I agree this is the best proof I've seen on you-tube. When people draw circles with unit 1 radius to recreate that same digram, you 'can't see the woods for the trees' . You drew one triangle on top of the other and all of a sudden it is clear as day.
It's great to finally see a video that demonstrates several trig formulas in one diagram only. Not even to mention that if one masters that he can hence easily get a great bunch of related trig identities
I have a good maths teacher but his explanation was really confusing but your maths vid broke it down into tiny steps which made it so much easier to digest thank you👍
I don't care about proofs, but I needed to understand visually why adding or subtracting angle ratios is done thusly and I looked at the proof somewhere else but couldn't understand it... Now I understand exctly and what's happening and I don't even need to memorize anything about compound angles.. I can draw it on the spot and derive what I need even 10 years from now!! Thank you Erik!
nice, I watched other proof with matrix, but I already forgot how to multiplicate with one. This one is good for me, because its geometry based proof 😅
This is fine for 0 < A+B < 90 degrees, but what about obtuse angles when we have to use the circle definitions for trig functions. Do we need another proof or is this one enough?
First I would just remark that this proof by diagram works as long as both A and B are acute, even if 90 < A+B < 180. Next, if you have either or both A and B greater than 90, you can define A' = A - 90n and B' = B - 90m to get A+B = A'+B' + 90(n+m) where A' and B' are both acute and n+m = 1, 2, or 3. (Any multiples of 360 can be subtracted out since the sine and cosine won't change.) Now, sin(C+90) = cos(C), sin(C+180)=-sin(C), and sin(C+270)=-cos(C) and similarly for cosine. (You can verify these by coordinate geometry without having to use the addition formulas.) So, we've reduced the problem of sin(A+B) to -sin(A'+B') or ±cos(A'+B') for which the proof works. Definitely messier, yes. But I don't think you have to rely on a completely different proof.
@@erikthered109 I am glad the proof works out to the appropriate result but I think it needs to be justified why you can place the top triangle on the hypotenuse of the bottom, it's just not clearly justified in your video why the hypotenuse hit the lower triangle should equal cosine of B, if both of these triangles are on the unit circle they should share a hypotenuse of one. so if that middle line was equal to one that would be clear, I'm just confused on this, but as I said the result works out so clearly I'm missing something I just wish you had explained the diagrammatic construction but then again I'm no expert on math or making videos
Hi Anthony, thank you for your comment. I think I can understand why there might be some confusion. Only the top triangle has a hypotenuse of 1; the end of the lower triangle's hypotenuse is not on the unit circle. You may be wondering: given any right triangle on the bottom, how can I then draw a triangle with a hypotenuse of 1 on top of it, and the answer is, I can't always do that, unless the hypotenuse of the lower triangle is less than 1 unit in length. But, if the bottom triangle is too big, I can always scale down the entire diagram until the upper hypotenuse is 1; the new triangles are similar and the angles remain the same. I hope that helps clear things up.
There is a saying that Mathematics is not a hard subject. It is you who is taught by an idiot teacher. And today I found this proof in video. 3 weeks straight I couldn’t figure out this but this single video has eradicated all of the doubts and allowed me to fully underatand this. Thanks
If we're just adding angles, the original and final x,y should be at the same distance from the origin, right? But the second triangle has a hipotenuse of 1 while the first has a hipotenuse of 1cos. Why is that?
really glad I'm not the only one questioning this because the proof ends up working out to the appropriate answer but I also don't understand why cosine of B is the hypotenuse to the lower triangle, if the lower triangle is on the unit circle it's hypotenuse should be equal to one
the reason why is the following. Imagine you want to use this to find the rotation of a vector. Lets say that the vector is the line that the guy from the video said has length cos(B). Forget about cos(B) for now. Lets say that this vector has a length of 69 for example. The vector has an angle of A degrees in respect to the X axis. What we want to find now is what the vector will be if we rotate it by B degrees until we reach an angle of A+B degrees. So, if you look at the unit circle (circle of radius 1) then we can start working from that line that has a length of 1 in this guy's video. What we are looking for when we draw the line that is perpendicular to the line with length 1 is a line that will cut with our original vector which had a lenght of 69 units. The point in which it cuts is obviously shorter than 69 units. What will be the length of the segment that we have from the origin all the way to the point where the cut happens? well, since we're working on the unit circle so that the second line has a length of 1, then the cut happens at a length of cos(B). That is why that line has that length, because it doesnt actually reach all the way to the edge of the unit circle. If we were working with a line of any other length, then the cut would happen at a different point. Imagine that the length was h, then the cut would happen at h*cos(B) which would scale the rest of the operations done in this video by h, which doesnt affect our final result but it makes working with those values more cumbersome until we reach the answer, which is why he chooses to use a length of 1. Hope this wall of text was somewhat understandable and sort of made sense.
The best and most easy and logical proof!!! Thank you 😊
i agree.
I agree this is the best proof I've seen on you-tube. When people draw circles with unit 1 radius to recreate that same digram, you 'can't see the woods for the trees' . You drew one triangle on top of the other and all of a sudden it is clear as day.
It's great to finally see a video that demonstrates several trig formulas in one diagram only. Not even to mention that if one masters that he can hence easily get a great bunch of related trig identities
How clear and beautiful. Thank you!
This was the best video I have seen about this proof. Thank you.
I have a good maths teacher but his explanation was really confusing but your maths vid broke it down into tiny steps which made it so much easier to digest thank you👍
Very good Proof, thank you. I prefer this one rather than the one will a lot of fractions and assumptions, this one is much cleaner and well defined.
I don't care about proofs, but I needed to understand visually why adding or subtracting angle ratios is done thusly and I looked at the proof somewhere else but couldn't understand it... Now I understand exctly and what's happening and I don't even need to memorize anything about compound angles.. I can draw it on the spot and derive what I need even 10 years from now!! Thank you Erik!
It's the simpliest proof I've ever seen. Thank you so much!
That's the best explanation I've seen in years.
Completely brilliant. Thank you so much!
If there are talent of teaching, this is what it is. Thank you so much for easy explanation.
This guy is great
this guy is good. big ups man. this is the simplest proof i've ever seen on this topic
Wow, that's hella intuitive. So intuitive that it was forever written into my gene!
really simple diagram-much easier to understand than other diagrams-thank you
That's so beautifull proof! I thank you for this explanation. I haven't found this evidence in my native language. 🤗
This is easiest way of this proof...you are genuine! Live long!
This channel deserve millions of followers
I went through 4-5 videos ...Yours is BEST
Hey, you were that smart kid in my Calculus 101 class in Ithaca 40 years ago. I see that you found your calling! Well done.
That's me!
Very clear and concise - thank you!
You made my life easier. Thank you.
Simple and very clear. Appreciated
Thank you so much it is so clear and easy to understand
As someone who was struggling to figure out this shit, this video is a godsend for me.
Thank you, Mind-blowing explaination,very clear 😊
Sir this is a very, and I mean very sophisticated proof that's been made so easy to understand, I thank ye
this is a very clever proof thank you
really exceptional n can be a very effective methods all thanks to ERIK. best wishes from NEPAL
best and most easy proof !
super video! thumbs up.
thank you for sharing.
Wonderfull proof wonderfully explained. Thanks
Thank you very
Just a basic knowledge of complex numbers and Euler’s formula makes this proof almost trivial. But fascinating see the traditional real method.
The best video, thank you!
This is brilliant, thanks so much!
Very helpful! Thank you
Your the best maths teacher ever
Perfect explanation!
wonderful proof, thank you.
Good job . Thank you
omg thank you all the other videos complicate it so much!!
Math proofs - the ones I can comprehend anyway ;-) - beautiful; and t/y for this one.
I feel illuminated, thank you sir😁
Very nice.Thank you
Thank u so much.. its very clear..... Awesome 👍
we love you Dr. J
Brilliant🍒💚
Fantastic ,
great job! I appreciate it.
Very nice!
Thank you so much for this awesome video!💯👍🏻
Thank you sir. So clear and beautiful..
It was really cool, THANKS
Thanks a million!
Wow a filty amature like myself could even understand this. great job
Thank you this is helpful
Clear, brilliant.
Thanks a lot!
thank you sir. ➕♾ you made my day.
Awesome.
Thank you very much sir ❤
Clear and concise, thank you
God bless your soul
Excellent
GOAT
Thank you so much my friend.
nice, I watched other proof with matrix, but I already forgot how to multiplicate with one. This one is good for me, because its geometry based proof 😅
Damn he's good.
This is fine for 0 < A+B < 90 degrees, but what about obtuse angles when we have to use the circle definitions for trig functions. Do we need another proof or is this one enough?
First I would just remark that this proof by diagram works as long as both A and B are acute, even if 90 < A+B < 180. Next, if you have either or both A and B greater than 90, you can define A' = A - 90n and B' = B - 90m to get A+B = A'+B' + 90(n+m) where A' and B' are both acute and n+m = 1, 2, or 3. (Any multiples of 360 can be subtracted out since the sine and cosine won't change.) Now, sin(C+90) = cos(C), sin(C+180)=-sin(C), and sin(C+270)=-cos(C) and similarly for cosine. (You can verify these by coordinate geometry without having to use the addition formulas.) So, we've reduced the problem of sin(A+B) to -sin(A'+B') or ±cos(A'+B') for which the proof works. Definitely messier, yes. But I don't think you have to rely on a completely different proof.
@@erikthered109 I am glad the proof works out to the appropriate result but I think it needs to be justified why you can place the top triangle on the hypotenuse of the bottom, it's just not clearly justified in your video why the hypotenuse hit the lower triangle should equal cosine of B, if both of these triangles are on the unit circle they should share a hypotenuse of one. so if that middle line was equal to one that would be clear, I'm just confused on this, but as I said the result works out so clearly I'm missing something I just wish you had explained the diagrammatic construction but then again I'm no expert on math or making videos
Hi Anthony, thank you for your comment. I think I can understand why there might be some confusion. Only the top triangle has a hypotenuse of 1; the end of the lower triangle's hypotenuse is not on the unit circle. You may be wondering: given any right triangle on the bottom, how can I then draw a triangle with a hypotenuse of 1 on top of it, and the answer is, I can't always do that, unless the hypotenuse of the lower triangle is less than 1 unit in length. But, if the bottom triangle is too big, I can always scale down the entire diagram until the upper hypotenuse is 1; the new triangles are similar and the angles remain the same. I hope that helps clear things up.
thank you so much
THANJ YOU SO MCH !!!!!!
Thank you so much this is amazing
There is a saying that Mathematics is not a hard subject. It is you who is taught by an idiot teacher. And today I found this proof in video. 3 weeks straight I couldn’t figure out this but this single video has eradicated all of the doubts and allowed me to fully underatand this. Thanks
Ultimate clear explanation
Really helpful
Love it!
Nice explanation ...thank you sir❤❤
Fantastic...
really good and short and clear
Ooooh thankyou King 🙏🤝🤝just Subscribed❤❤❤
thanks a lot
Thanks for this wonderful video sir
excellent!
OMG 😢 Love you sir
If we're just adding angles, the original and final x,y should be at the same distance from the origin, right? But the second triangle has a hipotenuse of 1 while the first has a hipotenuse of 1cos. Why is that?
really glad I'm not the only one questioning this because the proof ends up working out to the appropriate answer but I also don't understand why cosine of B is the hypotenuse to the lower triangle, if the lower triangle is on the unit circle it's hypotenuse should be equal to one
the reason why is the following.
Imagine you want to use this to find the rotation of a vector. Lets say that the vector is the line that the guy from the video said has length cos(B). Forget about cos(B) for now. Lets say that this vector has a length of 69 for example. The vector has an angle of A degrees in respect to the X axis. What we want to find now is what the vector will be if we rotate it by B degrees until we reach an angle of A+B degrees. So, if you look at the unit circle (circle of radius 1) then we can start working from that line that has a length of 1 in this guy's video. What we are looking for when we draw the line that is perpendicular to the line with length 1 is a line that will cut with our original vector which had a lenght of 69 units. The point in which it cuts is obviously shorter than 69 units. What will be the length of the segment that we have from the origin all the way to the point where the cut happens? well, since we're working on the unit circle so that the second line has a length of 1, then the cut happens at a length of cos(B). That is why that line has that length, because it doesnt actually reach all the way to the edge of the unit circle. If we were working with a line of any other length, then the cut would happen at a different point. Imagine that the length was h, then the cut would happen at h*cos(B) which would scale the rest of the operations done in this video by h, which doesnt affect our final result but it makes working with those values more cumbersome until we reach the answer, which is why he chooses to use a length of 1. Hope this wall of text was somewhat understandable and sort of made sense.
Of all the proofs this has to be the easiest to follow and understand.
Elegant.
Thanks a lot sir
Very nice
thanks bro
thanks!!!!!!!!!!
Nice
thx
It's very easy trick😊😊
nice!
Thenks sir I am Indian
Love uuuuuuu so much sir
Hoped to see more turnips in this video but nice- cleared it up from class
Marshall Graves Chunky Turnips!!!!!