the sine triangle problem

Рет қаралды 86,210

8 ай бұрын

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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log triangle problem: • solution to the logari...
exp triangle problem: • solving an exponential...
triple angle identity for sine: • sin(3x) in terms of si...
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Пікірлер: 302

@blackpenredpen 8 ай бұрын

Check out the log triangle problem: kzfaq.info/get/bejne/ebOUfbOoqsipaHU.html

@mrchin7562 8 ай бұрын

I like the sin triangle way better

@rajeevmishra2912 8 ай бұрын

Please make video a day life of yourself

@mr.d8747 8 ай бұрын

*You should do a Lambert W triangle where the sides of the right triangle are W(x), W(2x) and W(3x).*

@dolos9250 8 ай бұрын

try cos triangle

@dolos9250 8 ай бұрын

@@mr.d8747 its not possible to do it algebraically

@zlam332 8 ай бұрын

The hardest part of maths is to explain why we like it.

@Owen_loves_Butters 8 ай бұрын

Seriously. People ask me all the time why I like math so much. I can never give an answer that I'd consider satisfactory.

@ac8210 8 ай бұрын

I’ve never agreed with a statement so much

@hybmnzz2658 8 ай бұрын

The dopamine of understanding. The structures and surprises which build on simple rules.

@TheBeautyofMath 8 ай бұрын

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.

@hareecionelson5875 7 ай бұрын

@@hybmnzz2658 the kick in the discovery ~ Richard Feynman

@qihengng5993 8 ай бұрын

This is like ASMR math, just slowly solving the problem and appreciating its elegance ❤

@canyoupoop 8 ай бұрын

This is softcore ASMR 3b1b is heavy hard core ASMR💀

@blackpenredpen 8 ай бұрын

Glad you enjoy it!

@cjfool5489 8 ай бұрын

@@canyoupoop😂

@Jack_Callcott_AU 8 ай бұрын

@@blackpenredpen And the triangle itself turns out to be 30°, 60°, 90° right triangle.

@RithwikVadul 8 ай бұрын

@@Jack_Callcott_AUguess check is ez

@archierm 8 ай бұрын

Sudden existential crisis?? Actually yeah, it's super cool.

@guy_with_infinite_power 8 ай бұрын

At the end, Bro was wondering if it was him who did all those things on board😅😂

@suyunbek1399 8 ай бұрын

heartaches😃🤤

@Mr23143sir 8 ай бұрын

was something wrong there or what was that ?

@guy_with_infinite_power 8 ай бұрын

@@Mr23143sir nothing was wrong, he just had some different outro plan

@Mr23143sir 8 ай бұрын

Oh, thanks for clarification then @@guy_with_infinite_power

@danielcingari5407 8 ай бұрын

This man just went ('-') /|\.

@tobybartels8426 8 ай бұрын

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).

@vinijoncrafts7213 8 ай бұрын

I love how he's just so mesmerized he couldn''t talk at the end of the video lmao

@randomcoder5 7 күн бұрын

he might have realised he could have just used the law of sines (sinA/a = sinB/b = sinC/c)

@LactationMan 8 ай бұрын

He was sad at the end, why?

@DavideCosmaro 8 ай бұрын

Bro at the end realized the meaning of the universe purely from math and had to run and tell someone else

@brololler 8 ай бұрын

what was that exit? anyway cool video

@grave.digga_ 8 ай бұрын

Nice video, you make math look so easy! Next do a tan(x), tan(2x) and tan(3x) triangle.

@jinhuiliao1137 8 ай бұрын

We can use law of sines. sinx/sin(A)=sin2x/sinB=sin3x/sin(90)

@gordonstallings2518 8 ай бұрын

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!

@flash24g 8 ай бұрын

@@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.

@gordonstallings2518 8 ай бұрын

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

@flash24g 8 ай бұрын

@@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.

@flash24g 8 ай бұрын

@@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.

@Johnny-tw5pr 8 ай бұрын

He had a stroke in the end

@hodossyb 2 ай бұрын

He always crosscheck the results.

@jan-willemreens9010 8 ай бұрын

... 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

@fizixx 8 ай бұрын

Fun problem, never thought about trying this with trig functions. Nice wall chart in the background.

@alexsokolov8009 8 ай бұрын

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)

@koioioioi 8 ай бұрын

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!

@hareecionelson5875 7 ай бұрын

Maths is just the best

@koioioioi 7 ай бұрын

@@hareecionelson5875 i would have to agree

@Ivan.999 7 ай бұрын

This was easier than expected. Really liked solving this question

@billprovince8759 8 ай бұрын

This was very satisfying!

@c4ashley 8 ай бұрын

That was truly beautiful.

@MrMasterGamer0 8 ай бұрын

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!

@TheBeautyofMath 8 ай бұрын

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.

@calculuslite5 8 ай бұрын

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.❤❤

@Leivoso 8 ай бұрын

Buddy lost his train of thought at the end 😢

@paul_c15 8 ай бұрын

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!

@proximitygaming8253 8 ай бұрын

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!

@prateeks6323 8 ай бұрын

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

@AlcyonEldara 8 ай бұрын

@@prateeks6323 it is, he just needs to reject the negative "solutions", like in the video (the part 2sin(x) + 1 = 9).

@proximitygaming8253 8 ай бұрын

@@prateeks6323 that's true, but it still finds one answer.

@alanclarke4646 8 ай бұрын

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.

@sethv5273 8 ай бұрын

Am I missing some easy way you got 4sinxcosxsin2xcos2x how is that much simpler

@MeQt 8 ай бұрын

What happened at the end

@pietergeerkens6324 8 ай бұрын

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.

@cybersolo 4 ай бұрын

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.

@hiwhoareyou01 6 ай бұрын

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.

@robertsellers1153 8 ай бұрын

super cool!

@vishalmishra3046 7 ай бұрын

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* .

@MusicCriticDuh 8 ай бұрын

what happened in the last 10 seconds? he looks visibly upset... 🥺🥺

@General12th 8 ай бұрын

So good!

@acuriousmind6217 8 ай бұрын

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

@blackpenredpen 8 ай бұрын

Ah! I can’t believe I didn’t see that even I worked out those values at the end. Nice!

@fisimath40 8 ай бұрын

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

@hiimgood 8 ай бұрын

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.

@acuriousmind6217 8 ай бұрын

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.

@TheRenaSystem 4 ай бұрын

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??

@joshuahillerup4290 8 ай бұрын

You're killing me with leaving that 4 in the front so long 😂

@o_s-24 8 ай бұрын

The most useless number in the equation

@normanstevens4924 8 ай бұрын

But if 4 equals 0 we have another solution.

@TundeEszlari 8 ай бұрын

You are a very good KZfaqr.

@calculuslite5 8 ай бұрын

He is not a KZfaqr but also he is a mathematician professor 😮

@johnathaniel11 8 ай бұрын

Literally just rewatched the log triangle video yesterday

@romanbykov5922 8 ай бұрын

dude, you're great, even tho you forgot what you wanted to say in the end :)

@blackpenredpen 8 ай бұрын

Lol thanks!

@kristofersrudzitis727 8 ай бұрын

@@blackpenredpenI thought you said "because i..." to say that we may have some complex number solutions, haha

@MichaelDarrow-tr1mn 8 ай бұрын

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.

@lightxc5618 6 ай бұрын

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.

@AbouTaim-Lille 8 ай бұрын

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 .

@ogxj6 7 ай бұрын

That is a great triangle!

@davidcroft95 8 ай бұрын

"I didn't know this was so cool, because..." *stares into the endless void* *leaves without answering*

@JohnAbleton 8 ай бұрын

Just gives up at the end😂😂

@Starchaser41817 8 ай бұрын

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.

@dolos9250 8 ай бұрын

@kornelviktor6985 8 ай бұрын

I waited for the: "But we are adults now so say pi over 6"😂😂

@joshcollins7771 8 ай бұрын

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

@Starchaser41817 8 ай бұрын

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.

@muntasirmahmud3349 7 ай бұрын

Very nice problem

@thirstyCactus 8 ай бұрын

damn, can't leave me hanging like that, at the end!

@det-tn5qf 8 ай бұрын

can we get a closer look the trig idenities

@carly09et 8 ай бұрын

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

@Queenside_Rook 8 ай бұрын

as soon as i got it to a quadratic form i just plugged and chugged the quadratic formula

@rogerkearns8094 8 ай бұрын

That's something interesting that I never knew about my set square.

@3hustle 6 ай бұрын

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

@yigit819 8 ай бұрын

the end 😂

@kmathmedia 7 ай бұрын

At the end bro was wandering 🤣🤣🤣🤣🤣🤣🤣🤣

@fedzhuhray 8 ай бұрын

Hello from Russia. this problem so looks simply and so beatifull. we need more triangle problem

@niranjanjwarrier731 4 ай бұрын

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!

@yaboy919 8 ай бұрын

I also got this question on my inverse trigonometry exam today

@blackpenredpen 8 ай бұрын

Exactly like this?

@tylercampbell2147 7 ай бұрын

I can only assume man was ingulfed in new thoughts looking at the sick math he just spit out.

@thatomofolo452 8 ай бұрын

Adjacent/OPP

@Qwentar 8 ай бұрын

"Enjoy the moment" 😂🤣😂🤣

@yuukitakanashi4506 8 ай бұрын

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.

@richardbraakman7469 7 ай бұрын

He rejected that solution because it makes the sin(2x) edge have length zero

@bol9332 8 ай бұрын

Trig is so satisfying

@yenimath 7 ай бұрын

Tüm durumlar için sanmıştım tarım açı misali bir formül bekliyorsum . Pi/6 için özel bir durumla karşılaştım.Güzeldi.

@funterive5132 8 ай бұрын

Hey man, any idea how to prepare for the IMO?

@Wandering_Horse 8 ай бұрын

What CCC you teaching at? I want to take your math courses. For real, currently at vccd and ready for a change!

@Medoet 8 ай бұрын

for next lets do tangent triangle!

@BadalYadav-wz3vq 7 ай бұрын

So good 👍👍👍👍

@johns.8246 8 ай бұрын

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?

@aquaticstarr4607 8 ай бұрын

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! 😅

@TheAmorchef 8 ай бұрын

is there a relationship that the coefficients of the angles multiply to 6 or 1 2 3 are factors of 6

@kobey3044 7 ай бұрын

in the ENDing.. LOLZ

@luvvluma 3 ай бұрын

i honestly relate too much to the ending

@swapnarajmohanty6698 8 ай бұрын

ending man 😂😂

@li-ion6333 7 ай бұрын

can we substitue sin^2x as a t, and use horners method for solving polynome?

@_QWERTY2254 8 ай бұрын

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

@ABHIGAMING-yo9my 7 ай бұрын

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😎😎

@blackpenredpen 7 ай бұрын

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)

@illumexhisoka6181 8 ай бұрын

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

@powerllesss2672 7 ай бұрын

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!

@ore_wa_nagi 8 ай бұрын

Sir can I use the

@a.xaberof948 5 ай бұрын

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

@garythesnail8674 8 ай бұрын

Bro didn't feel like talking anymore. Been there😂

@Getsomewaterplease 8 ай бұрын

Can you prove without calculator that e^3 is bigger than 20?

@zeno1402 8 ай бұрын

where is angle x located in the problem picture?

@AlmostMath 7 ай бұрын

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

@marceliusmartirosianas6104 8 ай бұрын

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

@Regularsshorts 8 ай бұрын

This is like a proof for the Law of Sines.

@giuseppemalaguti435 8 ай бұрын

(sinx)^2=0,1,1/4

@agsantiago22 7 ай бұрын

I did it using Euler’s identity.

@crochou8173 7 ай бұрын

Solved this by tanx=sinx/2sinxcosx sinx=1/2 check sin3x. Just under a minute

@flintsparks8406 8 ай бұрын

What happened at the end there?

@jd9119 8 ай бұрын

What happened at the end?

@safriwildan6506 8 ай бұрын

now do tangent triangle => (tan x)^2 + (tan 2x)^2 = (tan 3x)^2 😁

@blackpenredpen 8 ай бұрын

I think that is going to be horrendous 😆

@aaryan8104 Ай бұрын

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???

@JonnyMath 8 ай бұрын

Hi professor!!! Your videos are amazing!!! I also make videos on KZfaq and I recently made a special Halloween video when I used integrals to find the area of a Jack-o'-lantern!!!😅 Thanks for making these videos!!! I wouldn't have started on KZfaq if it wasn't for your amazing videos!!!🤩🥳🤗