Timeline for A "known" tangent half-angle formula?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| S Jul 1, 2014 at 13:30 | history | suggested | Ben Barber | CC BY-SA 3.0 |
mend broken TeX
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| Jul 1, 2014 at 13:29 | review | Suggested edits | |||
| S Jul 1, 2014 at 13:30 | |||||
| Dec 14, 2012 at 18:05 | comment | added | Yoav Kallus | @Gerry: You are right. This is a proof that $\tan\gamma=\ldots$ if $\tan(\gamma/2)=\ldots$, rather than the converse. Assuming $\tan(\gamma)=\ldots$, we get merely that $f(\tan(\gamma/2))=f(\tan(\beta/2)\tan(\alpha/2))$, where $f(t)=2t/(1-t^2)$, as you note in your answer. | |
| Dec 14, 2012 at 17:37 | vote | accept | Michael Hardy | ||
| Dec 14, 2012 at 16:47 | comment | added | Michael Hardy | I probably ought to have said "if and only if" in my question, rather than just going one direction. | |
| Dec 14, 2012 at 16:47 | comment | added | Michael Hardy | @GerryMyerson : The one thing that's not quite reversible is this: You can't just let $\gamma=\arctan\dfrac{\sin\alpha\sin\beta}{\cos\alpha+\cos\beta}$; rather, you have to choose the approrpriate one of two points on the circle where the tangent has a given value. Although $\tan\gamma$ is the same regardless of which of those you pick, $\tan(\gamma/2)$ is not. That question need not be mentioned if you do the proof in the direction seen in this answer, but for the converse of that, the issue comes up. | |
| Dec 13, 2012 at 22:52 | comment | added | Gerry Myerson | Technically, this is proving $A$ implies $B$, where the question asked for a proof that $B$ implies $A$. But I suppose all the steps are reversible. | |
| Dec 13, 2012 at 18:46 | history | edited | Michael Hardy | CC BY-SA 3.0 |
align
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| Dec 13, 2012 at 4:56 | history | answered | Yoav Kallus | CC BY-SA 3.0 |