tan(x) + 1/tan(x) = 8/3

[@11en21]

I agree!!😅 The feeling of "I understand it all, but I couldn't do it by myself" gotta be one of the most frustrating things about math problems

[@gabrielygoncalves7934]

I’m brazilian and I understand all of it!! Tks so much❤️

[@Lucas-sf1be]

É a braba esqueça

[@laniakeasupercluster9440]

Your video was amazing, Teacher!

[@GeneralWinter9]

Thanks great explanation!

[@cewkins721]

What a fun question! personally i took the easy way of working with the original expression to solve for the angles then plugging them back into the sin(x) + cos(x) expression, turning the equation into a quadratic is also a very useful method that is used a lot in this level of math and above. In my opinion the "elegant solution" is probably the best because it eliminates the need to evaluate any angles and simply adding numbers which gives the exact answer of sqrt(7)/2, convenient and quick. transforming to trig ratios is also a good idea. This question is very interesting because it can be solved using identities and these methods taught in most trig courses which makes it good for practice.

[@Pericles_Vicente]

I used y+1/y=8/3, to begin with, then the result follows. sin(arctan x)=±x/√(1+ x²), cos(arctan x)=1/√(1+x²).

[@julianamaths]

That’s exactly what I did, I just didn’t call it y 😉

[@user-ct1ns6zw4z]

I solved this using something similar to your “elegant solution” by remembering that sec^2(x) = 1 + tan^2(x) (because they are both useful ways to write d/dx tanx, the second way helps me remember d/dx arctan(x)). That gave me sec^2(x)/tan(x) = 1/sin(x)cos(x) = 8/3, then the rest was the same. I guess it’s just a slightly more convoluted way doing it. Wouldn’t have thought to set up a quadratic equation

[@julianamaths]

Ahhh I like your sec thing, but for the second half of your comment, you're saying the sentence wrong! It should be "next time, setting up a quadratic equation will be one of the options that I have", not "I didn't think of it this time and therefore I will continue to never think of it"!

[@user-ct1ns6zw4z]

@@julianamaths good attitude 😁

[@vipindia]

Thank you Angelina Jolie of Mathematics ❤😊

[@AmlanSarkar-wr2pr]

Basically what I do is tan²x+1/tanx = 8/3 1/sinxcosx=8/3 8sinxcosx = 3 Now, 2sinxcosx=3/4 Add 1 both on both sides, (sinx + cosx)² = 7/4 sinx + cosx = +- √7/2 [Note:- I did it by watching just the image of video which is usually seen in any video.I don't even know what the value is.Please let me know mam if I am right or if wrong which step I did the mistake.I have also decided I will not watch the video until I get any feedback mam.After the response I will watch the video.]

[@julianamaths]

Your solution is one of the possibilities that I show on the video, however, I spend more time on a different one that I think may be easier for more people to find :)

[@AmlanSarkar-wr2pr]

@@julianamaths definitely I am going to watch it mam the easier version. Love from India 🇮🇳🇮🇳 ❤️❤️

[@bobross7473]

Honestly I feel like both methods require the same amount of clever thinking. This is one of those problems that’s just like “wtf, how am I supposed to guess that’s the way it’s done?” That’s how I feel about these types of problems. Unfortunately, these sorts of problems that are are kind of just brute force or trial and error do appear in school sometimes. I know that the solution only utilizes basic concepts, but there are a number of different ways the student could go about solving the problem, some of which include utilizing other trig identities besides the Pythagorean identity. It’s essentially just “hope you get lucky and the lock breaks open.”

[@julianamaths]

Obviously this is going to be, to some extent, a matter of personal opinion. But I do feel that multiplying both sides by tan(x) in order to get rid of a denominator does not qualify as “clever”, in the sense we are using that word here. On the other hand, I really don’t like the other trig identities, so usually my first instinct is to change everything to sines and cosines, I honestly don’t know why this time I didn’t find that solution before the other one. So I guess you are arguing that they are BOTH clever and I am arguing that NEITHER of them are… 😂

[@MarleneWalker-su8ku]

​@@julianamathsKeep them coming, Juliana you are a fabulous teacher, greetings from the UK.

[@mouthiknaradas962]

This is an easy one. If tan x +1/ tan x = 8/3 (sin x / cos x) + (cos x / sin x )= sin²x + cos²x/ cos x sin x = 8/3 1/cos x sin x = 8/3 cos x sin x = 3/8 (sin x + cos x)² = 1+6/8 = 7/4 sin x + cos x = √7 / 2

[@julianamaths]

Yes, that is an easy solution if you think of it :-) What I am mostly talking about in the video is that there are other ways, that involve more work, to also find the same answer even if you don't happen to have this best idea when you need it.

[@Zaasz_]

TE AMO

[@timewaster0123]

Great video 👍

[@Hjominbonrun]

at minute 0, that looks like a trinomial with u = tanx, now I will watch. Also, this is stuff we saw often in Math papers in South Africa so it cannot be a 22 minute video.

[@julianamaths]

Well, they are pretty common, yes, and I bet your teacher also took longer than 22 minutes to teach the class how to solve them, so :) But yes, the tan (x) = u is my favorite way to solve it, too.

[@Hjominbonrun]

@@julianamaths yeah, maybe this is a niche concept and I have just been trained to spot hidden quadratics very well. My teacher just said, now that you know how to solve a polynomial, lets look at this, and he asked us to spot it, with many examples, and it came quickly. I think Mathematics has a lot to do with spotting patterns. So when you see 2^2x+2*2^x +1 =0, then you should already be trying to spot the polynomial pattern. Maybe a video on how to spot hidden patterns would be great, as opposed to solving one entire problem from start to finish could be a good idea. Of course, you make the videos, so only you know how much pain it is to do, and what I might think is easy might be a pain.

[@royalroamer1]

Hello, great video once again ❤ (1 question, how did you make the observation that both tanx and 1/tanx are solutions? Did you use the sum of roots rule?

[@julianamaths]

No, it's because tan = 1/(1/tan)). I said that in the video by calling it p, that p = 1/(1/p). So if you turn it upside down they just change places with each other, but are still added, so it doesn't matter.

[@AnUnknownSentience]

When solving with the second solution, (4-sqrt(7))/3, we surprisingly get -sqrt(7)/2 despite using the triangle and assuming that the angle is in the first quadrant. The reason I am confused is because at the end you mention that we 'lost' the (+or-) because we used the hypothetical triangle to find other trig ratios. But that doesn't seem to be the case when solving with the second solution of the quadratic equation. It seems that we do get both the solutions in both the quadrants when we use both the solutions of the quadratic equation, and that the triangle isn't what is causing us to 'lose' the (+or-). That at least seems to be the case because as you mentioned, both solutions of the quadratic equation are positive and when using the triangle, we are taking an acute angle, which means we should get the first quadrant solution in both cases, but we are not. I am confused at to why this is? Edit: We certainly don't need to use both solutions of the quadratic equation when we understand that the answer would be in first and third quadrant and that the soutions of the quadratic are reciprocals of each other, however that doesn't mean that we aren't 'losing' the (+or-) by choosing one solution or the other.

[@julianamaths]

I honestly haven't thought too much about it, I just assumed it would have been the triangle. As for x, we should have 4 solutions and not 2, being 2 in the first quadrant and 2 in the third, with both of the solutions for tangent in both of the quadrants. I'm not getting the same numbers you're getting, though. I followed the final steps here now with (4-sqrt(7))/3 for a tangent in the triangle, continued the step to the end, and I am still getting the same positive solution.

[@AnUnknownSentience]

@@julianamaths I agree that x would have four solutions as there would be four different angles to choose from and they would all work in the equation. As for using (4-sqrt(7))/3, we get a triangle with sides 3 and (4-sqrt(7)), which gives us the hypotenuse as sqrt(32-8sqrt(7))=P (for simplicity). Which then gives us sin(x)=(4-sqrt(7))/P and cos(x)=3/P. Adding the two we get: sin(x)+cos(x) = (7-sqrt(7))/P Then using the method you showed for P: sin(x)+cos(x) = (7-sqrt(7))/2(1-sqrt(7)) Rationalising then gives us: sin(x)+cos(x) = (6sqrt(7))/2(-6) Which simplifies to: sin(x)+cos(x) = -sqrt(7)/2 That was what I worked out, however I'm not sure if there has been a mistake or not. I'm not exactly sure why the second solution directly gives us the third quadrant answer for sin(x)+cos(x) rather than the first quadrant solution. I would assume it is because they are conjugate surds and we get the quadrant with the opposite sign, but that isn't a proof of any kind. But aside from that, this was a great video. I especially loved the simplifying of the double radical, it's a very intuitive version of completing the square and I had never seen it before. Definitely added to my toolbox.

[@mathrodriguezschwartzmann]

senx/cosx+cosx/senx=8/3 )/senx.cosx=8/3 (sen²x+cos²x)/(senx.cosx)=8/3 senx.cosx=3/8 But, sen²x+cos²x=1 (senx+cosx)²-2.cosx.senx=1 (senx+cosx)²-2.8/3=1 (senx+cosx)²=7/4 (senx+cosx)=±√7/2

[@xadrikxaulxu]

Good.

[@JoaoGabriel-ow4ue]

Se cair na ESA vai ser uma das mais fáceis

[@wassi_7511]

Nice teacher. I'm just wondering where are you from ? You have a very perfect and charming english accent

[@andrade7859]

Brazil

[@bobmarley9905]

First off I'll just say I really like how you structured/organized this video! However, I'm confused on how once you arrive at the solution for sinx + cosx, why we don't need to check if this solution makes sense and doesn't cause any sort of contradiction.

[@julianamaths]

Why would it? What sort of contradiction could it cause?

[@bobmarley9905]

@@julianamaths I mean intuitively it feels like there shouldn't be a contradiction, but I'm just not sure how to prove there isn't. That's my main question/confusion here.

[@bobmarley9905]

@@julianamaths To better elaborate, let's 1st just look at the elegant solution: What it showed is that tanx + 1/tanx = 8/3 iff sinx + cosx = +/- sqrt(7)/2 (this is because we can prove both the forward and reverse implication by starting with tanx + 1/tanx = 8/3 or sinx + cosx = +/- sqrt(7)/2, respectively). What this means is that both of these statements must have same truth value, but then how do we know if either of these statements is possible/true or not?

[@julianamaths]

That is the thing that I don't get about your question. You clearly understand what it means to get from one to the other and vice versa - that they are equivalent statements - and that was what the question was. From one, get to the other. Even if you wanted to question whether such angles x do exist in order to make both statements true, the non-elegant solution explicitly finds these angles so yes, they do exist. But even if we were to only look at the elegant solution, I don't quite understand questioning their existence (if that is indeed what you are doing, which is not clear to me). There are no restrictions on the range of tangent, so 8/3, as well as any other number, is the tangent of somebody, and it's not much different for tan + 1/tan.

[@julianamaths]

Even as I typed that answer it started to make more sense to me. As a function of tan(x), tan(x)+1/tan(x) does have a minimum value (let's consider only positive numbers for simplicity as the other side is symmetrical), which is 2 when tan(x)=1. So when I said 8/3 could be "any" number, I was thinking of a possible issue with large numbers which there isn't, but there actually is an issue with small numbers. We couldn't solve the problem for, say, 5/3. But 8/3 is fine. Thank you for that! Another layer to the problem that I had not considered :) Another way we could see it is if we tried to carry out the elegant solution for 5/3 instead, we get as an answer sin+cos=sqrt(11/5), which is impossible as the maximum value of the function sin+cos is sqrt(2) when x=pi/4.

[@ciaopeople9664]

Sinrah Cosnor ? Yes. Tan ... Tan ...

[@erimosini]

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