Wow, just wow. I have never before seen anything like that. You explained everything so nicely. Btw, there is another way to solve cubic equations, and it is called cardano's method. I think you would find it pretty interesting.
And the final answer comes out in terms of closed form functions, which might just give some insight into the physical meaning of the solution in some lucky cases. Would it be possible to alternatively use the Euler formula to solve this using the complex plane?
Yes in fact it works for any nth degree polynomial because the n*theta formulas describe polynomials of the nth degree. But if the degree is higher than five you might not be able to simplify the higher order equation because of the unsolvibility of quintics.
@@dominicellis1867 For higher degree equations there is hypergeometric functions, theta function , etc I solve quartics using difference of two squares like Ferrari did but in modern version of his method but method for cubics can be generalized to quartic (In Euler's book this generalization can be found)
@@karunk7050 Yes, I figured out. While the method is versatile to be sure, there are more cases where this method is really cumbersome. The example in the video is an example where the numbers are chosen "favorably"
because the cubic function and sin function are single function that means f(-x) = -f(x) sin(-x) = - sin (x) always sin refers to the cubic function and cos refers to the quadratic function
What is so hard about it? I think it is beautiful and simple. Have you ever solved a cubic algebraically and noticed how nasty the solutions look… Of course this method is limited to cubic equations with all real solutions, but I think he could possibly generalize it using De Moivre’s theorem, since the real numbers are a subset of the complex numbers.
@@charlesquinn8485 Solving cubic the algebraic way just like we do for quadratic is the most beautiful way of solving it you dont need any theorem when doing so..its so clean getting 3 roots
@@charlesquinn8485 nobody knows the algebraic way actually that’s why they are forced to use theorems :) but I can find the formulas of the 3 roots easily the algebraic way!!
You do not understand how long I've tried to find a video that explains this clearly! Thank you my good Sir!!
Glad it helped! 😊
I know just enough math to be very impressed with this guy, but that's about all.
I've used Cardano's Method, but this trig method I've never seen or heard of before. Thank you!
THAT IS THE CUTEST MATH TEACHER I HAVE EVER SEEN IN MY LIFE
I used cardanos method to find an explicit expression for cos(theta/3) to find the value of cos or sin at pi/9 and pi/18 but this is so brilliant.
But you did not find those values, did you?
@@user-gr5tx6rd4h there’s no explicit expression for cos(pi/9) believe me I’ve tried to find one
@@dominicellis1867 I know, since the angles (10 and 20 degrees) can not be constructed with compass and ruler.
Thank you bro. Best method so far to resolve such equations.
Fantastic video showing how to use trigonometry to solve polynomial equations. Great topic.
The last thing I expected to see when clicking on this vid, was a super jacked dude.
This was just fucking amazing and so much clarity here!Love your lecturing style bro!
Wow, just wow. I have never before seen anything like that. You explained everything so nicely. Btw, there is another way to solve cubic equations, and it is called cardano's method. I think you would find it pretty interesting.
Yes😊
this is a real gem
And the final answer comes out in terms of closed form functions, which might just give some insight into the physical meaning of the solution in some lucky cases. Would it be possible to alternatively use the Euler formula to solve this using the complex plane?
Thanks a lot
Glad you found it useful :)
Amazing!!!
I wonder if you could do this for quartics 🤔
Yes in fact it works for any nth degree polynomial because the n*theta formulas describe polynomials of the nth degree. But if the degree is higher than five you might not be able to simplify the higher order equation because of the unsolvibility of quintics.
@@dominicellis1867 For higher degree equations there is hypergeometric functions, theta function , etc
I solve quartics using difference of two squares like Ferrari did but in modern version of his method
but method for cubics can be generalized to quartic (In Euler's book this generalization can be found)
noice * clicks *
Thanks 😊
Namaste🙏 sirji💐.
Why this method won't work for:
X^3-3x^2-3x-1=0
By letting u=x-1
u^3-6u-6=0
The cos 3 theta value > than 1
Because there are two complex roots in your equation. This method works for cubic equations with three real roots.
@@karunk7050 And even then, it's still possible to find theta.
I didn't understand how he has found cos^3θ-3/4cosθ-1/4cos(3θ). Can anyone explain it?
Are you still interested?
The formula says
cos3θ = 4cos³θ-3cosθ
If you divide both sides with 4, you get 1÷4×cos3θ = cos³θ-3÷4×cosθ
@@A_Random_Ghost Nah I understood it alone some time ago
@@20icosahedron20 Thanks, even though I understood it in the end alone
Where would this method "break down" if the cubic only has one real root? I admit, I have not tried it out
|cos theta| would be > 1
@@karunk7050 Yes, I figured out. While the method is versatile to be sure, there are more cases where this method is really cumbersome. The example in the video is an example where the numbers are chosen "favorably"
may sin or cos be not real ? imaginary for example !!!!!
How come y has to be positive?
Also, why specifically cos? Does sin work?
It's about the cosine because of the trig identity
because the cubic function and sin function are single function that means
f(-x) = -f(x)
sin(-x) = - sin (x)
always sin refers to the cubic function and cos refers to the quadratic function
@@kareemkoka1747 f(-x) = -f(x) has nothing to do with "single function(?)"
@@azzteke
What do you want to say ?
@@kareemkoka1747 What is a "single function" supposed to be??
Hi!
Why? Is it hard to solve it the algebraic way? like quadratic equations? why do it the hard way?
What is so hard about it? I think it is beautiful and simple. Have you ever solved a cubic algebraically and noticed how nasty the solutions look… Of course this method is limited to cubic equations with all real solutions, but I think he could possibly generalize it using De Moivre’s theorem, since the real numbers are a subset of the complex numbers.
@@charlesquinn8485 Solving cubic the algebraic way just like we do for quadratic is the most beautiful way of solving it you dont need any theorem when doing so..its so clean getting 3 roots
@@TruthOfZ0 if you say so 👏 Not everything is as you see it.
@@charlesquinn8485 nobody knows the algebraic way actually that’s why they are forced to use theorems :) but I can find the formulas of the 3 roots easily the algebraic way!!
@@TruthOfZ0 what is the algebraic way?
Indian students also learning from uhh😊
Nah that's too clever dude!
I fell asleep during the intro
The visibility is poor. Your writing on the board