An Interesting Trigonometric Equation | cos(x)=i

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cos(x)=i
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[@chrisjuravich3398]

X=arccos(i). 🎉😂

[@lucas0_03]

Xddd

[@SyberMath]

That's what WA says 😁😜

[@s1ng23m4n]

Ingeniously)))

[@chahitbhardwaj3075]

I think, x = arccos[e^(iπ/2)]😅

[@user-lu6yg3vk9z]

@@SyberMath would replacing sinx with cos(pi/2-x) since sine and cosine are complementary angles

[@matematiqueiro]

This class is absolutely awe-inspiring! Your teaching style is incredibly captivating, making it impossible not to soak up knowledge.

[@SyberMath]

Wow, thank you! 🤩

[@slr150]

Thank you for the video!

[@SyberMath]

My pleasure! Thank you! 🤩

[@mathisnotforthefaintofheart]

Alternatively, if cosx=i then by Pythagoras we have sinx=SQRT{2} and then use Euler's identity....goes very quick

[@s1ng23m4n]

Rewrite cosine with complex exponents. Then make quadratic equation relative to e^ix and solve.

[@SyberMath]

hmm

[@s1ng23m4n]

​@@SyberMath (e^ix + e^-ix) / 2 = i
(e^2ix + 1) / 2 = i * e^ix
e^2ix + 1 = 2i*e^ix
(e^ix)^2 - 2i(e^ix) + 1 = 0 (its quadratic relative to e^ix)
(e^ix - i)^2 + 1 + 1 = 0
e^ix = i +- sqrt(-2)
On your video all the rest)

[@franciscook5819]

I haven't viewed the full solution but ...
cos x = i = e^i(π/2 + 2nπ)
cos²x + sin²x = 1 (even for complex numbers) => sin x = ±√2
e^(ix)=cos x + i sin x = i ± i√2 = i(1 ± √2) = e^i(π/2 + 2nπ)(1 ± √2) then take ln
ix=i(π/2 + 2nπ)+ln(1 ± √2) then times -i
x=π/2 + 2nπ - i ln(1 ± √2)

[@Blaqjaqshellaq]

Hyperbolic trigonometry!

[@juanuribebravo6196]

Yo amo todas las ecuaciones con imaginaros, saludos desde Colombia, i love tahat your videos ❤❤

[@SyberMath]

Thank you!!!

[@matthewfeig5624]

At 7:55, you lost the plus-or-minus in front of sqrt(2). The minus sign leads to a second set of solutions, x = -pi/2 + 2*pi*n + ln(1+sqrt(2))*i.
These solutions are the opposites of the ones you already listed, which fits with cosine being an even function. (In simplifying, it helps to note that 1-sqrt(2) = -1/(1+sqrt(2).)

[@SyberMath]

I just did one of them. I think I said that

[@DonEnsley-yi2ql]

problem
cos x = i
By complex cosine definition (Euler's equation derived)
cos x =(e^ix + e^-ix)/2
(e^ix + e^-ix)/2 = i
{e^ix + 1/[e^ix}/2 = i
Let u = e^ix.
u + 1/u = 2i
u²-2iu + 1 = 0
u = { 2i ± √(4i²-4) } / 2
= { 2i ± 2i√2 } / 2
= i ± i√2
= i (1±√2)
Use the positive value since u = e^ix must be positive.
e^ix = i (1+√2)
Take ln both sides.
ix = ln [i (1+√2)]
= ln(1+√2)+ ln i
= ln(1+√2)+ [ ln 1 + i (π/2+2kπ)]
For integer k.
x = { ln(1+√2)+ [ i π(1/2+2k)] } / i
x = π(1/2+2k) - i ln(1+√2)
k ∈ ℤ

[@JefiKnight]

If you like your angles in degrees, that would be (90+360n-(180/π)ln(1+√2)i)°

[@georget8008]

haven't you forgotten to provide the solutions for x when t=(1-sqrt(2))*I?

[@tunistick8044]

why couln't we write it as:
x=-i(ln((1+√2)i))?

[@BlaqRaq]

But we have 2i before you even started.
I like how rigorous you approach the problem.

[@Qermaq]

Is i the only number where the character is discontinuous?

[@lawrencejelsma8118]

In Electrical Engineering we don't have the cosine and sine of waves having imaginary phase results but rather real angular actual phase. I think Physics problems also. Unless I see a real world application where we aren't using hyperbolic function solutions on e^u where u is not imaginary as real phase change calculations I'll suspend judgement of this mathematics as just exercise of the imaginary beyond. 🤯😂🤣

[@KennethChile]

I need to study. I didn't get it :(

[@zawatsky]

Мнимая единица подразумевает, что квадрат отношения сторон будет равен -1. Для этого одна из сторон должна иметь комплексную длину. Косинус 1 или -1 у углов, где катет равен гипотенузе - 0°, 180°, 360°. Соответственно, косинус будет равен i, если катет iy, а гипотенуза y. Тогда iy/y=i. Второй случай, y/iy=1/i, не подходит. Для 0° и 360° b=c=y. Для 180° b=-y, c=y. b/c=-iy/y=-i=i³=1/i - возвращаемся к неприемлемому варианту. В случае, если b=-y, c=iy, cos(x)=1/(-i)=i⁴/i³=i. Таким образом, это будут некие комплексные углы, соответствующие 0° и 360° в целых числах, а для 180° комплексной должна быть гипотенуза.

[@s1ng23m4n]

Ты такую портянку непонятную накатал))) Реально ничерта не понятно. Может по-проще переформулируешь свои размышления?)))

[@zawatsky]

@@s1ng23m4n попроще это тебе на концерт Милохина, в другую дверь. )

[@s1ng23m4n]

Не увлекаюсь российской эстрадой и понятия не имею о ком ты говоришь. Жаль, что ты все так "в штыки" воспринял. Мог бы потратить 5 минут и объяснить то, что написал и все были бы довольны.

[@s1ng23m4n]

Нагуглил милохина. Видел только его клип с Басковым. Вполне годно. Но остальное творчество наркомана-сироты желания смотреть нет)

[@zawatsky]

@@s1ng23m4n я недоволен. Этого достаточно. Как мудро подметил Кирилл Юрьевич Лавров, если хоть один не согласен - консенсуса нет. Так что упс.

[@rainerausdemspring3584]

At 5:05 you say "Some quadratic equations (polynomials) are not factorable". I assume you are joking, right?

[@SyberMath]

I meant into factors with integer coefficients 😉

[@abdesselambassou3618]

Hi. It will be nice if you consider studying the following function: f(x)+f(T-x)=1 with x in [0,T]

[@Sparky1_1]

Are You Alright Sir You seem a little sick?

[@SyberMath]

Thanks for asking. I sounded weird because of recording 5-6 videos and realizing that the mic was turned off, this was my attempt around 3 am 😜

[@ayuvell4790]

@@SyberMath I assume you usually cut coughing on montage, but I guess you were understandably sleepy to remember it this time. Was the coughing just over dust/dryness or..? Your health is important to us)

[@khundeejai7945]

I stopped at e^ix = (1+-sqrt(2))i 🥴

[@marianne-wt8it]

I stopped at cosx=i 💀💀

[@sriprasadjoshi3036]

@@marianne-wt8it I stopped at 'An interesting trigonometric equation'

[@SyberMath]

I would stop at youtube.com but something kept me going...😜

[@TheNarukman]

You seem to be sick. Please look after yourself.

[@SyberMath]

Thank you! It's more like exhaustion. I don't know if you heard me saying I forgot to turn on the mic when I was recording 5 videos