tag:blogger.com,1999:blog-5116217195460851233.post6663116983558575638..comments2016-08-11T22:47:22.757+02:00Comments on Nemderogatorius' blog: nemderogatoriushttp://www.blogger.com/profile/11479519156593310864noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-5116217195460851233.post-22775062888732123542016-08-11T22:47:03.913+02:002016-08-11T22:47:03.913+02:00Wow, you are awesome! Nice use of the tg(3x) formu...Wow, you are awesome! Nice use of the tg(3x) formula there, it's pretty advanced stuff (even I didn't know about it, though I think it can be proved by complex unity vectors). I guess, Sunset wasn't so well prepared from trigonometry. As you said, she's just a silly horse after all :)nemderogatoriushttps://www.blogger.com/profile/11479519156593310864noreply@blogger.comtag:blogger.com,1999:blog-5116217195460851233.post-89579449619617181322016-08-11T20:49:41.548+02:002016-08-11T20:49:41.548+02:00Proof on point.
Also, tell you what, the law of ...Proof on point. <br /><br />Also, tell you what, the law of sines would've indeed been much more straightforward here. Write the law out for all inner triangles within ADEB, reduce the system via sequential substitutions, and this particular set of magic numbers boils the problem down to something like<br /><br />sin(x+50)/sin(x) = sin(20)*sin(40)*sin(70) / (sin(10)*sin(30)*sin(60)).<br /><br />This is already a step away from the general answer as an arctangent. Prettifying this is somewhat messy, but some time and a handful of shenanigans later, I got to<br /> <br />ctg(x) = ctg(10)ctg(50)tg(30),<br /><br />and this thing gets simplified instantly (and amusingly) if you expand the triple tangent argument in tg(30) (as in tg(3x) = tg(x)tg(Pi/3-x)tg(Pi/3+x)).<br /><br />The result is simply ctg(x)=ctg(20), so there's that.<br /> <br />Why didn't Sunset choose to boil that problem down to a trigonometrical charade, we will most likely never know. Probably because she's a silly horse.Bruse Hosshttps://www.blogger.com/profile/17000725941432226714noreply@blogger.com