Whenever I'm feeling burnt out from university I always come here, your enthusiasm is contagious!! Thank you so much for all your videos!!

Haven't had the chance to work this out for myself, but I'm curious to get feedback about using this approach: The LHS of the equation is merely a linear combination of a sine and a cosine with the same angular frequency. As such, it can be represented as one sinusoid with an appropriate phase shift as follows: sin(x)+cos(x) = sqrt(2)*sin(x+pi/4) Then solve sqrt(2)*sin(x+pi/4) = 2 using the complex exponential method that was so eloquently derived in this video.

In my head, I'm thinking: cos(x) + sin(x) =2 cos²(x) + 2cos(x)sin(x) + sin²(x) = 4 2cos(x)sin(x) = 3 sin(2x) = 3 2x = arcsin(3) x = 0.5arcsin(3) I hope this is consistent with the video. I also don't know what complex thing arcsin(3) is equal to, but I'm leaving it like this because I can claim it's exact form.

It's somewhat fitting that the real part of the solution involves π/4 + 2πn, since it's these are the real maxima points of the function

14:46 oh so that's what complex exponentials sound like. Good to know!

We can go a little further. (√2 + 1)(√2 - 1) = (√2)^2 - 1^2 = 2 - 1 = 1 so (√2 - 1) = 1 / (√2 + 1) so ln(√2 - 1) = - ln(√2 + 1) so ln(√2 +/- 1) = +/- ln(√2 + 1) So the solution is: x = (π/4 + 2πM) +/- i ln(√2 + 1)

For √i, multiply 2/2 inside square root, you get √(2i)/√2 = √(1+i)^2/√2 = (1+i)/√2.

It can also be done by first squaring both sides, which leads to solving sin2z = 3, which is a straightforward complex trig equation. As a fun alternative to exponentials, one may also solve it using a combination of trig and hyp functions, given sin2z = sin(2x+ 2iy) and the addition formula.

Yes, I am using this at the moment in electronics, where it could be cos(x) + sin(x) = Ψ or cos(x) - sin(x) = Ψ, based on if I am working with the sum or difference of the signals. Where Ψ is the output waveform from two input sine waves at any point in time, using 0° and 90° offset that makes up the real and imaginary parts.

Love the clarity.

What about rewriting cosx+sinx as sqrt(2)cos(pi/4-x)?

حلقة ممتازة دكتور پايام وبارك الله فيك

Same Spiel at 1:38. 😂 Many greetings from Austria.

Complex solutions can make life easier, even it is not clear at first glance. 🙂

I never thought that equation in such a clever way, I study math by my own and this motivates me a lot, thanks and greetings from Perú!!!

I loved this video, I'm excited to see more videos! Lots of thanks for making people to enjoy maths :)

I have no idea how I ended up here, but your captivating energy kept me here for the entire video, and I loved every second of it

the lazy rooster at the end won me over ;-)

More and more crazy. And this is really what I love in maths. Thank you for your enthusiasm

Your energy is so contagious!! I love it

cos(x)+sin(x)=2 1+sin(2x)=4 sin(2x)=3 u=2x sin(u)=3 (e^(iu)-e^(-iu))/2i=3 (e^(iu))^2-6i(e^(iu))-1=0 e^(iu)=i(3(+ or -)2sqrt(2)) iu=i×pi/2+ln(3(+ or -)2sqrt(2)) u=pi/2-i×ln(3(+ or -)2sqrt(2)) x=pi/4-i/2×ln(3(+ or -)2sqrt(2))

I've always wondered how one would define cos and sin for a generic vector. But it's actually pretty obvious, you just need to define the trig of vector a for the vector b. I feel like it would just end up at something we already know and just call by a different name though xD.

Haha 11:47, LN the-generous. BRILLIANT!

In (-1+i)/(1+i) you could also factor out i from the numerator to get (i(1+i))/(1+i) which simplifies to just i

Thanks sir. You always motivate me

this just showed up and i already love ur channel, congrats

Very nice & clear development

I have tried to graph the function and the result is really surprising.. sin((π/4)-(ln(sqrt(2)+-1)i)+(2πD))+cos((π/4)- (ln(sqrt(2)+-1)i)+(2πD))

Sir, this was really great, literally blew my mind!! it was like lambasingi !!!

To solve this problem, I would first divide by sqrt(2), to give sin(x)cos(pi/4)+cos(x)sin(pi/4)=sqrt(2) since sin(pi/4)=cos(pi/4)=1/sqrt(2). Hence sin(z)=sqrt(2) where z=x+pi/4. Thus cos²(z)=1-sin²(z)=-1, and hence, exp(iz)=cos(z)+i sin(z)= i(sqrt(2)+1) or i(sqrt(2)-1). It follows that x=z-pi/4=-i[ log(sqrt(2)+1) + i pi/2 + 2 N pi] - pi/4 =-i log(sqrt(2)+1) + (2 N + 1/4) pi or x=z-pi/4=-i[ log(sqrt(2)-1) + i pi/2 + 2N pi] - pi/4 = -i log(sqrt(2)-1) + (2 N + 1/4) pi for any integer N. Now, (sqrt(2)-1)(sqrt(2)+1)=2-1=1, and hence x = (2N +1/4) pi (+/-) i log(sqrt(2)+1), for any integer N. These are complex conjugate solutions.

Great job Dr. Peyam!👍

I wish my university lecturers had a half of your enthusiasm and energy 🤩

"How could it have a solution?" "Yes"

I think this suggests that a treatment of what the algebra means in terms of geometry would be of great interest.

You can also notice by simple operations that ln(1+sqrt(2))i=ln(1+sqrt(1^2+1))i=arsh(1)i=arcsin(i). So the answer is pi/4+2pim-arcsin(i). Pretty nice, huh. Maybe you have a bit different sign for the arc functions, but I'm russian.

once again i think theres likely an easier way to d this using hyperbolic functions and the identities that relate them to the regular trig functions. not exactly sure what the working out would look like off the top of my head

Dr. Peyam, this was a fun derivation. It is one of those problems that would not normally be considered even in math studies. The idea of using a complex number solution for the equation is novel and your development was interesting. Your later comments about digital signal processing was also a nice addition. This problem would be an interesting question in a PhD oral exam or a MS oral exam to assess how the candidate would solve the problem by casting the solution as possible only with a complex value for “x”. Very nice video. Thank you.

Here is also a nice way to avoid having to deal with logarithms and exponentials. cos(z) + sin(z) = 2 where z = x+iy, x and y element of real numbers. cos(x+iy) + sin(x+iy) = cos(x)cosh(y) - isin(x)sinh(y) + sin(x)cosh(y) + icos(x)sinh(y) = 2 by trig sum identity. cos(iy) = cosh(y) by setting x = iy for cos(x) = 1/2(exp(ix) + exp(-ix)) and same for sin(iy). Factoring, we get cosh(y)(cos(x) + sin(x)) + isinh(y)(cos(x) - sin(x)) = 2 As the right hand side has nothing in terms of i then we know sinh(y)(cos(x) - sin(x)) = 0 (eq1) and cosh(y)(cos(x) + sin(x)) = 2. (eq2) If we set y = 0 then eq1 would be true but eq2 cannot as there is no x that can satisfy that equation as cosh(0) = 1. Therefore we need to find a value of x so eq1 holds which would be x = nπ/4 where n is an integer. Plugging in x = nπ/4 into eq2 we get sqrt(2)cosh(y) = 2 cosh(y) = 2/sqrt(2) y = arccosh(2/sqrt(2)). Therefore, z = nπ/4 + iarccosh(2/sqrt(2)) where n is an integer is the solution.

Perhaps others have written this already, but you can also use the substitution t = tan(x/2). Then cos(x) = (1 - t²) / (1 + t²) and sin(x) = 2·t / (1 + t²). The equation for t is quadratic: 3·t² - 2·t + 1 = 0, whose solutions are t = (1 +- sqrt(2))/3. Then x = 2*arctan(t). Three steps from start to solution.

In your first video in this year you talked about properties of the number 2021. Will you do so with the number 2022 beginning the next year?

Can you help me with a physics equation I've been seeking for years and don't have the math language to create? Starts off simply as follows. A circle around a line where the circle has a point rotating that dictates the tangent of the angle of the line through the center. The rotation of the circle is caused by a 1 dimensional object (just +) interacting with a 2 dimensional (dipole of +&-) object as such that this causes this energetic system to rotate. Now the line would be just a 2 dimensional object (+&-) by itself. When you combine the two, the 3-brane system (1brane+2brane) acts on the 2-brane system which adjusts the vector angle at the center point of the circle. (To what degree can be played with). The line is obviously in some sort of Sin wave and the circle by itself in a void is more like a corkscrew but ends up stablelizing when it has the line through the center. The line I call Es, the 2 I call Et, and the 1 I call Ep. The entire system together is Esys. So that Esys=Ep+Es+Et What I am looking for is the mathematical modal of the path of Es, and the overall vector modal of Esys. It's like a little dimensional piston basicly. Its been driving me crazy for years.

awesome.....the real challenge is to imagine the problem in the first place......the solution is easy once the complex form is used. Another beauty by Doc the Guru 👋 🤪

Well done Dr!

Very helpful indeed. Hope we had this 15 years ago

Hey I signed up of real analysis in the spring and was wondering what it was about. Is it anything like abstract algebra? Thanks guys!!

Cos(x) +sin(x) =sin(x+pi/4)sqrt(2) =2 So Sin(x+pi/4)=sqrt(2) This has solutions in complex numbers

Cos(x) +sin(x) =2 1/2 (cos(x)+sin(x))=1 rad2 /2 (sin(x) + cos(x)= rad 2 Sin(x +pi/4) =rad 2 X +pi/4 = indefinit

In physics we use exponentials (exp(ix) for instance ) because they are easier to manipulate but whenever we want the final answer or something meaningful , we must take the real part of the expressions This idea is used throughout any wavy thing

So cool Really Really good stuff 🙌

"You could use the quadratic formula everywhere, but it's more stylish not to." 🤣🤣🤣

Brilliant!!!

Very Wholesome Video...

wow that great Dr peyam

Square both sides to get 1+sin2x = 4, so sin2x = 3 and work from there. Much simpler.

Yo diria senx^2 + cosx^2 = 1 jajaja, no me imaginaba resolver esto, pero escribiendo al seno y coseno en su forma exponencial salen muchas posiblidades

Dear Sir: At time 7:10 when you completed the square..Could you add a few more steps I do not follow what you did.. Thanks Richard

What an enjoyable problem to see get worked out.

Amazing

How do you solve X raised to the X power equals a particular number?

@14:39 - Where do I learn about how complex exponentials are used in sound processing? I must know. The brain can process sound, so does that mean the brain is processing complex numbers?

5:54 if im wrong someone correct me but i believe Minus i square equals plus one !! He said it equals minus one

What I did was sinx+cosx=2 square both sides (sinx+cosx)^2=2^2 (sinx)^2+(cosx)^2+2sinxcosx=4 using trigonometric identities, 1+sin(2x)=4 sin(2x)=3 x=(arcsin3)/2 I find using the identities simpler to change forms x=-i*ln(3i+2i(sqrt2))/2 x= -i*(ln(i)+ln(3+2sqrt2))/2 x= -i* (i*pi/2+ln(3+2sqrt2))/2 x= (pi/2-i*ln(3+2sqrt2))/2 x= pi/4- i*ln(3+2sqrt2)/2

just saw this on yt home, first reaction: he is a younger version of electroBOOM bruh btw cool video

But why Peyam...because I say so...hilarious, that made me laugh out loud, thank you 🤣

U missed ln(root2+-1)i-pi/4+2pim

Great job, and good "hook" too - "How can (cos(x)+sin(x)==2?????")

This is a nice problem. But the solution is very lengthy and complicated. We may try using half angle formula for solving the equation in few steps as follows: Substitute t = tan(x/2); so that we will end up with the simple Quadratic Eqn 3t^2 - 2t + 1 = 0 and the two solutions will be x = 2arctan ((1 ± isqr2)/3)

i really liked the "but yyyyy peyam...bcs i say so "😬😂😂

could you not solve this with t results or auxiliary method ?

Very Very Interesting!

Dr. Peyam tks for your classes. I would like to share with you an thought about numbers. Root(2) has a representation in a geometric plane but is transcendental mathematicaly speaking. This implies, to me, that the nature of all numbers is exponencial (there is no endding) and our discretization of reality is just special cases, an kind of eigenvalues that not changes after a certain scale. PS: root(2) is irracional, I know, but the thought is the same for transcendental.

"multiples of myself" 2 PI M i. That's a good line

Thank you very much

kuch samjha nahi lekin sunke acha laga :)

The field behinds every clickbait title: complex numbers

It is really great mathematician

cos x + sin x = √2 cos (x-π/4) = 2, cos (x-π/4) = √2 cos x = cosh ix x = ±i(arccosh √2 + π/4)

We're solving sqrt(2) sin(z + pi/4) = 2 or sin(z + pi/4) = sqrt(2). Using arcsin w = - i log(iw +- sqrt(1 - w^2)), this gives z = - pi/4 - i log (i sqrt(2) +- i) = - pi/4 - i [log(i) + log (sqrt(2) +-1)] = - pi/4 + pi/2 + log(sqrt(2) +- 1) = pi/4 + ln(sqrt(2) +- 1) + 2pi m.

Thanks!

14:39 This is why i love maths

Helo Dr Peyam,im 67 years old,from Buenos Aires;Argentina,it is rather difficult for me not only because of my poor english but my brain is going hard rock,but i try ;but i try; i can get no satisfaction,as old rockers Rollings,says. Really,is very nice the maths you teach us. I would like to learn some tutorial of you telling us some of problems resolved for Dr Ramanujuan. Thanks for all ;Dr Peyam ;and what work gave you the DOCTOR state.

Bravo!

very clear

Incorrect solution as x is meaning less. Can you provide x as some range in degrees? (0 - 360)? Logs can't be negative and hence it's very close to 45 degree.

7:21 I think you meant y^2-2(1+i)y+i=0 -> (y-(1+i))^2-(1+i)^2/2+i , NOT (y-(1+i))^2-(1+i)^2+i . Correct me if I'm wrong please

very very cool!!!

When I tried to solve it i didnt use periods because with periods exp and ln are no longer inverses

Is it possible for cos(x)+sin(x)=i? If possible could you show how to do it? Thanks! I love watching your videos.

double angle formula Rsin(alpha+b)=2

Can it reach any complex number?

Would have been easier to use the fundamental equality of trigonometry that sq of cosine plus sq sin of cosine is zero to transform everything into a simple quadratic equation where x is either sin or cosine….

The Complex world is actually complex hahaha nice video Dr Peyam, I liked it a lot!

You could use complex plane to calculate root i but I guess we could do it this way lol

You teach so nice

At angle 45° Sin45°+ cos45°=2 1/√2 +1√2=2 2/√2×√2×√2 2√2/2

Why can't we square LHS and RHS? The procedure can be much simplified in terms of no. of steps...

Maths is more interesting with ppl like u👌🙏❤️

Amazing sir 😘

In Vietnam,I was study cos,sin,tan,co tan ad thk for teach me

Read the thumbnail as sin^2(x)+cos^2(x)=2 and I was like >:( no Fascinating video though. Love complex numbers!