Hello again Graeme, thanks for posting these videos! Nowadays I've been busy with tons of things so I hadn't had much time for maths, but after this week, I'll be jumping head into it. I'm currently working on a task, given by my PhysChem teacher, and that is deriving Heron's formula of triangle area from thin air. I've been battling with it for a while, trying all sorts of trigonometric laws and all that. I hope I'll be solving the problem soon. Best wishes, Steve.
+The Skooma Cat G'day, Steve, and lovely to make contact again. You obviously have a bit 'on your plate,' but I hope these recent videos give you some good food for thought. I like Heron's Formula and remember deriving it when I was at school, too ... just privately though, as it was never taught or referred to in my courses. It is a fiddly proof, but start by dividing your triangle into two ... let's say, in the triangle ABC, drop a perpendicular line from C to side c and call it h. The point of intersection divides c into two parts. Call them x and (c - x). Now, write Pythagoras' Theorem for each of the two smaller triangles, eliminate h and then find x in terms of a, b, and c. This should allow you to find h in terms of a, b, and c (this was your ultimate goal ... to create the two triangles with h and x and then eliminate h and x to find further relationships). You can see that the use of Pythagoras' Theorem could, conceivably, give rise to a radical/root sign. Find the area using A = hc/2 and then you are in for a bit of clever rearranging of your algebra. It is a challenging exercise, but the rewards are worth it ... a great feeling of achievement and of treading in the footsteps of the great. It is good to learn that some of their discoveries were not trivial! Best wishes to you, friend.
As these t-Formulae refer to trigonometric functions somehow I find the derivation #2 usig half angle triangle the most accessible and the easiest one to understand. Nevertheless, great job, Graeme ! Valuable explanation of some very, very useful formulae !
+MrVoayer Hello friend, and thank you. They are, indeed, very useful formulae. I like all three derivations and probably favour converting directly from the double angle formulae because I am so familiar with them. But, they say that variety is the spice of life ^^. I think, being able to derive or prove any mathematical principle in a variety of ways gives one a more rounded 'feel' for the material. Warm regards to you!
You are very kind, Danil. Thank you. Unfortunately, however, I have not produced any new videos for at least two years as both my parents fell ill and died ... and a number of other things have been happening also. It will be about another year before I can resume making videos, so there is no urgency. I would be very happy for you to share with your friends. Please don't use artificial means because someone did that to one of my videos some years ago, and UA-cam shut it down :-(. It is lovely meeting kind people via the Internet. Very kind regards to you ... and "thank you." Graeme
very sorry to hear that... your power to kindle fires of people’s minds is absolutely amazing, so i will be patiently waiting for you to come back. i think a lot of us will do) all the best wishes to you and your beloved ones, Danil
When you decide to travel,firt of all you beguin, step by step, then the way appear, it 's your own way, maybe a sky's gift accompan you for a while,but no one will acheive your goal instead of you. thanks lot my friend.
I very much valued your comment on my other video, friend. Would you mind being more explicit about the identity that you would use here, please? Were you suggesting dividing by sin²(θ/2) + cos²(θ/2) and then dividing numerator and denominator by cos²(θ/2)? I agree that that would be a very good way to proceed. Thank you.
@@CrystalClearMaths Yes that is what I mean You have sin(x) = 2sin(x/2)cos(x/2) and we know that 1 = cos^2(x/2)+sin^2(x/2) so let's divide it by this version of one to get 2sin(x/2)cos(x/2)/(cos^2(x/2)+sin^2(x/2)) now we can divide numerator and denominator by cos^2(x/2)
+Syafiq Abdillah Colleges are not always the best places in which to learn, but they provide the certificates and degrees to demonstrate to others (future employers, for example) that we have learned. That is why you pay for college. I hope you continue to find helpful resources on UA-cam to help you through your courses, Syafiq! It is a hard slog, but I wish you well with your studies.
This is a beautiful video and the only video I could see and use for the proof. Thanks for making the proof Crystal Clear😉😉
You are very welcome, Rosette.
So much helpful, you've illuminated my mind on this topic. Thank you!
You are very welcome, UA Games. Thank you for letting me know.
This has been very helpful Sir. Thanks.
I am delighted to learn that my video has helped you. Very kind regards to you, Uzair.
Well done my friend
Thank you, friend.
Hello again Graeme, thanks for posting these videos!
Nowadays I've been busy with tons of things so I hadn't had much time for maths, but after this week, I'll be jumping head into it. I'm currently working on a task, given by my PhysChem teacher, and that is deriving Heron's formula of triangle area from thin air. I've been battling with it for a while, trying all sorts of trigonometric laws and all that. I hope I'll be solving the problem soon.
Best wishes, Steve.
+The Skooma Cat G'day, Steve, and lovely to make contact again.
You obviously have a bit 'on your plate,' but I hope these recent videos give you some good food for thought.
I like Heron's Formula and remember deriving it when I was at school, too ... just privately though, as it was never taught or referred to in my courses.
It is a fiddly proof, but start by dividing your triangle into two ... let's say, in the triangle ABC, drop a perpendicular line from C to side c and call it h. The point of intersection divides c into two parts. Call them x and (c - x). Now, write Pythagoras' Theorem for each of the two smaller triangles, eliminate h and then find x in terms of a, b, and c. This should allow you to find h in terms of a, b, and c (this was your ultimate goal ... to create the two triangles with h and x and then eliminate h and x to find further relationships). You can see that the use of Pythagoras' Theorem could, conceivably, give rise to a radical/root sign.
Find the area using A = hc/2 and then you are in for a bit of clever rearranging of your algebra. It is a challenging exercise, but the rewards are worth it ... a great feeling of achievement and of treading in the footsteps of the great.
It is good to learn that some of their discoveries were not trivial!
Best wishes to you, friend.
Thank you!
You are welcome, Chloe!
As these t-Formulae refer to trigonometric functions somehow I find the derivation #2 usig half angle triangle the most accessible and the easiest one to understand. Nevertheless, great job, Graeme ! Valuable explanation of some very, very useful formulae !
+MrVoayer Hello friend, and thank you.
They are, indeed, very useful formulae. I like all three derivations and probably favour converting directly from the double angle formulae because I am so familiar with them. But, they say that variety is the spice of life ^^.
I think, being able to derive or prove any mathematical principle in a variety of ways gives one a more rounded 'feel' for the material.
Warm regards to you!
So great, appreciate very much what you are doing!!!
You are welcome, Danil.
I am glad that you are enjoying these videos. Best wishes from Australia!
Graeme
Going to help you spread your influence here in Russia!!!:)
You are very kind, Danil. Thank you.
Unfortunately, however, I have not produced any new videos for at least two years as both my parents fell ill and died ... and a number of other things have been happening also. It will be about another year before I can resume making videos, so there is no urgency.
I would be very happy for you to share with your friends. Please don't use artificial means because someone did that to one of my videos some years ago, and UA-cam shut it down :-(.
It is lovely meeting kind people via the Internet. Very kind regards to you ... and "thank you."
Graeme
very sorry to hear that... your power to kindle fires of people’s minds is absolutely amazing, so i will be patiently waiting for you to come back. i think a lot of us will do)
all the best wishes to you and your beloved ones,
Danil
Thank you very much, Danil.
I greatly appreciate your kind thoughts and hope to "come back" soon.
In the meantime, best wishes to you!
Graeme
When you decide to travel,firt of all you beguin, step by step, then the way appear, it 's your own way, maybe a sky's gift accompan you for a while,but no one will acheive your goal instead of you. thanks lot my friend.
Thank you, Mus kamel Ogbi. Kind regards to you.
6:57 In my opinion it would be better if you divide it by one but this one should come from Pythagorean identity
I very much valued your comment on my other video, friend.
Would you mind being more explicit about the identity that you would use here, please?
Were you suggesting dividing by sin²(θ/2) + cos²(θ/2) and then dividing numerator and denominator by cos²(θ/2)?
I agree that that would be a very good way to proceed.
Thank you.
@@CrystalClearMaths
Yes that is what I mean
You have sin(x) = 2sin(x/2)cos(x/2)
and we know that 1 = cos^2(x/2)+sin^2(x/2)
so let's divide it by this version of one to get
2sin(x/2)cos(x/2)/(cos^2(x/2)+sin^2(x/2))
now we can divide numerator and denominator by cos^2(x/2)
@@holyshit922 Thank you, hs.
I appreciate your clarification. It is an excellent suggestion.
why do i pay for college :(
+Syafiq Abdillah Colleges are not always the best places in which to learn, but they provide the certificates and degrees to demonstrate to others (future employers, for example) that we have learned. That is why you pay for college. I hope you continue to find helpful resources on UA-cam to help you through your courses, Syafiq!
It is a hard slog, but I wish you well with your studies.