - \tan A}{1 + \tan A} &=& \frac{\cot A - 1}{\cot A + 1}\end{eqnarray}$$
The main reason for doing it this way is to avoid introducing non-reversible steps that break the proof (like multiplying or dividing by a term that might be equal to zero).
ConMan
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Another way of doing it:
LHS: multiply numerator and denominator by $\cos A$ to get $$\dfrac {\cos A - \sin A}{\cos A + \sin A}$$.
RHS: multiply numerator and denominator by $\sin A$ to get $$\dfrac {\cos A - \sin A}{\cos A + \sin A}$$.
Since $LHS = RHS$, we are done, unless you want to simplify it further...
$$\dfrac {\cos A - \sin A}{\cos A + \sin A}\cdot \dfrac {\cos A - \sin A}{\cos A - \sin A} \\ \dfrac {\cos^2 A - 2 \sin A \cos A + \sin^2 A}{\cos^2 A - \sin^2 A} \\ \dfrac {1-\sin 2A}{\cos 2A} \\ \sec 2A - \tan 2A$$
bjcolby15
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