Last year when our physics teacher taught us the definition of sine and cosine (he did it before we learn it in math and he only taught the definitions), the first test we get is to prove these two laws
I love the way you teach things! I accept the rules when I am taught them but proving them makes it soooo much better for learning maths. Thank you so much BPRP
You wanted us to prove the British flag theorem after the video. Here is my attempt. Let the vertical line passing through the red point have a lower segment with length h1 and an upper segment with length h2, such that h1 + h2 = h, and let the horizontal line passing through that same point have a left segment with length w1 and a right segment with w2, such that w1 + w2 = w. (w1)^2 + (h1)^2 = d^2 (w1)^2 + (h2)^2 = a^2 (w2)^2 + (h1)^2 = c^2 (w2)^2 + (h^2)^2 = b^2 a^2 + c^2 = (w1)^2 + (h2)^2 + (w2)^2 + (h1)^2 = (w1)^2 + (h1)^2 + (w2)^2 + (h2)^2 = d^2 + b^2 = b^2 + d^2. Therefore, a^2 + c^2 = b^2 + d^2. Q.E.D.
Your Law of Cosines proof is much, much more simple than the Stewart textbook, thank you! Edit: I am struggling to do this for the other sides, but will get some rest and try again.
Personally, I find BPRP's Law of Cosines proof rather fiddly. Here's my (I think simpler) version: Instead of b1 and b2, notice that b1 is c∙cos A, meaning b2 is (b - c∙cos A) Now apply Pythagoras to each of the two smaller triangles to give the height² : h² = c² - c²∙cos² A = a² - (b - c∙cos A)² = a² - b² + 2bc∙cos A - c²∙cos A The -c²∙cos A term cancels from each side; solve the rest for a²: c² = a² - b² + 2bc∙cos A ⇒ a² = b² + c² - 2bc∙cos A Your mileage, as they say, may vary.
Oops. It might have been clearer if I'd solved for c², as BPRP did - sorry. In that case, you start with b2 = a∙cos C and b1 = (b - a∙cos C) and proceed as before. Sorry.
Be careful, the height can be outside of the triangle. Si you have 2 cases to study. For the 2nd theorem, you can use the scalar product by writing BC^2 = (BA + AC)^2 a^2 = c^2 + b^2 + 2BA.AC a^2=c^2 + b^2 + 2c.b.cos(BA,AC) a^2=c^2 + b^2 + 2c.b.cos(pi - a) Then a^2=c^2 + b^2 - 2c.b.cos(a) Quite easy isn't it? Hello from France, keep going!
actually it comes down to the same equation. if you let b2 to be equal to b+b1, where b1 is the segment outside of the triangle, then you get b1=b2-b => (b1)^2=(b2-b)^2 which is equal to (b-b2)^2 used in the original proof.
I love how mathematics although theoretically universal still has dialects. My teachers would have had a fit of I'd use "sin A", because A is a Point, not an Angle. The angle had to be labeled independently, mostly α for the angle near A, β near B etc.
Here, it's pretty much a convention to use capital letters for angles and small corresponding letters for the sides opposite them in a triangle. This was exactly how we would have done it too. Though ig there's no harm in using alpha, beta either.
I really loved how you gave that extra note that proving Law Of Cosines from pythagorean theorem is not circular reasoning because phythagorean already has like some 1000 proofs. Say a set of theorems, X, is used to prove Pythogorean. So X=> Pythogorean => Law of cosine => Pythogorean
I came here to learn the derivations of these two laws, because I was lazy when I was in class and i missed what my teacher had said about this. Thanks teacher!
Laws of sine/cosine are frequently used in calculus if you take higher course than pre-cal courses so it would be very useful to have some understanding about the proof of them as bluepenredpen illustrates in the video right here. Great representation! Basic geometrical understanding is also a plus to understand this proof!
A very well explained video. I tended to explain things qualitatively as most students have no patience for proofs when they learn this for the first time.
Really cool video. I originally proved law of sine the same way you presented, but in fact our text book provides method that also shows sinA/a = 1/2R where R is radius of a circle with points ABC on its perimeter(I don't know the word for it, English is not my first language). Btw. I enjoyed the bonus exercise. Keep up the good work. Cheers.
Perimeter sounds understandable. Though, maybe you could also just say 'lying on the circle', or speak of a circle going through those three points (after all, a circle is perfectly defined, if three or more points on it are known).
Circumscribed circle or circumcircle. Showing a/sin(A) = D is very simple it clearly doesn't matter which angle you use. I was very pleased with myself when I proved it. I wasn't very old :-)
My favorite proof of the Law of Sines is this: Let BX be a diameter of the circumcircle of triangle ABC. Now consider triangle BXC. It is a right triangle with BC = a, BX = 2R (where R is the radius of the circumcircle), and either angle BXC = A or angle BXC = 180 - A, depending on whether A is acute. Either way, sin(A) = sin(BXC) = a/2R, giving us **2R = a/sin(A)**. By symmetry, we also have b/sin(B) and c/sin(C) equaling 2R. Hence they equal each other.
technically different diagrams need to be used if angles are obtuse. The formulae end up being the same though. for the sine rule: Draw triangle ABC such that angle BAC is obtuse (implying the other two angles are acute) and make point C due east of point A. Draw a perpendicular line from point B down to point D such that points D, A, and C are collinear. Connect D to A. Now we have created another triangle ABD such that angle ADB is a right angle and angle DAB is 180 - angle BAC (supplementary angles on straight line CD) sin(angle BAD) = BD/AB BD = AB*sin(angle BAD) BD = AB*sin(angle BAC) , since sine is positive in the second quadrant Notice we have a bigger triangle: triangle BDC Using this triangle to find sin(angle BCD) sin(angle BCD) = BD/BC BD = sin(angle BCD) * BC => sin(angle BCD)*BC = AB*sin(angle BAC), both were equal to BD Let angle BCD = C Let side AB opposite the angle C be c Let angle BAC = A Let side BC opposite the angle A be a => Sin(C)/c = Sin(A)/a to find a relation to sin(angle ABC), draw a line from point A to a point E such that point E is on the line BC and that AE is perpendicular to BC. let angle ABC = B let side AC opposite angle B be b This implies that in triangle AEC Sin(C) = AE/b , remember b = AC AE = sin(C)*b and in triangle AEB sin(B) = AE/c , remember c = AB AE = sin(B)*c therefore sin(C)*b = sin(B)*c since they're both equal to AE therefore Sin(C)/c = sin(B)/b therefore Sin(A)/a = sin(B)/b = Sin(C)/c given that angle A is obtuse. -------------------------------------------------------------------------------------------------------------- for the cosine rule: use the same construction from sine rule cos(angle BAD) = AD/AB (in triangle BAD) -cos(angle BAC) = AD/AB, (cosine in second quadrant is negative) AD = -AB*cos(A) , to be used for subsitution later on pythagoras on triangle BAD BD^2 + AD^2 = AB^2 BD^2 + (-AB*cos(A))^2 = AB^2 , substituting AD BD^2 + AB^2*(Cos(A))^2 = AB^2 BD^2 = AB^2 - AB^2(Cos(A))^2 , making BD^2 the subject BD^2 = AB^2(1-(Cos(A))^2) , factorising AB^2 as a common factor BD^2 = AB^2*(Sin(A))^2 , unit circle identity (BD to be used as substitution later on) pythagoras on triangle BDC DC^2 + BD^2 = BC^2 (AD+AC)^2 + BD^2 = BC^2 , the whole line DC is comprised of the two lines AD and AC added together (or DA and AC) AD^2 + 2AD*AC + AC^2 + BD^2 = BC^2 , binomial expansion (-AB*cos(A))^2 + 2(-AB*cos(A))*AC + AC^2 + AB^2*(Sin(A))^2 = BC^2 , substituting AD and BD AB^2(cos(A))^2 -2*AB*AC*Cos(A) + AC^2 + AB^2*(Sin(A))^2 = BC^2 AB^2((cos(A))^2 + (Sin(A))^2) - 2*AB*AC*Cos(A) + AC^2 = BC^2 , grouping like terms AB^2 together and factorising BC^2 = AB^2 + AC^2 - 2*AB*AC*Cos(A) , swapping LHS and RHS to make BC^2 the subject as well as rearranging terms. a^2 = b^2 + c^2 - 2bcCos(A) , provided that angle A is obtuse. (AB = c, AC = b, BC = a)
One way to remember the law of cosines: if C is 0, cos C is 1, and c²=(a-b)², because your triangle is lines a and b on top of each other meeting at C and c is the line between the other two ends, so its length is the absolute value of the difference between the other two lengths. So the law of cosines is the square of a difference with a cosine thrown in.
Thank God for this man, I had the hardest time trying to prove the sine law as I couldn't figure out how to relate the top peak of the triangle or "B" with the other angles as h1 wasn't opposite to the angle. My biggest realization was that I should of drawn another parallel line and work out the two sines and relate them back to each other, thus proving the Sine law. :) happy to find this channel and may he keep uploading these awesome videos.
The triangle is drawn nicely so that more than one right angles can be made … In other drawing it’s not so obvious that another right angle can be made with a perpendicular line
F(x) = sum (an(x-x0) ^n) All you need to do is find an. If you differentiate once and evaluate in x0 you find out that a1= f'(x0). Differentiate again and you find out a2=f''(x0) /2. Differentiate again and a3=f'''(x0) /6 and 6=3!. Keep going till a certain number until you notice an=f^n(x0) /n! Substitute that in the original sum and you found out that the coefficient an which best approximates the power series is that one. Plug it in and you got your "MacLovin" series 😂
12 Semitones The law of sines can be reexpressed as a/sin(A) = b/sin(B). Let d = a/sin(A) = b/sin(B). Hence d·sin(A) = a, and d·sin(B) = b. The above implies that (a - b)/(a + b) = [d·sin(A) - d·sin(B)]/[d·sin(A) + d·sin(B)] = [sin(A) - sin(B)]/[sin(A) + sin(B)]. By the sum-to-product identity, sin(A) + sin(B) = 2·sin[(A + B)/2]·cos[(A - B)/2], and this can be proven from the addition of angles identities for sines and cosines and inverting the equations. An analogous identity exists for subtraction. Thus, [sin(A) - sin(B)]/[sin(A) + sin(B)] = (2·sin[(A - B)/2]·cos[(A + B)/2])/(2·sin[(A + B)/2]·cos[(A - B)/2]) = (sin[(A - B)/2]/cos[(A - B)/2])/(sin[(A + B)/2]/cos[(A + B)/2]) = tan[(A - B)/2]/tan[(A + B)/2] = (a - b)/(a + b), or alternatively, tan[(A - B)/2]/(a - b) = tan[(A + B)/2]/(a + b)
Hi, blackpenredpen Thanks a lot for this vid!!! You are Awesome!!! Keep uploading the amazing vids I wanted to ask you If you have any tips for improving My ability to Prove theorems and formulas as I am very weak in the topic!!! Thanks in advance Yours sincerely, A random Subscriber :D
We proved the Law of Sine in a much different and easier way at school. First we used the formula to obtain the measurement of a chord inside a circle through the sinus of the angle it formed multiplied by the diameter itself ( 2r*sin(A)), then as a second proof we used the theorem of the surface of a triangle ( 1/2 * c*b*sin(A) where A is the angle between c and b) and using this formula with all the other sides of the triangle, since the surface is the same, you end up with the law of sine. For the Law of cosine we used a further proof that isn't easy to explain.
Well done. When I was at school the text book showed how to get (SinA)/a = (SinC)/c but did not show us how to get the last part (and the teachers didnt know either) YET the reality is (as you have shown) that it is dead easy ! It goes to show you that a little bit of lateral thinking can sovle many a problem. EDIT : on second thought you have not demonstrated that a perpendicular from (or passing through) AB actually passes through angle C - rather you have merely drawn it that way. (try drawing AB as is but making AC longer and you dont get a perpendicular from AB passing through angle C and hence you dont get two right angled triangles in solving the second part when you extend it to solviong for (Sin B)/b ADDITIONAL EDIT : are the lenghts of AB and AC identical ? And if so then does this work for an iososceles triangle ? EDIT 3 It looks like your method should also work for equilateral triangles (but not for scalene triangles) ?
I feel like the proof for the law of sines is incomplete. Not all triangles have more than one altitude, yet the law of sines holds for all triangles. I think the area proof is much more compelling.
Thanks for sharing! I posted a short video on deriving the law of cosines, It should apply for all angles (acute, obtuse, reflex, negative). Hope to get your thoughts.
For Law of Sine, you proved the equality for any two arbitrary sides. Wouldn't you automatically get that the equality is true for all 3 sides of the same triangle implicitly, given its true for any pair of sides in the same triangle?
By reason of symmetry there is no need to repeat for h2 what was proved for h1. Just relabel letters, or permute them, or use (the groupe) axial symmetry. Not using arguments of symmetry is wrong, it makes the proof heavy, longer, inelegant, redondant and akward. Its a kind of violation of the "Principle of Least Action" (economy). Arguments of symmetry are extremely important since in real problems the complexity gets almost always so strong that there is no chance to solde the problem without using the helpful simplifications braught by the symmetries of the problem.
Sir please can you solve for me: Two friends visit a restaurant during 5pm to 6 pm . Among the two ,whoever comes first waits for 15 min and then leave . The probability that they meet?
![Pythagoras Would Be Proud: High School Students' New Proof of the Pythagorean Theorem [TRIGONOMETRY]](http://i.ytimg.com/vi/p6j2nZKwf20/mqdefault.jpg)
The most power step in geometry/trigonometry: drop the perpendicular line!
I don't think so....
I love the damn video title
@@godson200 it's mean you know nothing about math
The question is: how do you get a perpendicular line through a given one at an arbitrary point with ruler and compass?
@@xCorvus7x I know the answer.....you are not capable of understanding a mathematician like blackpenredpen
Last year when our physics teacher taught us the definition of sine and cosine (he did it before we learn it in math and he only taught the definitions), the first test we get is to prove these two laws
Damn that's depressing
If you thought dropping the bass was cool, just you wait until you drop that perpendicular line.
Exactly!
You know nothing
@@al-quran9845 why man
I love the way you teach things! I accept the rules when I am taught them but proving them makes it soooo much better for learning maths. Thank you so much BPRP
So true BPRP is a brilliant teacher. The best on UA-cam as far as I can see.
You wanted us to prove the British flag theorem after the video. Here is my attempt.
Let the vertical line passing through the red point have a lower segment with length h1 and an upper segment with length h2, such that h1 + h2 = h, and let the horizontal line passing through that same point have a left segment with length w1 and a right segment with w2, such that w1 + w2 = w.
(w1)^2 + (h1)^2 = d^2
(w1)^2 + (h2)^2 = a^2
(w2)^2 + (h1)^2 = c^2
(w2)^2 + (h^2)^2 = b^2
a^2 + c^2 = (w1)^2 + (h2)^2 + (w2)^2 + (h1)^2 = (w1)^2 + (h1)^2 + (w2)^2 + (h2)^2 = d^2 + b^2 = b^2 + d^2.
Therefore, a^2 + c^2 = b^2 + d^2. Q.E.D.
Angel Mendez-Rivera perfect, as always!
I solved it the same way as you did, but you solved it with less steps. Good job, man!
Ah yes, PERFECT timing. Currently learning trigo in school, and I am pretty confident this video will be a great resource for my class!
Hahaha, thank you!
Your Law of Cosines proof is much, much more simple than the Stewart textbook, thank you!
Edit: I am struggling to do this for the other sides, but will get some rest and try again.
Personally, I find BPRP's Law of Cosines proof rather fiddly. Here's my (I think simpler) version:
Instead of b1 and b2, notice that b1 is c∙cos A, meaning b2 is (b - c∙cos A)
Now apply Pythagoras to each of the two smaller triangles to give the height² :
h² = c² - c²∙cos² A = a² - (b - c∙cos A)² = a² - b² + 2bc∙cos A - c²∙cos A
The -c²∙cos A term cancels from each side; solve the rest for a²:
c² = a² - b² + 2bc∙cos A ⇒
a² = b² + c² - 2bc∙cos A
Your mileage, as they say, may vary.
Oops. It might have been clearer if I'd solved for c², as BPRP did - sorry. In that case, you start with b2 = a∙cos C and b1 = (b - a∙cos C) and proceed as before. Sorry.
Be careful, the height can be outside of the triangle.
Si you have 2 cases to study.
For the 2nd theorem, you can use the scalar product by writing BC^2 = (BA + AC)^2
a^2 = c^2 + b^2 + 2BA.AC
a^2=c^2 + b^2 + 2c.b.cos(BA,AC)
a^2=c^2 + b^2 + 2c.b.cos(pi - a)
Then
a^2=c^2 + b^2 - 2c.b.cos(a)
Quite easy isn't it?
Hello from France, keep going!
actually it comes down to the same equation. if you let b2 to be equal to b+b1, where b1 is the segment outside of the triangle, then you get b1=b2-b => (b1)^2=(b2-b)^2 which is equal to (b-b2)^2 used in the original proof.
Very helpful - I have a math exam tomorrow! Thank you so much!
I love how mathematics although theoretically universal still has dialects. My teachers would have had a fit of I'd use "sin A", because A is a Point, not an Angle. The angle had to be labeled independently, mostly α for the angle near A, β near B etc.
Here, it's pretty much a convention to use capital letters for angles and small corresponding letters for the sides opposite them in a triangle. This was exactly how we would have done it too. Though ig there's no harm in using alpha, beta either.
E Rynn what if you used sin(/_ A)? You know, A with the angle symbol?
just had a exam with this ( pre cal , so no proof or anything ) but now know where it comes from
I really loved how you gave that extra note that proving Law Of Cosines from pythagorean theorem is not circular reasoning because phythagorean already has like some 1000 proofs.
Say a set of theorems, X, is used to prove Pythogorean. So X=> Pythogorean => Law of cosine => Pythogorean
I came here to learn the derivations of these two laws, because I was lazy when I was in class and i missed what my teacher had said about this. Thanks teacher!
Laws of sine/cosine are frequently used in calculus if you take higher course than pre-cal courses so it would be very useful to have some understanding about the proof of them as bluepenredpen illustrates in the video right here. Great representation! Basic geometrical understanding is also a plus to understand this proof!
Gonna watch all of your videos in this quarantine
I really love how you say little a little b and little c 😄😄😄
A very well explained video. I tended to explain things qualitatively as most students have no patience for proofs when they learn this for the first time.
INSANE PERPENDICULAR PRANK (GONE WRONG, GONE S*XUAL, POLICE, ALMOST SH*T)
hey man, im from brazil. im watching your calculus's playlist and enjoying it. thanks for your videos, they really help me comprehend all the concepts
Really cool video. I originally proved law of sine the same way you presented, but in fact our text book provides method that also shows sinA/a = 1/2R where R is radius of a circle with points ABC on its perimeter(I don't know the word for it, English is not my first language).
Btw. I enjoyed the bonus exercise.
Keep up the good work. Cheers.
Perimeter sounds understandable.
Though, maybe you could also just say 'lying on the circle', or speak of a circle going through those three points (after all, a circle is perfectly defined, if three or more points on it are known).
Circumscribed circle or circumcircle.
Showing a/sin(A) = D is very simple it clearly doesn't matter which angle you use. I was very pleased with myself when I proved it. I wasn't very old :-)
love your vidoes! You inspired me to make my own maths content, keep up the good work
Absolutely love the enthusiasm 👏🏾thank you for the help sir
I love the daily uploads!
Great video my dude!!! I love watching videos where people can perfectly explain the pythagorean thereom and I learned a lot about law of sine/cosine
This video is awesome. It helped me so much. Thank you. Now I understand it better☺🙏
My favorite proof of the Law of Sines is this: Let BX be a diameter of the circumcircle of triangle ABC. Now consider triangle BXC. It is a right triangle with BC = a, BX = 2R (where R is the radius of the circumcircle), and either angle BXC = A or angle BXC = 180 - A, depending on whether A is acute. Either way, sin(A) = sin(BXC) = a/2R, giving us **2R = a/sin(A)**. By symmetry, we also have b/sin(B) and c/sin(C) equaling 2R. Hence they equal each other.
I like women.
technically different diagrams need to be used if angles are obtuse. The formulae end up being the same though.
for the sine rule:
Draw triangle ABC such that angle BAC is obtuse (implying the other two angles are acute) and make point C due east of point A.
Draw a perpendicular line from point B down to point D such that points D, A, and C are collinear.
Connect D to A.
Now we have created another triangle ABD such that angle ADB is a right angle and angle DAB is 180 - angle BAC (supplementary angles on straight line CD)
sin(angle BAD) = BD/AB
BD = AB*sin(angle BAD)
BD = AB*sin(angle BAC) , since sine is positive in the second quadrant
Notice we have a bigger triangle: triangle BDC
Using this triangle to find sin(angle BCD)
sin(angle BCD) = BD/BC
BD = sin(angle BCD) * BC
=> sin(angle BCD)*BC = AB*sin(angle BAC), both were equal to BD
Let angle BCD = C
Let side AB opposite the angle C be c
Let angle BAC = A
Let side BC opposite the angle A be a
=> Sin(C)/c = Sin(A)/a
to find a relation to sin(angle ABC), draw a line from point A to a point E such that point E is on the line BC and that AE is perpendicular to BC.
let angle ABC = B
let side AC opposite angle B be b
This implies that in triangle AEC
Sin(C) = AE/b , remember b = AC
AE = sin(C)*b
and in triangle AEB
sin(B) = AE/c , remember c = AB
AE = sin(B)*c
therefore sin(C)*b = sin(B)*c since they're both equal to AE
therefore Sin(C)/c = sin(B)/b
therefore Sin(A)/a = sin(B)/b = Sin(C)/c given that angle A is obtuse.
--------------------------------------------------------------------------------------------------------------
for the cosine rule:
use the same construction from sine rule
cos(angle BAD) = AD/AB (in triangle BAD)
-cos(angle BAC) = AD/AB, (cosine in second quadrant is negative)
AD = -AB*cos(A) , to be used for subsitution later on
pythagoras on triangle BAD
BD^2 + AD^2 = AB^2
BD^2 + (-AB*cos(A))^2 = AB^2 , substituting AD
BD^2 + AB^2*(Cos(A))^2 = AB^2
BD^2 = AB^2 - AB^2(Cos(A))^2 , making BD^2 the subject
BD^2 = AB^2(1-(Cos(A))^2) , factorising AB^2 as a common factor
BD^2 = AB^2*(Sin(A))^2 , unit circle identity (BD to be used as substitution later on)
pythagoras on triangle BDC
DC^2 + BD^2 = BC^2
(AD+AC)^2 + BD^2 = BC^2 , the whole line DC is comprised of the two lines AD and AC added together (or DA and AC)
AD^2 + 2AD*AC + AC^2 + BD^2 = BC^2 , binomial expansion
(-AB*cos(A))^2 + 2(-AB*cos(A))*AC + AC^2 + AB^2*(Sin(A))^2 = BC^2 , substituting AD and BD
AB^2(cos(A))^2 -2*AB*AC*Cos(A) + AC^2 + AB^2*(Sin(A))^2 = BC^2
AB^2((cos(A))^2 + (Sin(A))^2) - 2*AB*AC*Cos(A) + AC^2 = BC^2 , grouping like terms AB^2 together and factorising
BC^2 = AB^2 + AC^2 - 2*AB*AC*Cos(A) , swapping LHS and RHS to make BC^2 the subject as well as rearranging terms.
a^2 = b^2 + c^2 - 2bcCos(A) , provided that angle A is obtuse. (AB = c, AC = b, BC = a)
Great video! You should do the nth derivative of f(x)g(x). The answer is really nice!
Thank you for these videos, they are really helpful and you explain everything so well.
Very good and simple explanation.
Thank you from Belgium !
Egill Andersson thank you, too! : )
One way to remember the law of cosines: if C is 0, cos C is 1, and c²=(a-b)², because your triangle is lines a and b on top of each other meeting at C and c is the line between the other two ends, so its length is the absolute value of the difference between the other two lengths. So the law of cosines is the square of a difference with a cosine thrown in.
While im watching your 6 hour video of 100 integrals you drop another video??? i have so much to watch haha i love it
Algebra and Geometry: Who are you?
Trigonometry: I am you but stronger
Thank you, sir.
Love from Bangladesh,
Proof was amazing
I'm from Colombia, i love your videos
Thank God for this man, I had the hardest time trying to prove the sine law as I couldn't figure out how to relate the top peak of the triangle or "B" with the other angles as h1 wasn't opposite to the angle. My biggest realization was that I should of drawn another parallel line and work out the two sines and relate them back to each other, thus proving the Sine law. :) happy to find this channel and may he keep uploading these awesome videos.
You are great sir. Thanks a lot!
love those proof videos, keep doing this stuff please
Fact: Cosine rule helps in proving Heron's Formula.
I like that bpbp outro song! So cool! 😎🤘
Thank you!!
The triangle is drawn nicely so that more than one right angles can be made …
In other drawing it’s not so obvious that another right angle can be made with a perpendicular line
Sir I love you thank you so much
Method El Kashi is best too.
Merci mr Black
Thank you so much!
Love that ending song!
Love the unit circle shirt!
Mathematics - the UNIVERSAL CONSTANT!
Thank you 🙏
這兩個證明如果是鈍角三角形，圖形會有所不同，但算式會一樣。正弦定理我會用三角形面積1/2ab sin C=1/2bc sin A=1/2ca sin B，再同除以abc便得。
I never noticed that before, but yes. Elegant.
Hi,
Please....
.........maclaurin series proof
Ddp What exactly are you asking for?
F(x) = sum (an(x-x0) ^n)
All you need to do is find an.
If you differentiate once and evaluate in x0 you find out that a1= f'(x0). Differentiate again and you find out a2=f''(x0) /2. Differentiate again and a3=f'''(x0) /6 and 6=3!.
Keep going till a certain number until you notice an=f^n(x0) /n!
Substitute that in the original sum and you found out that the coefficient an which best approximates the power series is that one. Plug it in and you got your "MacLovin" series 😂
no, I know,,,
But why is it so....
How that general function comes....
Great explanation!
After hearing Presh's gougu theorem bullshit, it's refreshing to finally hear about Pythagoras again
Nailed it🙌💞
Nice.
Thanks you sir
I hope you will make video soon Thank you.
Please do the proofs for others properties of triangles, like..orthocentre, pedal triangle , circumcentre...etc..
I love the amazing work! I have just recently learned this in my precalculus class! And I am very happy to see the proof and how it all fits together.
nice video
好懷念啊
高二數學的餘弦定理
Can you do the law of tangents and law of cotangents?
12 Semitones The law of sines can be reexpressed as a/sin(A) = b/sin(B). Let d = a/sin(A) = b/sin(B). Hence d·sin(A) = a, and d·sin(B) = b.
The above implies that (a - b)/(a + b) = [d·sin(A) - d·sin(B)]/[d·sin(A) + d·sin(B)] = [sin(A) - sin(B)]/[sin(A) + sin(B)].
By the sum-to-product identity, sin(A) + sin(B) = 2·sin[(A + B)/2]·cos[(A - B)/2], and this can be proven from the addition of angles identities for sines and cosines and inverting the equations. An analogous identity exists for subtraction. Thus, [sin(A) - sin(B)]/[sin(A) + sin(B)] = (2·sin[(A - B)/2]·cos[(A + B)/2])/(2·sin[(A + B)/2]·cos[(A - B)/2]) = (sin[(A - B)/2]/cos[(A - B)/2])/(sin[(A + B)/2]/cos[(A + B)/2]) = tan[(A - B)/2]/tan[(A + B)/2] = (a - b)/(a + b), or alternatively, tan[(A - B)/2]/(a - b) = tan[(A + B)/2]/(a + b)
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I wanted to ask you If you have any tips for improving My ability to Prove theorems and formulas as I am very weak in the topic!!! Thanks in advance
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very nice sir
Nice one
thanks for the amazing content
Keep dropping those perpendicular bombs brah
Area of triangle ABC = 1/2 ab SinC =1/2 bc SinA= 1/2 ca Sin B
Divided all by 1/2 abc we
SinA/a= SinB/b= SinC/c
We proved the Law of Sine in a much different and easier way at school. First we used the formula to obtain the measurement of a chord inside a circle through the sinus of the angle it formed multiplied by the diameter itself ( 2r*sin(A)), then as a second proof we used the theorem of the surface of a triangle ( 1/2 * c*b*sin(A) where A is the angle between c and b) and using this formula with all the other sides of the triangle, since the surface is the same, you end up with the law of sine. For the Law of cosine we used a further proof that isn't easy to explain.
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nice video.....
Great
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i like the outro song btw
this is great! xD
Well done.
When I was at school the text book showed how to get (SinA)/a = (SinC)/c but did not show us how to get the last part (and the teachers didnt know either) YET the reality is (as you have shown) that it is dead easy !
It goes to show you that a little bit of lateral thinking can sovle many a problem.
EDIT : on second thought you have not demonstrated that a perpendicular from (or passing through) AB actually passes through angle C - rather you have merely drawn it that way. (try drawing AB as is but making AC longer and you dont get a perpendicular from AB passing through angle C and hence you dont get two right angled triangles in solving the second part when you extend it to solviong for (Sin B)/b
ADDITIONAL EDIT : are the lenghts of AB and AC identical ? And if so then does this work for an iososceles triangle ?
EDIT 3 It looks like your method should also work for equilateral triangles (but not for scalene triangles) ?
This method is generalized for all triangle
In the case of an obtuse triangle, some of the altitude line goes outside of the vertex of triangle
How small board he used to prove two laws
Do the inverse cosine law (inverse pytagorean of non-right triangles)
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I feel like the proof for the law of sines is incomplete. Not all triangles have more than one altitude, yet the law of sines holds for all triangles. I think the area proof is much more compelling.
True that. And that’s why I said this is only for acute angles.
Thanks for sharing! I posted a short video on deriving the law of cosines, It should apply for all angles (acute, obtuse, reflex, negative). Hope to get your thoughts.
this is a chapter named Properties of triangles in class 11 ISC book::: India this many logical question come from this chapter
ok
For Law of Sine, you proved the equality for any two arbitrary sides. Wouldn't you automatically get that the equality is true for all 3 sides of the same triangle implicitly, given its true for any pair of sides in the same triangle?
There are "weak" and "strong" induction proofs; is there something similar with the Law of Sines?
Could you do it in spherical triangle too?
In NCERT EXEMPLAR book both were proved by vector
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By reason of symmetry there is no need to repeat for h2 what was proved for h1. Just relabel letters, or permute them, or use (the groupe) axial symmetry. Not using arguments of symmetry is wrong, it makes the proof heavy, longer, inelegant, redondant and akward. Its a kind of violation of the "Principle of Least Action" (economy). Arguments of symmetry are extremely important since in real problems the complexity gets almost always so strong that there is no chance to solde the problem without using the helpful simplifications braught by the symmetries of the problem.
Find integral of 1/(cosx+x)
Correct me if I’m wrong. Can the British Flag Theorem be proven with just Pythagoras’ Theorem after dropping some perpendiculars?🤔
Yes
Thanks...The proof was so simple that I almost fooled myself😂.
It means all height of thid triangle no matter what angle is the starting point are equals to each other?
Alkashi was the first mathematician who demonstrated it
Sir please can you solve for me: Two friends visit a restaurant during 5pm to 6 pm . Among the two ,whoever comes first waits for 15 min and then leave . The probability that they meet?
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What about the radios in the boht laws ??
Sir I want Napler analogy one sir plzz tell me that one
I would like to know why is that there is a different result when using both sin and cos law in the same triangle problem