Timeline for Any hints on how to prove that the function $\lvert\alpha\;\sin(A)+\sin(A+B)\rvert - \lvert\sin(B)\rvert$ is negative over the half of the total area?
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| S Aug 18, 2021 at 8:01 | history | suggested | Buzz | CC BY-SA 4.0 |
fixed ambiguous notation
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| Aug 18, 2021 at 0:16 | review | Suggested edits | |||
| S Aug 18, 2021 at 8:01 | |||||
| Aug 17, 2021 at 22:38 | comment | added | user44143 | On the first question, I am not saying that the two areas represent the same region, but that they have the same measure. I implicitly used the fact that $\{A \in [0,\pi]: f>-2\cot A\}$ is the same region as $\{A \in [0,\pi]: f>2\cot(\pi-A)\}$ and therefore has the same measure as $\{A \in [0,\pi]: f>2\cot A\}$. On the second question, it’s only after evaluating the integrals that I can see that $dA(\alpha)/d\alpha=0$. Did you get something other than $\pm\log((1+\alpha)/(1-\alpha))/(2\alpha)$ when you did the integrals? | |
| Aug 17, 2021 at 6:08 | history | answered | user44143 | CC BY-SA 4.0 |