Dear subscribers and viewers, I am continually asked why the right angled triangle OPQ has a hypotenuse 1. My response to this is... Even if it was a 'k' for constant, you'd still end up with the same formula. I guess the people or person who discovered this formula (via the diagram above) recognised this, so out of convenience and for future proofs - they just decided to give the triangle OPQ a hypotenuse 1. And anyway... You've got to remember, when the hypotenuse is 1 ----> sin(A+B)=sin(A)cos(B)+cos(A)sin(B) still pops out. That's still pretty relevant. If you don't believe that you'll end up with the same formula even if you change OPQ's hypotenuse to 'k', try it out... I have done so at least 3 times - therefore in my mind - I'm not worried about OPQ's hypotenuse at all. I hope I've put this issue to bed! :-D Regards, Tiago.
Here is the proof... sin(α+β)=sin(α)cos(β) + cos(α)sin(β). Also know that: sin(-λ)=-sin(λ) And: cos(-φ)=cos(φ) In this case, α=A and β=-B. This means that: sin(A+(-B)) = sin(A-B) = sin(A)cos(-B) + cos(A)sin(-B) = sin(A)cos(B) + cos(A) * {-sin(B)} = sin(A)cos(B) - cos(A)sin(B) Other proof coming shortly...
These are proofs that I had already written down a long time ago. Finding them isn't much of a problem when you understand the proof provided in the video.
Bro thank you. You have somehow explained this better than anyone alive. And I am glad that I actually understand what I have been taught instead of regurgitating rote memorization.
Guys, once again I'd like to give you a big thankyou for all the positive responses. You've motivated me to continue making videos. :-) If you need more maths proofs and would like to continue communicating with me, please visit my "Maths Proofs" group on Google + communities. It's full of diagrams and interesting proofs you won't find in ordinary A Level Mathematics Edexcel books. All the best!
+Saptadip Roy There really is nothing to it... Just pause the video and copy the diagram. The right angles included in the diagram will provide you with clues as to which lines are perpendicular to one another.
+Adorablekyungsoo Hi and thanks for stopping by. sin(A+B) is equal to the length PR (because sin(Ѳ)=O/H) and the length PR is equal to the sum of the lengths PT and TR. Now cos(B)=A/H=PT/PQ=PT/sin(A), therefore sin(A)*cos(B)=PT. PR=PT+TR and this is why an expression for PT is required to produce the famous trigonometric identity sin(A+B)=sin(A)cos(B)+cos(A)sin(B). If you've got any more questions, feel free to leave them below. All the best! :-)
+Adorablekyungsoo What you're essentially trying to do here is prove that PR=PT+TR. You prove this using SOH CAH TOA. Also remember that the length TR is equal to the length QS. This means that PR=PT+QS too. If PR=PT+TR and PR=PT+QS, PT+TR=PT+QS therefore TR=QS.
This has already been explained in many of the comments sections. If you want an even deeper explanation and something to download and keep, you can get my e-book: mathsvideos.net/magick-6-6-fundamental-trigonometric-identities-their-proofs/
This is the only proof I have. It's the one I like most. I would recommend watching the video a few times to let it sink in. If it sinks in properly, you will find other proofs very easy to understand. It's worth the wait and effort. :-)
***** Hi. I don't if there are proofs similar to this one that contain angles greater than 90 degrees. Even if there were you wouldn't require such a proof. You only have to know that sin(-x)=-sinx, cos(-x)=cosx and tan(-x)=-tanx to get useful trigonometric identities.
***** Here is my video about transforming trigonometric identities of the type (A+B)... Transforming Trigonometric Identities Of The Form (A+B) Into (A-B) - Valuable Tip
+Jo n , hi, thanks for stopping by. OQ isn't equal to 1 because the triangle OPQ is a right angled triangle. OQ, logically speaking, cannot be as great in magnitude as OP. Please note that (OQ)^2 + (PQ)^2 = 1^2, therefore the magnitude of OQ is less than 1.
It doesn't matter what you change the hypotenuse to. You'll still get the same result. The hypotenuse is 1 just for convenience. If you'd like to, you can change the value of the hypotenuse and perform an experiment like I did 2 years ago. If you go to the bottom of the page then search for my comments in response to @Sunny Nikhil, you will find the results to that experiment.
If you don't have the time to perform such an experiment, I'll simply say that you end up getting the same formula. Nothing changes. I hope that puts your mind to rest.
triangle OSQ has a hypotenuse of length 1 right? If this is so then shouldn't triangle OPQ be an isosceles triangle with two sides of length 1? That would also mean it's no longer a right angle triangle and everything is now messed up
Hi... Triangle OSQ doesn't have a hypotenuse that is equal to 1. We discovered that its hypotenuse is cos(A). Please look at the diagram carefully once again (2:56). The right angled triangle that DOES have a hypotenuse of 1 is OPQ. If you have any more doubts, feel free to comment.
Please check out my new sin(A-B) proof here... ua-cam.com/video/sTOnQK7NfD0/v-deo.html. You will find that the diagram related to this proof contains a length of 1 which can be found in a different place. I'd say that this is my most detailed proof to date.
This is the sin(A+B) proof, not the sin(A-B) proof. If you want the sin(A-B) proof, click on this link: ua-cam.com/video/sTOnQK7NfD0/v-deo.html. And please note, in the other video OS=1.
And when you're looking at proofs, don't ask why a certain length is what it is. Don't try to control what you perceive. Instead look at what is there. If it turns out that a length is equal to 1, you accept it. That's mathematics, that's science. Now if you're wondering why this proof is what it is, the answer is: It is what it is. Someone performed an experiment and out popped the proof.
This is what is remarkable about maths... People perform these geometrical experiments and proofs pop out of nowhere. Mathematical rules just emerge. Why is it so?? Have you ever thought about that??
Guys i have a question, i totally understand almost everything related to these kind of concepts ONLY when i see it !! Do real mathematicians come up with these kind of axioms ( Like Euclid did in his book Elements of Geometry ) from inside of their heads ?? So the idea is should i feel demoralized because i am not able to interweave these axioms into one another and get the whole picture ? :/
Can Coteli I wouldn't say that mathematicians just come up with these axioms. Axioms usually come about due to our necessity to find them. These theorems come about naturally as we're constantly trying to figure out how the world around us operates. We can trip over discoveries accidentally. Some of the best theories that we have were found accidentally and by coincidence.
Maths Videos For Web Mathematical frameworks continue to evolve and to be adjusted. The maths we use today may change tomorrow. Nothing is set in stone. Mathematics as we know it may become as irrelevant as numerology in the future. We may discover better axioms (easier to understand and far more simple to work with) in the future. That's the beauty of science.
Can Coteli One more piece of advice... Don't do maths in your head. Let your ruler and compass guide you. Let nature reveal to you its secrets. Patterns and mathematical laws will reveal themselves to you if you allow them to. Observe, observe and experiment. Memorising formulas and plugging numbers into them isn't what maths is all about. Maths is more about making discoveries by observing nature and patterns.
There are plenty of calculus and differentiation videos on my channel, but of course, I will make more - so stay tuned!! BTW, check out my playlists and also my website www.mathsvideos.net. There you'll find other proofs and lots of useful information. ;-) "Time is what prevents everything from happening at once." - John Archibald Wheeler
There is another proof of this identity but it 's bit more complicated , i would be greatly grateful of you explain it as it's almost essential for me cuz its the required one in our maths program
You may have to watch the video more than once. Extra tips have also been provided on this page. Please read all the comments on this page carefully, as some of my responses could answer some of your pressing questions. :-)
Hello guys, for those of you who want this proof without the hypotenuse as '1', please download my new document: mathsvideos.net/magick-6-6-fundamental-trigonometric-identities-their-proofs. It costs $2.99 to purchase and some of the proceeds from sales will help me maintain this project and work on more proofs. The more books I can purchase, the more I can develop this channel and MathsVideos.net. BTW, this document is being trialled. If you do purchase it, be sure to leave me some feedback!! Cheers!! :-D Also, if for any reason you are unsatisfied with it - I offer a 30 day money back guarantee.
Aditya Naik Hi, that's an interesting question, however, it has been asked already. You can find my answer to this question below Sunny Nikhil's response on this page. Alternatively, you could re-produce this diagram but give the hypotenuse of the upper right angled triangle your desired value. If you were to do this however, you'd produce exactly the same proof. sin(A+B)=sinAcosB+cosAsinB would pop out once again. :-)
Just watch the video a few times to let it sink in. If you haven't seen many proofs, it may take a while for you to understand why the angle B was used for triangle PTQ and triangle OQS. Remember that: sin(A)=O/H = PQ/1 = PQ.
Most proofs aren't really understood at the first time of asking. That is completely normal. If you haven't understood it yet, well, you're like 99.5% of people. Just be patient with yourself, especially if you aren't accustomed to seeing proofs. That's my advice really. :-)
+govind kurup That isn't true. Angle P and O aren't both equal to B. Firstly, the angle P consists of an undefined angle and the angle B. Secondly, the angle O consists of the angle A and B. You've framed a question which cannot be answered because the assumptions you had made when structuring it are incoherent.
I think what he means is how does angle QPT = B, as well as angle SOQ = B. I've been wondering the same thing. How are they related? Usually I look for parallel lines or other hints, alternate angles, corresponding, cointerior.. What is the relationship/rule that ties these two angles together making them both angle B? You can sort of tell intuitively looking at it but how/why? Thank you !! :)
Angles in a triangle must add up to 180 degrees. If you've got a right angle in a triangle, that is already 90 degrees taken, which leaves only 90 degrees left. If another angle inside this triangle is B, then the last angle inside this triangle has to be 90 degrees minus B, logically speaking. Any more questions?
Also, vertically opposite angles must be equal to one another. If you'd like to know why, please watch my video: Proving that opposite angles are the same in 2 dimensions {ua-cam.com/video/TY49WCEB1BA/v-deo.html}. This video was created on the 3rd January 2014.
More Updates: If you want to be able to derive all the trigonometric identities from scratch, you can now use this bumper playlist: ua-cam.com/play/PLFM03zQeSz2MVj_4R9SZifMTL0re19LhK.html Everything is all in one place. Don't miss it!!
All thing was easy but you forget to prove that how Angle QOR = Angle PTQ = B But i proved it by doing Triangles OHR And PHQ similar by AA rule And rather taking hypoteneus 1 i took x which is cancelled at last THANKS SIR
Hi koteswararao gudla , thanks for commenting. All feedback is welcome. I can derive the formula for you in my next video, however, it would be important to take into account that the most fundamental and deepest mathematical truths are found using proofs. Derivations on the other hand emerge out of a set of predetermined axioms carved out of proofs.
There are other ways to find out what sin(A+B) is without a proof, but in order to construct such a derivation you'd have to recycle a set of predetermined axioms found in a trigonometric proof. If you were to state or claim that your derivation of sin(A+B) {constructed without using a trigonometric proof} is a proof, you'd be committing a "circular cause and consequence" fallacy.
GEOMETRICAL PROOF FOR SIN(A-B) IS NOW ONLINE. YOU WON'T BE ABLE TO FIND A PROOF LIKE THIS ANYWHERE ON THE INTERNET, AS FAR AS I'M AWARE... ua-cam.com/video/sTOnQK7NfD0/v-deo.html
These things usually happen by accident... A mathematician is experimenting with something, and suddenly a geometrical proof pops out. Once a proof is discovered, it then gets shared across the world. I don't know who discovered this geometrical proof, but I would bet good money that it was discovered either via experimenting or just good luck. Hope that helps!
The assumption you've made is incorrect. Point P doesn't have an angle B. B is only part of the angle at point P. You have to look at the diagram carefully otherwise you'll reach the wrong conclusions. Take good care. If you have any more questions, do not hesitate to ask.
The hypotenuse was given the value 1 for the purpose of this proof. It makes the proof possible. The hypotenuse (with the value 1) is related to the right angled triangle OPR and the right angled triangle OPQ.
Other rules I should mention: SOH CAH TOA. sin(x)=O/H, cos(x)=A/H, tan(x)=O/A. If you have any more questions, please feel free to leave your comments below.
Song An I've never come across a different proof for this formula so I won't be able to tell you whether that would be the case. I'd just be speculating if I were to say yes. Why not give it a shot to see what you'd get?
Song An I've tested it out for you. You'd get a similar formula if you were to create such an experiment. You'd get: 2sin(A+B)=2sinAcosB+2cosAsinB=2(sinAcosB+cosAsinB).
if only my teachers could explain it like you. explains thoroughly and doesn't skip any steps.
Dear subscribers and viewers, I am continually asked why the right angled triangle OPQ has a hypotenuse 1. My response to this is... Even if it was a 'k' for constant, you'd still end up with the same formula. I guess the people or person who discovered this formula (via the diagram above) recognised this, so out of convenience and for future proofs - they just decided to give the triangle OPQ a hypotenuse 1. And anyway... You've got to remember, when the hypotenuse is 1 ----> sin(A+B)=sin(A)cos(B)+cos(A)sin(B) still pops out. That's still pretty relevant.
If you don't believe that you'll end up with the same formula even if you change OPQ's hypotenuse to 'k', try it out... I have done so at least 3 times - therefore in my mind - I'm not worried about OPQ's hypotenuse at all.
I hope I've put this issue to bed! :-D
Regards, Tiago.
Brilliant and fairly easy to follow. Could you do the same for Sine (A -B) and Cos( A -B)?
Would have to invent a proof. I'll get back to you. You've given me something to do this week. Give me some time to think about it. :-)
Here is the proof...
sin(α+β)=sin(α)cos(β) + cos(α)sin(β).
Also know that: sin(-λ)=-sin(λ)
And: cos(-φ)=cos(φ)
In this case, α=A and β=-B.
This means that:
sin(A+(-B)) = sin(A-B) = sin(A)cos(-B) + cos(A)sin(-B)
= sin(A)cos(B) + cos(A) * {-sin(B)}
= sin(A)cos(B) - cos(A)sin(B)
Other proof coming shortly...
Also...
cos(α+β)=cos(α)cos(β) - sin(α)sin(β)
Also know that: sin(-λ)=-sin(λ)
And: cos(-φ)=cos(φ)
In this case, α=A and β=-B.
This means that:
cos(A+(-B)) = cos(A-B) = cos(A)cos(-B) - sin(A)sin(-B)
= cos(A)cos(B) - sin(A) * {-sin(B)}
= cos(A)cos(B) + sin(A)sin(B)
These are proofs that I had already written down a long time ago. Finding them isn't much of a problem when you understand the proof provided in the video.
Hey man I appreciate this a shit ton, this is all I've wanted to learn for a while but it's hard to find a teacher for it
Bro thank you. You have somehow explained this better than anyone alive. And I am glad that I actually understand what I have been taught instead of regurgitating rote memorization.
That's fantastic to hear. Glad I helped!
Thank you very much for posting, this is the simplest and therefore most elegant proof of this identity I have ever seen demonstrated.
+James Holbert Thank you sir. These positive comments mean the world to me.
Guys, once again I'd like to give you a big thankyou for all the positive responses. You've motivated me to continue making videos. :-) If you need more maths proofs and would like to continue communicating with me, please visit my "Maths Proofs" group on Google + communities. It's full of diagrams and interesting proofs you won't find in ordinary A Level Mathematics Edexcel books. All the best!
Here's its link: plus.google.com/communities/106007058741903558109.
thanks man i got a trig test tommorow atleast im not gonna complain (a+b) = sinacosb + sinbcosa for no reason
That's good. It's always good to know why these formulas are true. It's satisfying.
Love this video! Teachers told me to just remember the equation when i asked, now I'm definitely going to remember it.
easiest method of proof literally awesone
first time i watch it
and i have became your fan
Can you please explain the process of making the picture
+Saptadip Roy There really is nothing to it... Just pause the video and copy the diagram. The right angles included in the diagram will provide you with clues as to which lines are perpendicular to one another.
Quick question, to find CosB, why did you have to use a new triangle? couldn't one use OR or OS?
+Adorablekyungsoo Hi and thanks for stopping by. sin(A+B) is equal to the length PR (because sin(Ѳ)=O/H) and the length PR is equal to the sum of the lengths PT and TR. Now cos(B)=A/H=PT/PQ=PT/sin(A), therefore sin(A)*cos(B)=PT. PR=PT+TR and this is why an expression for PT is required to produce the famous trigonometric identity sin(A+B)=sin(A)cos(B)+cos(A)sin(B). If you've got any more questions, feel free to leave them below. All the best! :-)
+Adorablekyungsoo What you're essentially trying to do here is prove that PR=PT+TR. You prove this using SOH CAH TOA. Also remember that the length TR is equal to the length QS. This means that PR=PT+QS too. If PR=PT+TR and PR=PT+QS, PT+TR=PT+QS therefore TR=QS.
may I please ask what software or app you are using to create the video
Was using a Wacom graphics tablet. You can use any old program such as Microsoft paint to create videos as such.
Thank you
Why the hypotenuse is 1? where is it given?
This has already been explained in many of the comments sections. If you want an even deeper explanation and something to download and keep, you can get my e-book: mathsvideos.net/magick-6-6-fundamental-trigonometric-identities-their-proofs/
so simply done. Great explanation. thanks.
Cheers. Glad you liked it.
Sir,, who was 1st discovered this method ?
I don't actually know, but it's wonderful. :-)
Mathematics Proofs - GCSE & A Level its ok
Sir can you prove it by using unit circle ? More easily ?
This is the only proof I have. It's the one I like most. I would recommend watching the video a few times to let it sink in. If it sinks in properly, you will find other proofs very easy to understand. It's worth the wait and effort. :-)
I understood pretty well. Thank you very much. Waving from Mexico.
Thankyou. Have an awesome day. :-)
Proof made easy
Great going SIR
Like if you agree.
i finally can see this! hours wracking my brain, short and sweet from you = lightofday!
so thank you!!
is there a proof for this that involves angles greater than 90 degrees? or is the trigonometry the same in that situation?
***** Hi. I don't if there are proofs similar to this one that contain angles greater than 90 degrees. Even if there were you wouldn't require such a proof. You only have to know that sin(-x)=-sinx, cos(-x)=cosx and tan(-x)=-tanx to get useful trigonometric identities.
***** Here is my video about transforming trigonometric identities of the type (A+B)... Transforming Trigonometric Identities Of The Form (A+B) Into (A-B) - Valuable Tip
Excellent video! Your explanation is superb.
Cheers!
why is the hypotenuse 1 sir?
1 unit circle. if OP = 1, OQ should be 1 as well, correct? why don't replace OQ with 1 in the equation?
+Jo n , hi, thanks for stopping by. OQ isn't equal to 1 because the triangle OPQ is a right angled triangle. OQ, logically speaking, cannot be as great in magnitude as OP. Please note that (OQ)^2 + (PQ)^2 = 1^2, therefore the magnitude of OQ is less than 1.
Now, what OQ can be is cosA. This is because cosA=A/H=(OQ)/1=OQ
I am from india capital delhi could you explain cos(A-B ) ? Where from you?
Why did you consider,h( PO):Hypoteneus is 1; it is just assumptions or any valid logic.I hope that I will get reply from you soon.
It doesn't matter what you change the hypotenuse to. You'll still get the same result. The hypotenuse is 1 just for convenience.
If you'd like to, you can change the value of the hypotenuse and perform an experiment like I did 2 years ago.
If you go to the bottom of the page then search for my comments in response to @Sunny Nikhil, you will find the results to that experiment.
If you don't have the time to perform such an experiment, I'll simply say that you end up getting the same formula. Nothing changes. I hope that puts your mind to rest.
Thanks for sharing this, extremely interesting and helpful.
Cheers! Have an awesome day!
if A+B < 90 and A > 0 and B > 0 then this proof is ok but how about if A+B > 90 or A < 0 or B < 0 ?
Very simple and appropriate. Appreciate it 😀
You're welcome. :-)
triangle OSQ has a hypotenuse of length 1 right? If this is so then shouldn't triangle OPQ be an isosceles triangle with two sides of length 1? That would also mean it's no longer a right angle triangle and everything is now messed up
Hi... Triangle OSQ doesn't have a hypotenuse that is equal to 1. We discovered that its hypotenuse is cos(A). Please look at the diagram carefully once again (2:56). The right angled triangle that DOES have a hypotenuse of 1 is OPQ. If you have any more doubts, feel free to comment.
The right angled triangle OPR also has a hypotenuse of 1.
Mathematics Proofs - GCSE & A Level ahh right. sorry my bad. I thought OSQ was following the unit circle, thus the hypotenuse of 1.
No worries. :-)
Please check out my new sin(A-B) proof here... ua-cam.com/video/sTOnQK7NfD0/v-deo.html. You will find that the diagram related to this proof contains a length of 1 which can be found in a different place. I'd say that this is my most detailed proof to date.
can you explain why the length OQ is not 1
This is the sin(A+B) proof, not the sin(A-B) proof. If you want the sin(A-B) proof, click on this link: ua-cam.com/video/sTOnQK7NfD0/v-deo.html. And please note, in the other video OS=1.
And when you're looking at proofs, don't ask why a certain length is what it is. Don't try to control what you perceive. Instead look at what is there. If it turns out that a length is equal to 1, you accept it. That's mathematics, that's science. Now if you're wondering why this proof is what it is, the answer is: It is what it is. Someone performed an experiment and out popped the proof.
This is what is remarkable about maths... People perform these geometrical experiments and proofs pop out of nowhere. Mathematical rules just emerge. Why is it so?? Have you ever thought about that??
Great video, easy to understand. Thank you
Cheers!
how is angle opq b,Since angle tqo shd be b,Which means angle tqo be b,And why even draw this triangle ptq
very pretty. nice share.
Guys i have a question, i totally understand almost everything related to these kind of concepts ONLY when i see it !! Do real mathematicians come up with these kind of axioms ( Like Euclid did in his book Elements of Geometry ) from inside of their heads ?? So the idea is should i feel demoralized because i am not able to interweave these axioms into one another and get the whole picture ? :/
Can Coteli I wouldn't say that mathematicians just come up with these axioms. Axioms usually come about due to our necessity to find them. These theorems come about naturally as we're constantly trying to figure out how the world around us operates. We can trip over discoveries accidentally. Some of the best theories that we have were found accidentally and by coincidence.
Maths Videos For Web Mathematical frameworks continue to evolve and to be adjusted. The maths we use today may change tomorrow. Nothing is set in stone. Mathematics as we know it may become as irrelevant as numerology in the future. We may discover better axioms (easier to understand and far more simple to work with) in the future. That's the beauty of science.
Watch this scene about how displacement was discovered... ua-cam.com/video/OGKPmBtBpBo/v-deo.html
Can Coteli One more piece of advice... Don't do maths in your head. Let your ruler and compass guide you. Let nature reveal to you its secrets. Patterns and mathematical laws will reveal themselves to you if you allow them to. Observe, observe and experiment. Memorising formulas and plugging numbers into them isn't what maths is all about. Maths is more about making discoveries by observing nature and patterns.
wonderful bro never seen a easier proof
Cheers! I'm happy you like it. :-D
bro us ure r amazing
can u make videos on claculus and diffrentials
There are plenty of calculus and differentiation videos on my channel, but of course, I will make more - so stay tuned!! BTW, check out my playlists and also my website www.mathsvideos.net. There you'll find other proofs and lots of useful information. ;-)
"Time is what prevents everything from happening at once." - John Archibald Wheeler
This was actually really helpful thank you so much
You're more than welcome. Thanks for having stopped by. :-)
Ouststanding. Minor suggestion is to avoid calling both the adjacent side and the angle a by the same capital letter A. Thanks for the great video.
Beautiful proof thank you
There is another proof of this identity but it 's bit more complicated , i would be greatly grateful of you explain it as it's almost essential for me cuz its the required one in our maths program
You may have to watch the video more than once. Extra tips have also been provided on this page. Please read all the comments on this page carefully, as some of my responses could answer some of your pressing questions. :-)
Thank you very much sir....
We have considered OP as 1. I feel that result may vary if we take actual value of OP instead of considering 1.
You'll get the same result. You're speculating. If you want to see what happens, perform an experiment. Don't just make assumptions.
when you learn the area proof with sin a , you can easily proof this sin(a+b) thing but this one is good and rough.
Hello guys, for those of you who want this proof without the hypotenuse as '1', please download my new document: mathsvideos.net/magick-6-6-fundamental-trigonometric-identities-their-proofs. It costs $2.99 to purchase and some of the proceeds from sales will help me maintain this project and work on more proofs. The more books I can purchase, the more I can develop this channel and MathsVideos.net. BTW, this document is being trialled. If you do purchase it, be sure to leave me some feedback!! Cheers!! :-D Also, if for any reason you are unsatisfied with it - I offer a 30 day money back guarantee.
Wonderfully explained. Many thanks
Vishweshwer Mangalapalli Thankyou. :-)
If you need the sin(A-B)=sinAcosB - cosAsinB proof by any chance, please watch one of my latest videos... ua-cam.com/video/I_mGCUjCaQ8/v-deo.html.
What if the hypotenuse were to be greater than one?
Aditya Naik Hi, that's an interesting question, however, it has been asked already. You can find my answer to this question below Sunny Nikhil's response on this page. Alternatively, you could re-produce this diagram but give the hypotenuse of the upper right angled triangle your desired value. If you were to do this however, you'd produce exactly the same proof. sin(A+B)=sinAcosB+cosAsinB would pop out once again. :-)
Yes most certainly.Many thanks. ..
Brilliantly done!
Awesome, glad you liked it!
Very tough ! Is it not possible to more simply showing it ?
What's tough about it? I think most visitors enjoyed watching this video. Everything has already been simplified.
If there's something you haven't understood, feel free to pop me a question. :-)
This derivation is overly complex. Check out Linda Green's derivation. It is simple and it gives both sin and cos(a+B).
I am cofused that you have taken angle b in both trianle PTQ and triangle OQS
Just watch the video a few times to let it sink in. If you haven't seen many proofs, it may take a while for you to understand why the angle B was used for triangle PTQ and triangle OQS. Remember that: sin(A)=O/H = PQ/1 = PQ.
Most proofs aren't really understood at the first time of asking. That is completely normal. If you haven't understood it yet, well, you're like 99.5% of people. Just be patient with yourself, especially if you aren't accustomed to seeing proofs. That's my advice really. :-)
how does angle P and O both equal B?
+govind kurup That isn't true. Angle P and O aren't both equal to B. Firstly, the angle P consists of an undefined angle and the angle B. Secondly, the angle O consists of the angle A and B. You've framed a question which cannot be answered because the assumptions you had made when structuring it are incoherent.
I think what he means is how does angle QPT = B, as well as angle SOQ = B. I've been wondering the same thing. How are they related? Usually I look for parallel lines or other hints, alternate angles, corresponding, cointerior.. What is the relationship/rule that ties these two angles together making them both angle B? You can sort of tell intuitively looking at it but how/why? Thank you !! :)
Angles in a triangle must add up to 180 degrees. If you've got a right angle in a triangle, that is already 90 degrees taken, which leaves only 90 degrees left. If another angle inside this triangle is B, then the last angle inside this triangle has to be 90 degrees minus B, logically speaking. Any more questions?
Also, vertically opposite angles must be equal to one another. If you'd like to know why, please watch my video: Proving that opposite angles are the same in 2 dimensions {ua-cam.com/video/TY49WCEB1BA/v-deo.html}. This video was created on the 3rd January 2014.
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All thing was easy but you forget to prove that how Angle QOR = Angle PTQ = B
But i proved it by doing Triangles OHR And PHQ similar by AA rule
And rather taking hypoteneus 1 i took x which is cancelled at last
THANKS SIR
love you math
Thnk u very much undersandable
You're welcome. :-)
this is beautiful
I totally agree. Of course, I didn't come up with the proof; I'm just sharing it with the world. The way things click is beautiful.
yes it is a proof, actually we need derivation which is done earlier by the person who invented this formula, could you please derive this for us.
Hi koteswararao gudla , thanks for commenting. All feedback is welcome. I can derive the formula for you in my next video, however, it would be important to take into account that the most fundamental and deepest mathematical truths are found using proofs. Derivations on the other hand emerge out of a set of predetermined axioms carved out of proofs.
There are other ways to find out what sin(A+B) is without a proof, but in order to construct such a derivation you'd have to recycle a set of predetermined axioms found in a trigonometric proof. If you were to state or claim that your derivation of sin(A+B) {constructed without using a trigonometric proof} is a proof, you'd be committing a "circular cause and consequence" fallacy.
thank you
+ALLAH KI BATAIN You're welcome. Have a great day. :-)
It's too good
But why the angle QOS = angle QPT
I have an ebook that will explain everything. Let me know if you want the link to buy it. It's $2.99, 30 day money back guarantee.
GEOMETRICAL PROOF FOR SIN(A-B) IS NOW ONLINE. YOU WON'T BE ABLE TO FIND A PROOF LIKE THIS ANYWHERE ON THE INTERNET, AS FAR AS I'M AWARE... ua-cam.com/video/sTOnQK7NfD0/v-deo.html
how we got the triangle first
These things usually happen by accident... A mathematician is experimenting with something, and suddenly a geometrical proof pops out. Once a proof is discovered, it then gets shared across the world. I don't know who discovered this geometrical proof, but I would bet good money that it was discovered either via experimenting or just good luck. Hope that helps!
Thank you sir for your kind reply.
How if A+B>90??
+Lina Dwi Khusnawati Hi. In this diagram (A+B) isn't greater than 90 degrees. (A+B) is an acute angle. :-)
excellent video
Sophie NEGUS Thankyou!
Thank youu !!
why the hypotenuse is 1
Just out of convenience. This question has been answered twice already. :-) If you read the responses on this page, you'll get your answer. ;-)
Thanks a lot
Thank you!
Siebrand Romeyn You're welcome!
why angle p is B
The assumption you've made is incorrect. Point P doesn't have an angle B. B is only part of the angle at point P. You have to look at the diagram carefully otherwise you'll reach the wrong conclusions. Take good care. If you have any more questions, do not hesitate to ask.
Thanks
Cheers!
why you took hypotenus as 1
The hypotenuse was given the value 1 for the purpose of this proof. It makes the proof possible. The hypotenuse (with the value 1) is related to the right angled triangle OPR and the right angled triangle OPQ.
Other rules I should mention: SOH CAH TOA. sin(x)=O/H, cos(x)=A/H, tan(x)=O/A. If you have any more questions, please feel free to leave your comments below.
Maths Videos For Web So if the hypotenuse wasn't given the value 1, it would still be possible to prove the equation but just takes more time, right?
Song An I've never come across a different proof for this formula so I won't be able to tell you whether that would be the case. I'd just be speculating if I were to say yes. Why not give it a shot to see what you'd get?
Song An I've tested it out for you. You'd get a similar formula if you were to create such an experiment. You'd get: 2sin(A+B)=2sinAcosB+2cosAsinB=2(sinAcosB+cosAsinB).
oh youre genius!!!
I wish, but thanks. :-D
Vaasi👌
nice :)
Hi
Sir pls keep easy proof then this