@blackpenredpen hi. Would you have any interest in making a video explaining how Wang computers used a lookup table of a few logarithms to compute functions in their calculator from the 1960s? Asianometry made a video about Dr. Wang and his company but didn't discuss the maths. Ty, love your channel.
Wow man that is seriously a great proof. I actually did a proof of this couple weeks ago using the unit circle but gotta say this one’s way more simple and just beautiful overall. Great video man really
I see circular reasoning here. To proof the complex-exponential formulas for trigonometric functions, you need to know their Taylor expansions. To know the Taylor expansions, You need to know, what the derivatives of sin and cos are. To know the derivatives, you need to know at least the sin of the sum formula.
@@maciejkubera1536 NO !!!!!!!!!!! The formula is derivated from Taylor expansion of exp(i*x) and separation of the serie into 2 series composing the real part and imaginary part by grouping the terms with i. The real part corresponds to cos(x) and imaginary part to sin(x) ---> exp(i*x) = cos(x) + i*sin(x)
@@maciejkubera1536 I dont want to detail it because it is well know. Perform Taylor expansion of cos(x) and sin(x) and you will see that they correspond to the real part and the imaginary part. It is like that Euler did !!!
Yes, but how many students know rotation matrices before they know basic trig functions? The proof in the video is much more suitable for a first time learner.
Wow, this is such a beautifully explained proof! Such a nice geometric version of this proof your enthusiasm is contagious-it’s inspiring to see how much you love your work. Great job!
That is so good! I taught high school mathematics for 30 years and always dreaded having to do the traditional proof in front of the class. So messy, so many lines, so easy to make a slip! This de-clutters it. I think there may be a better way, using matrix multiplication, which is perhaps more general. But your neat little diagram is great.
Foarte foarte frumos explicat de D-ul profesor care trăiește intens prin predarea matematicii. Este o soluție foarte elegantă de deducere a acestor formule trigonometrice, explicată consider eu, foarte logic. Felicitări D-le profesor, sănătate și succes în continuare.
The two angle-sum formulas have an extremely familiar form to me, having dealt a lot with applying 2D rotation to a 2D vector, which results in an extremely similar formula (the x of the result being the sum of the original vector components multiplied by cos and sin, and the y of the result being the subtraction of the components multiplied by sin and cos). It's probably not a coincidence, but I can't immediately see the connection.
In the past I have tried to derive the a+b relations by starting with the same two triangles, but was unable to see past that. Thanks for showing something that should have been so simple for me to figure for myself.
You can do it by starting with a unit vector at alpha to the x-axis. The x coordinate is cos(alpha) and the y coordinate is sin(alpha). Now rotate the vector through beta degrees. The new x and y coordinates are cos(alpha +beta) and sin(alpha+beta). These coordinates can easily be calculated.
right yeah . at first the geometrically doable case in which lengths all are positive and angles are sharp. because elementary geometry doesn't work with negative lengths or negative values of angles . we remember the difficulties in a first time meeting the cosine rule in an obtuse angled triangle . think it was in third class of secondary school .
Thanks for posting this. This is the proof I give in my trig classes. It was the proof given to me. It is also a proof found on Khan Academy. I think it is the best geometric proof.
I really enjoy these diagram illustrations of the trig functions. Do you have one that explains the formulae for sin(3Theta), cos(3Theta) and Tan(3Theta)? Also is there a similar for the hyperbolic double angles? thanks muchly!
Hey there, been watching you for a while. I recently got interested in taking a function to another function power, I saw you showed the derivative for it, can you please make a video showing the indefinite integral of it? Thanks.
I remember my sixthform teaching this in y12(A-levels) telling us to memorise it just in case this topic comes up it only appeared once in aqa but never in edexcel.
yeah we could do this ourselves . consider that the picture will of course point in other directions , and that we are forced to calculate with negative lengths which is rather odd in visual (basic) geometry .
My teacher showed me this proof and told after that it is not valid for all alpha as alpha and beta can be larger then pi/2 then he told me that only circle one is valid
I like flexing what I've come up with the euler's formula e^[ix] = (cosx+isinx) --- 1 e^[iy] = (cosy+isiny) --- 2 1×2 e^[ix]×e^[iy] = (cosx+isinx)(cosy+isiny) => e^[i(x+y)] = (cosx+isinx)(cosy+isiny) => cos(x+y)+isin(x+y) = cosxcosy + icosxsiny +isinxcosy + i²sinxsiny => cos(x+y)+isin(x+y) = cosxcosy - sinxsiny + i(sinxcosy + cosxsiny) Comparing real and imaginary part: cos(x+y) = cosxcosy - sinxsiny sin(x+y) = sinxcosy + cosxsiny Fun fact, you can prove the Pythagoras theorem using it, making Pythagoras theorem proven using trigonometry
@Kero-zc5tc I had come up with this when my friend and I were trying to come up with a proof for double angle identify. I had just found out about Euler's identify so I was applying it everywhere lol.
I first saw the proof by using complex numbers, take two complex numbers in trigonometric form and rotate one by other by mutiplying, then compare coefficients. That was pretty cool to me!
But it is not truly a proof... It is a proof only if you take as assumptions either the exponential properties (and Euler Identity) or how to prove those by using power series methods (which is a pain). The only stand alone proofs requiring only geometry are these ones
@@josenobi3022 he means that those assumptions arent basic enough, you have to prove them with actual basic axioms, if i could assume anything, ill just assume x is true and then its true
He is constructing the rectangle and one part of that is to set that length to 1. You will notice that no lengths had any value before he set it to 1. If you still dont like it, you can set it an arbitrary "a" and still get the same proof, just scaled by "a".
Once it has been proven for alpha and beta between 0 and pi/2 (with alpha + beta = pi/2), it then holds for all complex alpha and beta. This is a consequence of the identity theorem in complex analysis
2017 version: th-cam.com/video/OcXqF8l2crI/w-d-xo.htmlsi=v9rleoZwaKVuCy8h
Time flies!
@@blackpenredpen it really does!
@blackpenredpen hi. Would you have any interest in making a video explaining how Wang computers used a lookup table of a few logarithms to compute functions in their calculator from the 1960s?
Asianometry made a video about Dr. Wang and his company but didn't discuss the maths.
Ty, love your channel.
@@JimmyMatis-h9y I also want to have a video like that, plz
Wow what a beautiful and easy to understand explanation.
Wow man that is seriously a great proof. I actually did a proof of this couple weeks ago using the unit circle but gotta say this one’s way more simple and just beautiful overall. Great video man really
Thank you!
My high school teacher proved it using the unit circle too, but I struggle remembering the proof. This is more straightforward to recall.
This is a classical proof.
Use of Euler formula :
exp(i*(a+b)) = cos(a+b) + i*sin(a+b) (1)
exp(i*(a+b)) = exp(i*a)*exp(i*b) = (cos(a) + i*sin(a))*(cos(b) + i*sin(b)) --->
exp(i*(a+b)) = cos(a)*cos(b) + cos(a)*i*sin(b) + i*sin(a)*cos(b) + i*sin(a)*i*sin(b)
exp(i*(a+b)) = cos(a)*cos(b) - sin(a)*sin(b) + i*((cos(a)*sin(b) + sin(a)*cos(b)) (2)
By equating the real parts and the imaginary parts of (1) and (2) ---->
cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b)
sin(a+b) = cos(a)*sin(b) + sin(a)*cos(b)
Yeah this is the only way I know.
I see circular reasoning here.
To proof the complex-exponential formulas for trigonometric functions, you need to know their Taylor expansions.
To know the Taylor expansions, You need to know, what the derivatives of sin and cos are.
To know the derivatives, you need to know at least the sin of the sum formula.
@@maciejkubera1536 NO !!!!!!!!!!!
The formula is derivated from Taylor expansion of exp(i*x) and separation of the serie into 2 series composing the real part and imaginary part by grouping the terms with i.
The real part corresponds to cos(x) and imaginary part to sin(x) --->
exp(i*x) = cos(x) + i*sin(x)
@@WahranRai so How did You know, that the Real part corresponds to cos(x), and the imaginary part to sin(x)?
@@maciejkubera1536 I dont want to detail it because it is well know.
Perform Taylor expansion of cos(x) and sin(x) and you will see that they correspond to the real part and the imaginary part.
It is like that Euler did !!!
I like using matrices. Multiply an angle transformation of alpha by a matrix with an angle transformation of beta.
I tried proving it myself and you're right, it's even faster than the one on the video
Yes, but how many students know rotation matrices before they know basic trig functions? The proof in the video is much more suitable for a first time learner.
the problem with matrices is that it abstracts things away too much -- people do not see the geometry from matrices
@@duckymomo7935 Watch 3blue1brown's Essence of Linear Algebra
@@duckymomo7935 Try watching 3blue1brown's Essence of linear algebra series
Wow, this is such a beautifully explained proof! Such a nice geometric version of this proof your enthusiasm is contagious-it’s inspiring to see how much you love your work. Great job!
Me doing this in a test cuz i can't remember the formulae directly:
Very nice proof. I only knew the proof with Complex Numbers before. Very cool
That's the best proof i've seen for that trigonometric relation!
Wow. Beautiful proof !!
We can also use the scalar and vector products to prove these results.put a video to prove these results by vector method.
That is so good! I taught high school mathematics for 30 years and always dreaded having to do the traditional proof in front of the class. So messy, so many lines, so easy to make a slip!
This de-clutters it. I think there may be a better way, using matrix multiplication, which is perhaps more general. But your neat little diagram is great.
This is beautiful. And if angles are over 90°, there is a way to work with 180° minus the angle.
yeah .
please could you show that way
?
Foarte foarte frumos explicat de D-ul profesor care trăiește intens prin predarea matematicii. Este o soluție foarte elegantă de deducere a acestor formule trigonometrice, explicată consider eu, foarte logic. Felicitări D-le profesor, sănătate și succes în continuare.
i thought this was a reupload lol, i definitely remember watching the 2017 version
Yeah me too
Very cool proof, thanks!
Great explanation, excellent👍
The two angle-sum formulas have an extremely familiar form to me, having dealt a lot with applying 2D rotation to a 2D vector, which results in an extremely similar formula (the x of the result being the sum of the original vector components multiplied by cos and sin, and the y of the result being the subtraction of the components multiplied by sin and cos).
It's probably not a coincidence, but I can't immediately see the connection.
This is the best proof I've ever found to tell 14 years old students in their first contact with trigonometry where does those formulas come from.
Felicitari maxime din ROMANIA!!!
Brilliant and original. 👍
Thanks for sharing. 😉
I needed it just 2 days before but YT recommendations are yk... Drunk .
our teacher taught it i love it
How is this extended to angles bigger than 90 degrees?
1)sin2t=2sintcost
2)cos2t=cos²t-sin²t
(x=y=t)
sin(x+y)=sinx×cosy+siny×cosx.
cos(x+y)=cosx×соsy-sinx×siny.
Very Very Nice, Thanks a lot
In the past I have tried to derive the a+b relations by starting with the same two triangles, but was unable to see past that. Thanks for showing something that should have been so simple for me to figure for myself.
Wow, brilliant proof, bravo, what a nice proof !!
In my opinion, you can use the ruler or metrics for your drawings... But it's a great work for proofing these equations or formulas
Very good explanation, it's actually godsent because I've been searching for a simple proof of this
You can do it by starting with a unit vector at alpha to the x-axis. The x coordinate is cos(alpha) and the y coordinate is sin(alpha). Now rotate the vector through beta degrees. The new x and y coordinates are cos(alpha +beta) and sin(alpha+beta). These coordinates can easily be calculated.
@@rogerphelps9939 Thank you, this one seems good aswell
¡muy buena y bella demostracion!
Great job. Thank you. ❤❤
This is awesome but it only covers restricted alpha and beta. Don't you need a generalisation proof?
right yeah . at first the geometrically doable case in which lengths all are positive and angles are sharp. because elementary geometry doesn't work with negative lengths or negative values of angles . we remember the difficulties in a first time meeting the cosine rule in an obtuse angled triangle . think it was in third class of secondary school .
Does that mean that that we can use this identity's only when alpha + beta < 90 deg because triangles need to fit inside a rectangle?
No, it is valid for any value of α and β but it can be proved by this method only when α+β
To be continued 🎉 if you change beta by -beta you will have two more trigonometric identities of sin (alpha-beta) and cos(alpha-beta) thanks
Thanks for posting this. This is the proof I give in my trig classes. It was the proof given to me. It is also a proof found on Khan Academy. I think it is the best geometric proof.
Tanke you!!!
U are a legend ❤❤
So amazing 🤩
i think this is my favourite proof now
I really enjoy these diagram illustrations of the trig functions. Do you have one that explains the formulae for sin(3Theta), cos(3Theta) and Tan(3Theta)? Also is there a similar for the hyperbolic double angles? thanks muchly!
Hey there, been watching you for a while. I recently got interested in taking a function to another function power, I saw you showed the derivative for it, can you please make a video showing the indefinite integral of it? Thanks.
Brilliant Bprp.
Great demonstration : I wish this was shown to me instead of having to learn it, and obviously to forget it !
the magic of choosing the right geometrical construction for the identity
What about angles larger than pi/2 ?
I also took the same approach but messed up in rectangle.
I'm sorry that this is not related to the video but is it possible to have a general formula for integral of 0 to a of sin/x dx (Not Si(a))
Great demo! IMHO since sin(pi-a)=sin(a), the proof stays valid for angles greater than 90°.
sin(pi-a) is only equal to a for a=0, so I guess you meant sin(pi-a)=sin(a).
@@Apollorion Yes, thanks! I edited my comment.
My favorite proof will always be the complex number one. Where you multiply 2 and compare real and imag parts to get the identities.
Beautiful!
Suuuuupppper
Thanks Teacher ❤
Very nice!
You're the best ❤❤❤❤
can you make next video to
solve 2^(1/3) without calculator (cube root of 2)
I remember my sixthform teaching this in y12(A-levels) telling us to memorise it just in case this topic comes up
it only appeared once in aqa but never in edexcel.
Very nice 👍👍
Ha wow, that's one of the best proof-by-picture proofs I've seen.
Does i = -i???
Very beautiful !
Indeed a beautiful proof. 😀
The "hypanuse".
Very nice proff
My teacher taught me this before only by Vectors
This was actually the proof I was shown first when being taught these formulae
11 days ago? Man what a coincidence, I really needed this for my finals lol
Amazing
Does this proof work if the sum of the angles is greater than 90 degrees?
yeah we could do this ourselves . consider that the picture will of course point in other directions , and that we are forced to calculate with negative lengths which is rather odd in visual (basic) geometry .
"Wait I've already seen this."
* reads description*
"Oh it's a remake nice."
By the way: I love this proof
@@cyorter7737 thank you!
Awesome man
All I remember from a previous bprp video is that the symbol for alpha equals fish
Do 100 trig equation and trig identities in a row
Very good!
Beautiful!
Awesome ❤
such a cool proof
Can anyone tell me hot to find the integral of (cot(sqrt(x)+tan(sqrt(x)) dx. I looked on internet but no solution is present there.
I remember this was the proof taught to us in our A level textbooks
Beautiful
My teacher showed me this proof and told after that it is not valid for all alpha as alpha and beta can be larger then pi/2 then he told me that only circle one is valid
I like flexing what I've come up with the euler's formula
e^[ix] = (cosx+isinx) --- 1
e^[iy] = (cosy+isiny) --- 2
1×2
e^[ix]×e^[iy] = (cosx+isinx)(cosy+isiny)
=> e^[i(x+y)] = (cosx+isinx)(cosy+isiny)
=> cos(x+y)+isin(x+y) = cosxcosy + icosxsiny +isinxcosy + i²sinxsiny
=> cos(x+y)+isin(x+y) = cosxcosy - sinxsiny + i(sinxcosy + cosxsiny)
Comparing real and imaginary part:
cos(x+y) = cosxcosy - sinxsiny
sin(x+y) = sinxcosy + cosxsiny
Fun fact, you can prove the Pythagoras theorem using it, making Pythagoras theorem proven using trigonometry
wow
You wouldn’t have happened to look at the other proof bprp linked in his comment would you?
@Kero-zc5tc ooh I just noticed now that you pointed out
@Kero-zc5tc I had come up with this when my friend and I were trying to come up with a proof for double angle identify. I had just found out about Euler's identify so I was applying it everywhere lol.
@@Amit_Pirate I understand the feeling totally of wanting to use it
That is beautiful
Bravo!
Amazing 🤩
That's the standard high school proof of these identities, as far as I'm concerned.
Interesting Proof.
Well Done. 🙂 👍
I first saw the proof by using complex numbers, take two complex numbers in trigonometric form and rotate one by other by mutiplying, then compare coefficients. That was pretty cool to me!
But it is not truly a proof... It is a proof only if you take as assumptions either the exponential properties (and Euler Identity) or how to prove those by using power series methods (which is a pain). The only stand alone proofs requiring only geometry are these ones
@@Cannongabang "It’s only a proof if you assume this and that" yeah that’s math
@@josenobi3022 he means that those assumptions arent basic enough, you have to prove them with actual basic axioms, if i could assume anything, ill just assume x is true and then its true
@@Cannongabang Still faster than the one on the video, and it's not like it's circular or inconsistent if you use Euler's formula/De'Moivre's theorem
@@Dravignor no he's saying if you do that then you have to prove demoivres theorem which is much harder than just doing it geometrically
Nice proof.
How can you just assume the hypotenuse is 1?
He is constructing the rectangle and one part of that is to set that length to 1. You will notice that no lengths had any value before he set it to 1. If you still dont like it, you can set it an arbitrary "a" and still get the same proof, just scaled by "a".
@mazterlith not that I don't like it but more like "I don't get it" but thank you for helping me out
It makes the math easy with 1. You could make it any arbitrary number and only add a few steps to make the value cancel out.
This would be the way Euclid or Pythagoras would use to prove this trig identity.
Thanks tuzz 🙏
MARAVILLOSO
Proofs through geometry 💘
Once it has been proven for alpha and beta between 0 and pi/2 (with
alpha + beta = pi/2), it then holds for all complex alpha and beta. This is a consequence of the identity theorem in complex analysis
This is exaclty how AOPS “precalculus” proves it 😭😭😭
Thanks bruzz 🙏
Genius
Actually my high school teacher has shown us this proof
Long live sir
this is how i learned it actually :>