the sine triangle problem
ฝัง
- เผยแพร่เมื่อ 2 พ.ย. 2023
- Can we find x so that we have sin(x), sin(2x), and sin(3x) on a right triangle? Yes! I actually have done the log triangle, exp triangle, so today this is the sine triangle for you! This is a perfect problem to challenge any precalculus or even calculus student!
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Thank you all!
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Check out the log triangle problem:
th-cam.com/video/CMdJPwEbE8A/w-d-xo.html
I like the sin triangle way better
Please make video a day life of yourself
*You should do a Lambert W triangle where the sides of the right triangle are W(x), W(2x) and W(3x).*
try cos triangle
@@mr.d8747 its not possible to do it algebraically
The hardest part of maths is to explain why we like it.
Seriously. People ask me all the time why I like math so much. I can never give an answer that I'd consider satisfactory.
I’ve never agreed with a statement so much
The dopamine of understanding. The structures and surprises which build on simple rules.
Math is a sandbox for logical reasoning. Unlike reasoning applied to philosophical questions(also an enjoyable endeavor) we can determine conclusively the accuracy of our reasoning in that the outcomes are known. One of the reasons why I like it. But it's a multifaceted appreciation for sure.
@@hybmnzz2658 the kick in the discovery ~ Richard Feynman
Sudden existential crisis??
Actually yeah, it's super cool.
At the end, Bro was wondering if it was him who did all those things on board😅😂
heartaches😃🤤
was something wrong there or what was that ?
@@Mr23143sir nothing was wrong, he just had some different outro plan
Oh, thanks for clarification then @@guy_with_infinite_power
This man just went
('-')
/|\.
This is like ASMR math, just slowly solving the problem and appreciating its elegance ❤
This is softcore ASMR 3b1b is heavy hard core ASMR💀
Glad you enjoy it!
@@canyoupoop😂
@@blackpenredpen And the triangle itself turns out to be 30°, 60°, 90° right triangle.
@@Jack_Callcott_AUguess check is ez
I love how he's just so mesmerized he couldn''t talk at the end of the video lmao
What's cool at the end is that the reference triangle you drew in the middle of the solution is actually the same as the triangle you were solving (well, up to a scale factor of 2).
what was that exit? anyway cool video
Bro at the end realized the meaning of the universe purely from math and had to run and tell someone else
He had a stroke in the end
He always crosscheck the results.
... Good day to you, At about time 9:03 you say that angle 5*pi/3 is an angle in the 3rd Quadrant, but 5*pi/3 is in the 4th Quadrant, however the sine is still negative, so it doesn't change anything ... best regards and thanks Steve, Jan-W
Nice video, you make math look so easy! Next do a tan(x), tan(2x) and tan(3x) triangle.
I got inspired by your video with log triangle and considered the problem e^x, e^(2x) and e^(3x):
e^(2x) + e^(4x) = e^(6x)
Changing to t = e^(2x) will give
t + t^2 = t^3
1 + t = t^2
Since t is positive, we have the only solution t = phi = (1+sqrt(5))/2, which gives x = 0.5 ln(phi). The Pythagorean triangle is therefore with sides sqrt(phi), phi and phi*sqrt(phi)
We can use law of sines. sinx/sin(A)=sin2x/sinB=sin3x/sin(90)
Exactly. 3x = 90 degrees and angle x is the left angle in the figure. Trig identity says sin(2x) = 2 sin(x) cos(x). But by the figure, cos(x) = sin(2x). So sin(2x) = 2 sin(x) sin(2x) which means that sin(x) = 1/2. Quick and easy!
@@gordonstallings2518 How do you know beforehand that 3x = 90 degrees?
It's true that one can set the common value of the three sides of the equation to be 1 and discover quickly that this solution works. But there's no obvious way to show that 1 is the only common value that works.
Sin(x) is opposite over hypotenuse. And the sine of the smallest angle is the upright divided by the hypotenuse, which is labeled "sin(x)". The law of sines says that the sine of an angle divided by the opposite side length makes the same ratio for all three angles. So sine of the smallest angle divided by length 'sin(x)' is the same value as sin(90) divided by sin(3x). sin(x)/sin(x) = sin(90)/sin(3x). So 3x = 90, x = 30. @@flash24g
@@gordonstallings2518 "And the sine of the smallest angle is the upright divided by the hypotenuse, which is labeled "sin(x)"." Nonsense. It's the length of the upright, not this divided by the length of the hypotenuse, which is labelled sin x. So this would only be valid if we knew that the hypotenuse is length 1, which we don't know yet.
@@gordonstallings2518 And where do you get sin(x)/sin(x) = sin(90)/sin(3x) from? What we have from the law of sines is
sin A / sin x = sin (pi/2) / sin 3x
where A is the smallest angle. We have not shown that A = x.
I found a much simpler way btw. If you rearrange so that (sin(3x))^2 - (sin(x))^2 = (sin(2x))^2, then use difference of squares and sum-to-product in each of the factors. You get 4sin(x)cos(x)sin(2x)cos(2x) = (sin(2x))^2. Let sin(x) cos(x) = sin(2x)/2 on the left then divide both sides by sin(2x), getting 2cos(2x) = 1, or cos(2x) = 1/2. Then we immediately get x=30 degrees!
no , because then u will get 2x=2nπ + π/3
x=nπ + π/6 this is not the answer for every case where n is odd
@@prateeks6323 it is, he just needs to reject the negative "solutions", like in the video (the part 2sin(x) + 1 = 9).
@@prateeks6323 that's true, but it still finds one answer.
It's much simpler than that. The vertical side if his triangle is obviously the sine of the left hand angle. The bottom side is, likewise, the sine of the top angle. Therefore the one angle is twice the size of the other, and the only right-angle triangles with this property have angles of 30, 60 and 90 degrees.
Am I missing some easy way you got 4sinxcosxsin2xcos2x how is that much simpler
The unit circle is the set of points such that x² + y² = 1. If we parametrize it, we get cos²(x) + sin²(x) = 1. So, keeping that in mind, if a triangle has one side as the perpendicular side with length sin(x), that would mean the other sides are cos(x) and 1. You can't scale any triangle in a way where the other sides are otherwise.
So, with that in mind, sin(3x) has to be 1. Therefore, arcsin(1) = π/2, and x = π/6.
Edit : This is not rigorous and just happened to work because of the assumption that x is the angle that the triangle makes with sin(3x) and the sin(2x), and one side is sin(x). Look at the comments below for more clarification as to why that is
Ah! I can’t believe I didn’t see that even I worked out those values at the end. Nice!
You are right in what you say, but at no time is it said that x is one of the angles of the triangle, it is true that the results coincide, but only by coincidence (proposed manipulation of the values) of what was stated. That is why x=π/3+2nπ is also a solution, since x has nothing to do with the angle of the triangle.
They coincide since if we call the angle of the left vertex ϴ then
sinx=sin3x*sinϴ
sin2x=sin3x*cosϴ
dividing
sinx/sin2x=sinϴ/cosϴ, this is possible if we assume ϴ=x
sinx=sinϴ, ok
sin2x=2sinxcosx=cosϴ, only possible if x= π/3.
If the hypotenuse had been changed to sin5x, a solution as you indicate would be x= π/10≈0.3141596
But an approximate solution for this case is x≈0.4234166058162681
Although this does work out, it is not necessary for the circle to be a unit circle. sin(x), sin(2x) and sin(3x) are just numbers in the context of this triangle and the parametrization of a unit circle you provided used a dummy variable x (you could have used theta or 'a' or alpha or anything), which is not necessarily the same as the one in the problem.
You could scale the triangle so it had a hypotenuse of 1 though, by scaling by 1/sin(3x), then it would be sin(x)/sin(3x), sin(2x)/sin(3x) and hyp 1. Then, for exists SOME value of alpha such that sin(alpha) = sin(x)/sin(3x) and cos(alpha) = sin(2x)/sin(3x). Not sure why would one do this though, since what @@blackpenredpen showed in the video is the "simplest" and pretty much the only way of doing this without unrigorous and baseless pattern matching.
Your solution is not "Simple," it's not rigorous -enough- *at all* and it just happened to work out. Also, adding to what @@fisimath40 said, sin(5x) is also just a number and in the example they provided, your method doesn't even work.
Thank you, @fisimath40 and @hiimgood, for your comments. This "method" does not work for other values for the hypotenuse, as @fisimath40 pointed out. It is only valid based on the assumption that x is one of the angles. I was considering deleting the comment since it can cause confusion, but I realized that it could actually help avoid the same mistake that I made.
Buddy lost his train of thought at the end 😢
What happened at the end
Fun problem, never thought about trying this with trig functions. Nice wall chart in the background.
This was easier than expected. Really liked solving this question
This was very satisfying!
That was truly beautiful.
Professor will always be like a professor. I dreamt to become a professor. Now I am a student and I learned a lot from you Sir.❤❤
Can you do a video of "100 of factoring polynominals of grad 3" (+-ax^3 +- bx^2 +- cx +- d) please? I would love to see that!
Even though I've only just started a-level maths and further maths i watch all of your videos and its great to see different types of math that just isn't on the curriculum and without these videos i'd never see. Great video as always!
Maths is just the best
@@hareecionelson5875 i would have to agree
So good!
On that last triangle you were testing reference angles and you said that one side couldn’t be negative after showing it with math. However, you showed it when you wrote -sqrt3 right above it!
super cool!
I liked the "do we have a triple angle identity for sine?" at 1:11 followed by the fast-forward replay to the conclusion that we do. Great idea.
dude, you're great, even tho you forgot what you wanted to say in the end :)
Lol thanks!
@@blackpenredpenI thought you said "because i..." to say that we may have some complex number solutions, haha
You're killing me with leaving that 4 in the front so long 😂
The most useless number in the equation
But if 4 equals 0 we have another solution.
Nice! Even cooler is the same ratio of sides with all three angles - alpha, beta, and gamma - undetermined.
BTW, 5 pi / 3 is in the 4th quadrant, not the third, so that solution is completely valid EVEN THOUGH IT GIVES A NEGATIVE LENGTH, considering the angle as - pi / 6..
Not all negative lengths are invalid in a geometry problem. On occasion, they generate additional valid and interesting solutions involving a reflection of the hypothesized problem. Here though it's just a duplicate of the given solution, except drawn underneath the x-axis.
Using tan(x) = opposite / adjacent and setting it equal to tan(x) = sin(x) / cos(x), then substituting cos(x) = adjacent / hypotenuse immediately gives you sin(3x) = 1 without all the algebra and trigonometric substitutions. Then you have x= pi/6 +2npi and you just need to rule out the n congruent to 1 or 2 mod 3 cases, which is easy enough to do as well since triangles have positive side lengths.
what happened in the last 10 seconds? he looks visibly upset... 🥺🥺
To compute sin(3*x) I started with e^(3*x*i). I got a different expression that finally completly simplifies to cos(x)^2 = 3/4.
Just apply sine rule in so many different ways to get the 3 angles (x, 2x ,3x) of the triangle from the opposite sides.
So, 3x = 90 (right angle is opposite to hypotenuse) or x + 2x = 90 (acute angles are complementary in a right triangle) or x + 2x + 3x = 180 (sum of angles of any triangle is 180).
All of them imply *x = 30 deg* .
I have shortest solution
sin^2(x)+sin^2(2x)=sin^2(3x)
Take sin^2(x) to RHS
sin^2(2x)=[sin3x-sinx]*[sin3x+sinx]
Then sin^2(2x)=sin(2x)sin(4x)
Cos(2x)=1/2
Hence x=pi/6
Solved😎😎
Unless I don’t see the steps you skipped but sin(3x)-sin(x) is not sin(2x). Likewise sin(3x)+sin(x) isn’t sin(4x)
You are a very good TH-camr.
He is not a TH-camr but also he is a mathematician professor 😮
"I didn't know this was so cool, because..." *stares into the endless void*
*leaves without answering*
0:00: 🔍 The video discusses how to find the value of x in a right triangle using trigonometric identities.
4:35: 🔢 The video explains how to factor a quadratic expression and find the solutions for a given trigonometric equation.
7:36: 📐 The video explains how to find the value of x in a trigonometric equation using reference triangles and the unit circle.
Recap by Tammy AI
I have a question.
let's say f(x) = e^(x pi/2)
As you repeat this function over and over, the value gets larger and larger.
Suppose you repeated it infinite times.
We know i = e^(i pi/2)
If we substitute into itself, we will find the same function as if we repeated f(x) infinite times.
Does f(x) tend toward infinity or i as it is repeated infinite times?
Edit: Solved my own problem using x=e^((pi/2)x), finding that x=-2(W(-pi/2))/pi, and both i and -i are solutions. Still not sure if infinity is a solution, though.
no
x can also equal to pi/2 and 0 right?
I got the same quadratic but instead used substitution to turn it into an easy cubic in terms of sinx. solving that, I got these 3 solutions
cool video!
Literally just rewatched the log triangle video yesterday
Very nice problem
That is a great triangle!
I tried this for base cos x, cos 2x, and hypotenuse cos 3x, but there don't appear to be any solutions. But for base cos 3x, cos 2x, and hypotenuse cos x, I did find some. Can you?
Hey man, any idea how to prepare for the IMO?
been watching your vids for years and rarely comment but i missed this when it came out, and seeing it now - good stuff as always, but the end has me absolutely dying from laughter and also a bit confused/concerned, were you ok??
I waited for the: "But we are adults now so say pi over 6"😂😂
Adjacent/OPP
So good 👍👍👍👍
What CCC you teaching at? I want to take your math courses. For real, currently at vccd and ready for a change!
Could you try solving arctan(x)=1/tan(x)? It looks simple like tan^-1(x)=tan(x)^-1, but obviously is harder than that
When you wrote tan^-1(x), are you referring to arctan(x)? If so, those are the exact same problem. Anyway, you can simplify that to x = tan(cot(x)), and you can use progressive calculations to find the solution, though it isn't very satisfying. Wolfram alpha doesn't have a solution.
He was sad at the end, why?
I did it differently. I used sin(x)^2=(1-cos(2x))/2, and then some algebraic manipulation. Then i tested 2x=60deg, and it worked, so x must be 30 degrees.
Using the Pythagoras theorem in classical Euclidean IR² space. And the trigonometric formulae of Sin nx. Where n=2,3 this is gonna be transformed into a classical linear equation of a degree 2x3 .
can we substitue sin^2x as a t, and use horners method for solving polynome?
where is angle x located in the problem picture?
can we get a closer look the trig idenities
Actually i think we can change sin^2(x) into 1/2(1-cos2x), likewise for sin^2(2x) and sin^2(3x). Then we can use the product formula and factor them together to get all the solutions.
is there a relationship that the coefficients of the angles multiply to 6 or 1 2 3 are factors of 6
Sin[pi/2] =>=90 pi/2 >> 3x so x>>pi/6
the hypotenuse is sin(3x) and is sin[right angle] a direct identity to solve for x
Can you prove without calculator that e^3 is bigger than 20?
what if we take (sin(x))^2 to the right side and use the difference of squares formula
we get smth like
(sin(2x))^2 = (sin(3x)+sin(x))(sin(3x)-sin(x))
using the formulas for sin(a) +- sin(b); sin(2x); and cancelling some terms
we get
sin(2x) = sin(4x)
sin(pi - 2x) = sin(4x)
pi - 2x = 4x
=> x = pi/6 + 2npi
i feel this is much shorter and easier to understand
and the formula for sin(3x) isnt that fun to use
Hi professor I’ve been wondering about the usage of dy=f′(x)dx
in my textbook.
There’s not a single justification of how it is proved and it just states that it is true.
Since dy/dx
can’t be assumed as a fraction, I’m guessing there’s more to it than just multiplying by dx
on both sides.
Are there any proofs to this equation?
Also with some research, I found this “proof”. Can it be done this way?
it's quite an abuse of notation i guess
That’s the def of a “differential”. You can also look up “total differential” in calc 3 to see the connection.
Hello from Russia. this problem so looks simply and so beatifull. we need more triangle problem
Not related but does deferent branches of the productlog have a closed elementary relationship
At least between productlog(-1,x) and productlog(0,x)
In other words is there an elementary function such as
f(productlog(-1,x),productlog(0,x))=0
Just a small correction, at 9:00 you said that 5pi/3 was in quadrant 3. It is in fact in quadrant 4. Great video though!
Hey bprp i dont know if this formula existed before, but i realized you can get any positive number multiple of 9 with this: a*(10^n)-a, where a and n are natural numbers. Do you know how to prove it?
Factor it:
a*(10^n - 1)
10^n - 1 will always equal something divisible by 9 because the number produced will only contain 9s. n 9s to be exact
@@strufullol it was easy, thanks
Sir can I use the
as soon as i got it to a quadratic form i just plugged and chugged the quadratic formula
Im wondeeing if we can solve it with the sine law? we already know one angle is 90 and the other two can be written as x and 90-x
Just gives up at the end😂😂
damn, can't leave me hanging like that, at the end!
At 3:05 when you move the +16sin^6 x to the other side of the equation shouldn’t it be a negative ?
He's moving everything to the right side of the equation, but then he rewrites it so that the x terms are on the left, but it makes it look like he is moving to the left. So the 16sin^6(x) did not move at all.
Alternative solution with parametrization of Pythagorean triples:
First, we shall determine the limits for the values of x. If x is a solution, then so are all numbers differing from it by an even multiple of π, as is addressed in the video. Thus we may assume 0
sin^2(x)+sin^2(2x)-sin^3(3x) is a third degree polynomial in sin^2(x), and it admits the roots sin^2(x)=0, sin^2(x)=1, as well as sin^2(x)=1/4, the last case is exactly x=π/6 up to an integer multiple of 2π. As the first two cases had been discarded, x=π/6 is the only solution if 0
I also got this question on my inverse trigonometry exam today
Exactly like this?
That's something interesting that I never knew about my set square.
At the end bro was wandering 🤣🤣🤣🤣🤣🤣🤣🤣
triangles ABC= AC=5 Bc=3 AB=2 sinx^2 +sinx = sinx^3]=[[[[ sin3x= 1-cos3x= 1cos3x[3x=8 x=5 x1=3 x2=2
for next lets do tangent triangle!
What happened at the end?
What happened at the end there?
So we know angles are x 2x and 3x,and 3x is 90(given)
so why cant we turn sin2x into cosx and directly get 1 upon squaring???
"Enjoy the moment" 😂🤣😂🤣
Trig is so satisfying
Hi, just found another solution
Lenght / sin(angle) is same for all sides for triangles, so
sin(2x)/sin(a) = sin(x)/sin(b)
a=2b
a+b=90
a=60
x=30
So what happened at 11:00? Obviously if it's cool 😎 then it is cool. So don't be ashamed of it! Unless something else happened that you lost it for a bit. Then you need to take it easy with all those math videos. But if not let's celebrate : th-cam.com/video/3GwjfUFyY6M/w-d-xo.html The top comment in that video says : "I just finished a math problem that took 4 hours" So there you go!
Tüm durumlar için sanmıştım tarım açı misali bir formül bekliyorsum . Pi/6 için özel bir durumla karşılaştım.Güzeldi.
i honestly relate too much to the ending
The thing is, this question has many solutions. Like when I solved it on my own (before seeing your answer) I got x = 2πn + π/2 (which is a correct solution). So there's multiple answers to this question.
He rejected that solution because it makes the sin(2x) edge have length zero
When I was calculating this, I forgot to square the expansion for sin(3x) after finding it was 3sin(x) - 4sin^3(x) and I arrived at the same answer. Luckily, all it did was not include any of the other false solutions! 😅
I can only assume man was ingulfed in new thoughts looking at the sick math he just spit out.
I did it using Euler’s identity.
in the ENDing.. LOLZ
This can be made easier:
sin(3x)=sin(x)*[3-4sin^2(x)]=sin(x)*[4cos^2(x)-1]
Therefore the equation can be written as
sin^2(x)+4sin^2(x)cos^2(x)=sin^2(x)*[4cos^2(x)-1]^2
and after discarding the solution sin(x)=0,
1+4cos^2(x)=[4cos^2(x)-1]^2
Now let t=4cos^2(x): we have
1+t=(t-1)^2 => 1+t=t^2-2t+1 => t^2-3t=0
The solution t=0 leads to sin(2x)=0, so we discard it and we are left with
t=3 => 4cos^2(x)=3 => cos(x)=+-sqrt(3)/2
ending man 😂😂
Why don't you first make substitution sin^2(x) = t and only then start simplifying?
Bro didn't feel like talking anymore. Been there😂
😂😂 why did you keep silent in the end of video ?
bro was done™
@@jbw416 nothing else was needed to say!!