cos(arcsin(x))
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- Опубликовано: 21 авг 2024
- Simplify cos(arcsin(x))
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I'm sorry but b/a=x/1 does not imply b=x and a=1.
Nevertheless, having b/a=x is enough to conclude.
Indeed, as a^2-b^2=c^2 => 1-(b/a)^2=1-x^2=(c/a)^2 ( a not equal to 0)
and thus cos(arcsin(x))=c/a=√1-x^2
Thank you! 😊
you are being a big help for me !!
thank you so much !
I wish the best for you !!
Thank you Izayoi, I'm very happy to know it :-D
thank you, this method is very useful because i don't understand how my teacher do
😊😊
Several mistakes:
1) Theta could be negative. You have to argue why you can suppose it to be non-negative.
2) Come on, b/a = x/1 does not always imply that b = x and a = 1 !!!!!!!!!!!!!!!!!!! You only can write that b = a.x ... Or you say that you can assume w.l.o.g. that you're working in a right-angled triangle with a = 1 and then obviously b = x.
You completely all right I mean look it this b\a= x\1 gives b=x a=1 that's crazy !!!!!!
Bien vu ! Mais ce qu'il a écrit est juste (vrai) UNIQUEMENT dans un cercle trigonométrique de rayon r=1 (r=hyp). Mais si r > 1, tout deviendra faux
صحيح ما كتبه عندما يكون نصف قطر الدائرة يساوي واحد
Thanks a bunch!! This pretty much explains how you can derive all the identities for trigonometric functions of inverse trigonometric functions :)
;-D Another way to do it can be:
1 = sin^2(theta) + cos^2(theta)
Let theta = arcsin(x) then:
1 = sin^2(arcsin(x)) + cos^2(arcsin(x))
1 = x^2 + cos^2(arcsin(x))
1 - x^2 = cos^2(arcsin(x))
sqrt(1 - x^2) = cos(arcsin(x))
@@IntegralsForYou before I saw this video , I tried this method. When you do the root, don't you have to take the positive and negative values? I stopped because I dind't know wich ones to take.
Sorry if I wrote something but I am italian.
@@piero4200 Ciao, Piero! Prima di tutto, il tuo inglese si capisce come il mio italiano :). Secondo, puoi trovare la risposta a la tua domanda qua => math.stackexchange.com/questions/1118400/trig-substitution-why-can-we-ignore-the-absolute-value
@@IntegralsForYou thank you very much
@@piero4200 You're welcome!
非常感谢您的分享!解决了我的疑惑!
You're welcome! ;-D
Dear god you just saved me from a calculus induced hernia. Thank you
😂😍😎
rly, good stuff
Thanks! 💪
Thank you ❣️❣️❣️
My pleasure! 💪💪
so nice thanks for help
You're welcome! 😊
Thank you, this is amazing. Helped so much❤️
Thanks, Varvara! 😉
thank you so much! It was so easy after all!!
;-D
thankyousomuch!!
You're welcome! 😊
Thank you 🇹🇷🤗
You're welcome, Esra! ;-D
Thanks man great video!
Thanks! 💪
Thanks
My pleasure! ☺
beautifully proved
Thanks! 😊
great help. I wish I found you sooner
Thanks, Jack Stolar! ;-D
thanks bro, i had that watch videos of other lenguague for learn xD, regards from chile!!
Gracias nicoisaac! Si tienes alguna duda siempre puedes preguntarme en el idioma que te apetezca ;-D
Español y audio
Thanks! 💪
Thank u
😊
what???
if we have 3/4 = 6/8 that means that 3=6 and 4 =8 ???
no, but you can fix any value, here a =1 (like the radius of the unit circle), and then the other value is determined
R
THx.....just
;-D
adam adam
❤️😎
وذندن
i can only see a hand