It was at exactly 16:07 I realised where the video was ultimately heading to, and lo and behold when I saw Heron's formula at the end it was very satisfying to know I saw it coming!
In the Heron formula calculation, ain't it simpler to go through: sin C = √sin² C = √(1 - cos² C) = √((1 + cos C)(1 - cos C)) = ... instead of the double-angle formulae?
Yes, this is much quicker and simpler! Originally, I thought about first deriving the formulae for sin(C/2) and cos(C/2) in terms of the semi perimeter as interesting results on their own (see e.g. here www.cuemath.com/jee/semiperimeter-and-half-angle-formulae-trigonometry/ ), then Heron's formula would follow pretty much immediately. But yes, this approach is quite inefficient, and doesn't make much sense without the extra results!
At 7:49 you give a formula for cos C, from the cosine rule, which is in fact the negative of cos C. Fortunately this didn't matter since you use it to find sin C as sqrt(1 - cos²C), thus squaring that negative out of harm's way. And of course sin C can only be the positive square root, since 0
the maths involved was spot on, no mistakes and clear on the board using recognised symbols, there was no ambiguity and presumably you understood what he meant? so no problem here apart from your own unnecessary criticism afaics
Almost everyone after their first hour or two of trig uses these standard shortened forms. Next you'll be wanting hyperbolic cotangent said instead of coth.
I was wondering if you’d derive Heron’s formula, very glad you got there at the end!
It was at exactly 16:07 I realised where the video was ultimately heading to, and lo and behold when I saw Heron's formula at the end it was very satisfying to know I saw it coming!
a masterful derivation of Heron's formula. well done Dr. Barker!
Excellent derivation of Heron's Formula!
In the Heron formula calculation, ain't it simpler to go through:
sin C = √sin² C = √(1 - cos² C) = √((1 + cos C)(1 - cos C)) = ...
instead of the double-angle formulae?
Yes, this is much quicker and simpler!
Originally, I thought about first deriving the formulae for sin(C/2) and cos(C/2) in terms of the semi perimeter as interesting results on their own (see e.g. here www.cuemath.com/jee/semiperimeter-and-half-angle-formulae-trigonometry/ ), then Heron's formula would follow pretty much immediately. But yes, this approach is quite inefficient, and doesn't make much sense without the extra results!
At 7:49 you give a formula for cos C, from the cosine rule, which is in fact the negative of cos C. Fortunately this didn't matter since you use it to find sin C as sqrt(1 - cos²C), thus squaring that negative out of harm's way. And of course sin C can only be the positive square root, since 0
Well-spotted! I'm very used to the standard labelling cos(A) = (b^2 + c^2 - a^2)/2bc, so switching the letters around wasn't a good idea!
It's nice to see that theres is always some basic math to be learned.
This Heron's formula works also for degenerated triangles
For segment length triplets which cannot form triangle this formula gives imaginary result
10/10 no notes:)
@Professor Barker
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👎🏽 for saying “coz” and “cot” instead of “cosine” and “cotangent”. Good video otherwise.
the maths involved was spot on, no mistakes and clear on the board using recognised symbols, there was no ambiguity and presumably you understood what he meant? so no problem here apart from your own unnecessary criticism afaics
Almost everyone after their first hour or two of trig uses these standard shortened forms.
Next you'll be wanting hyperbolic cotangent said instead of coth.
A frustrated linguist unleashing his intellect on high school trig. Hilarious.
Was I the only one who noticed that he said SEC at one point...."in a sec"......short for 'second' apparently😅
@@DBbbbbbbbbbbbb9248 Standard slang - certainly in the UK