Timeline for Inequality involving sine and cosine
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 4, 2019 at 12:13 | vote | accept | Loreno Heer | ||
| Sep 4, 2019 at 12:13 | history | bounty ended | Loreno Heer | ||
| Sep 3, 2019 at 9:08 | vote | accept | Loreno Heer | ||
| Sep 3, 2019 at 9:08 | |||||
| Sep 2, 2019 at 19:26 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
deleted 3 characters in body
|
| Sep 2, 2019 at 19:13 | comment | added | Loreno Heer | I think in your answer you meant, $\cos(B/2) = \sqrt{(1+x)/2}$. I understand the argument. I will definitely need to add $B+D \geq E$ as requirement. Hopefully that is the only one. | |
| Sep 2, 2019 at 15:34 | comment | added | Todd Trimble | @LorenoHeer I understand; I'm stating a general guideline, but you are free to continue as you see best. | |
| Sep 2, 2019 at 14:55 | comment | added | Loreno Heer | @Todd Trimble I will try not to change it anymore. However it is also inconvenient to ask a new question with only slightly changed conditions. | |
| Sep 2, 2019 at 14:51 | comment | added | Todd Trimble | @LorenoHeer It's generally not a good idea to keep changing the question as responses come in. This usually winds up being frustrating to the community. | |
| Sep 2, 2019 at 13:43 | comment | added | Loreno Heer | I am happy to give you the bounty, however I like to keep the question open with the additional condition as I try to find a way to get this inequality working. Not sure how this is usually done here. | |
| Sep 2, 2019 at 12:38 | comment | added | Loreno Heer | ok, that looks correct. I guess I will require another condition, $B+D \geq E$... | |
| Sep 2, 2019 at 12:09 | comment | added | Fedor Petrov | It is strict if $B\ne D$. | |
| Sep 2, 2019 at 11:31 | comment | added | Loreno Heer | I am not sure I fully understand your example. I see how you derive the first equality, however I believe it should be $\geq$ there. The last inequality does not contradict the required one as there could be equality... | |
| Sep 2, 2019 at 11:13 | comment | added | Fedor Petrov | Yes, but it is certainly not important: take very small $F$ if you do not like zero, and accordingly slightly change $B,D$. | |
| Sep 2, 2019 at 10:40 | comment | added | Loreno Heer | Please note that I specified $A,B,\ldots,F \in ]0,\frac{\pi}{2}]$. In particular $F=0$ is not possible. | |
| Sep 2, 2019 at 10:33 | history | answered | Fedor Petrov | CC BY-SA 4.0 |