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- Can't be negative - Can't be 0 - 72° > 60° so the sin has to be greater than √3/2, but √((5-√5)/8) is clearly less than √3/2 - Therefore A is the answer

Commenting before watching the video, I eliminated the negative because sin 72 is near 1 than zero, so the square root has to be near 1 so we must choose the bigger alternative

Nah, ain't no way. I just did this EXACT question at school today. I come home and, boom, there it is

6:55 37 appears everywhere, even in your jokes :)

Indian student do this is 11th class maths as using sin (90-12) =cos (12) and x=12 5x=90 3x=90-2x Cos 3x = cos (90-2x) 4cos^3 x -3cosx = sin 2x 4cos^3 x -3cosx =2.sinx.cosx 4cos^2 x -3 = 2sinx Covert cos^2 to sin^2 then use the quadratic formula u have your answer

I solved this question knowing that in a pentagram, if you cut a point, it will form an isosceles triangle with angles 72º, 72º and 36º. But, in a pentagram the aurum ratio is a rule, so if the base is equal to x, the equal sides are equal to x((5^0.5-1)/2). Applying the cos rule, you can find that cos72º = ((5^0.5+1)/4). Therefore, the fundamental law of trigonometry garantees the sin72º = ((5+5^0.5)/8)^0.5, as sinx >0 for 0>x>90º

golden triangle has 72°, 72° and 36° and sides a golden ratio (phi) larger than the base. Split the base in half to obtain the right triangle with 72° having a base of 1 and hypotenuse of 2 phi. The remaining far side of 72° will be phi*√(phi +2). Sin 72 = √(phi +2)/2 where phi is the positive golden ratio = (√5+1)÷2 √((√5+1)÷2+2)/2 = 0.951056

I love these kinds of admission test math problems. Would love to see more of your videos on this

It's not necessary to notice that 36° is also a solution. By inspection sin(72°) is pretty close to 1, approximately 0.95, so (sin(72°))^2 must be approximately 0.9. Now (5 - sqrt(5))/8 is much too small and (5 + sqrt(5))/8 is the correct answer.

Nice - I sorta got stuck on the last step, but your explanation made perfect sense.

Beautiful math excercise. ❤

Alternatively, consider sin(3x) + sin(2x) = 0, cancle one sin x after expanding and replace sin^2 by 1 - cos^2. We have a quadratic is cosine which is simpler to solve. Side note : Depending on what kind of education system you are in, you may have some trig values of 18 and 36 degrees memorized, in which case the question is trivialized. There also probably is a complex number solution but I am too lazy to find it.

Recall that sin(θ) is the perpendicular height to the x-axis from the point of intersection of the unit circle and the line making the angle θ. Clearly, this height is not 0, and also it is above the x-axis, so it is positive. We can therefore exclude B, C and E. We know that sin(45°)=sqrt(2)/2≈0.7, so sin(72°) must be greater that 0.7 But sqrt(5)≈2.2 So option A ≈ sqrt(7.2/8), and option B ≈ sqrt(2.8/8) From A and B, only A is greater than 0.7. As a multiple choice question, this is a one minute question.

I'm learning to like maths. It's slow but i'm beginning to appreciate it. That's after having my potential love of maths beaten out of me by a less than ideal experience of being taught it at school. My algebra is at the point where I followed everything that was done here without scratching my head. Although I probably could not have solved it on my own. Slow and steady. I'll get there in the end.

Very insightful

It’s worth noting that at 4:55 we arrive at the four choices which are given in the paper. We could have started there, and in the real exam, you should, as it saves time. Nonetheless, I appreciate seeing the derivation of the solution. It would also be interesting to see where the trig identity comes from. It comes from applying de Moivre’s theorem.

Very nice. I would never solve this myself but love seeing the magic happen haha

This is the Best content on KZfaq in the world 🙏🏿🌍!

Normal question: 95% time solving equation, 5% time choosing the correct answer This question: 50% time solving equation, 50% time choosing the correct answer I have never seen a question that takes a long time to filter and reject answers.

Facinating question

I used the same approach you did to calculate sin72 = x, obviously x cant be negative so we igonore that solution, now to determine if it sqrt((5 +/- sqrt5)/8, I set up a condition where sin72 lies between sin 60 and sin90 since sin is increasing from 0 to 90 degrees, we know that sin 60 is sqrt3/2 ~ 0.866 and sin 90 = 1 so when we approximate the two solutions, the one with + sign is ~0.951 and the one with negative sign is ~ 0.56 which does not satisfy our condition, there fore the one with + sign is the answer which is option A

From scratch it will be using similar and isosceles triangles Angle is acute so we can immediately eliminate (b) , (c) , (e) Sine is increasing in first quadrant so after comparing (a) and (d) , (a) is closer to one than (d) so we can suspect that (a) is correct but at this moment we can not to be sure that (a) is correct (In fact (a) = sin(72 degrees) and (d) = sin(36 degrees) )

do more calculus 3 question on you channel, i really like them

I compared the values with sin 45, to get to the answer. But ur argument is pure genious

I love your teaching so much ❤ form Cambodia teacher

Exelente Ejercicio. 😃

Very good video ! I have a question I would love you to answer : for any ineger n, what is the integral going from O to 1 of (lnx)^n

Teaching skill Love it

All the simple right triangles come from the √ 2, √3 and golden triangles. Triangles with sides 1:1:√2, 1:,√3,,2 and 1:2phi:X Which came from dividing the square, equilateral and golden triangle. Knowing this, then no trig formula is required.

Sir videos help me a lot to make my concept very strong in mathematics love from India❤❤❤❤

I guess you can also write the equation as 4^2.x^4-4.5.x^2+5 and transform the equation a quadratic in terms of 4. then you can use the quadratic formula to find out what equals 4 and pull out the x value from that. a lot more work but its kinda fun that you can write a quadratic in terms of 4

I had a similar question where we were supposed to find sin(72) on a complex analysis final.

At school I just did this question. When I saw 36 degrees, the golden triangle was the first thing that came to my mind. Then using sine rule to find a relationship between sin72, sin36, and (√5-1)/2. I got the same answer. However, I tried to remove the root sign, hoping that 5+√5 could be expressed as the form ()^2. Then one hour was wasted.

Tried and also done 👍🏻

Thanks for improving my "trigonometry" skills

Omg u just took me back to trigonometric algebra thnx 💀

In order to choose the right value between the + vs -sqrt5, you can compare to the value of sin45

very nice....i understand it, only i wouldn't know where to start

There are a lot of ways to approach this that I took in a vastly different way and still got the same results :3

I actually remember doing this question 2 years ago for the MAT

Before solving it, you can elliminate b, c and e because it is a first quadrant. Next, 72 is greater than 60, so the sine is greater than root 3 / 2. Root 5 is about 2.2 + 5 is 7.2 and root 8 is 2 x root 2. So, (root (7.2/2))/2 is (root (3.6))/2, so a is the answer

So it doesn't change the fact that I don't like minus and I remove it, and I am still correct

how i did this was just guessing game > neglect the negatives now A and C are left Sin(72) is much closer to Sin(90) which is 1 and if we just do some simple observations we can see A option is much closer to 1 and C is much closer to 0 so A is the answer :D

I've been your fan for so long.Could you please help me with this integral (Infinite Integral of xlnx/x+1)? Thanks in advance ❤

Based on the fact that it's multiple choice, I wonder if you could've cut straight to the analysis at the end, without doing the algebra. Like you could throw out all the non-positives right off the bat; then show that a and d both satisfies the given equation for theta = 72 deg; and then do a little geometry to show that answers a and d are actually talking about congruent triangles (a^2 + d^2 = 1); and the correct triangle is a.

can you do Singapore H3 A level math? been suffering through those although the threshold for distinction isn't that high

Please do integral of e^x/x dx, without Taylor or MacLaurin Series expansion. Thank you very muchA

Can you do integral of 1/(x^i)?

sir please make video for calculus 3 for multivariable case

You should try the 2011 IMO question number 2 it is very fun

You can approximate which one it is numerically, without using the identity they give you. And yes, you can do this without a calculator, remembering sin(60 deg) is approx 0.866. First, we can rule out the two negative choices, and zero. From our special angle sin(60 deg) = sqrt(3)/2, which is approx 0.866, we know the answer must be between this and 1. This narrows it down to option A and option D. sqrt(5) is approx. 2.25, which we can find by linearizing sqrt(x) at x=4 to get L4(x) = 1/4*(x-4) + 2. Plug in x=5 to L4(x). (5 + 2.25)/8 = 7.25/8 = 0.875 + 0.03125 = 0.90625. Linearize sqrt(x) at 0.81, to get L_0.81 (x) = 5/9*(x - 0.81) + 0.9. Plug in 0.9 to approximate the result. 5/9*0.09 + 0.9 = 0.95. This confirms option A meets the range where we expect sin(72 deg) to be. Now try option D: (5 - 2.25)/8 = 2.75 = 2/8 + 3/32 = 11/32 is approx 0.34 Use nearest number with rational square root of 0.36, to linearize. L_0.36 (x) = 5/6*(x - 0.36) + 0.6 L_0.36 (x) = 5/6*(-0.02) + 0.6 = 1/6 + 0.6, is approx 0.76667. This rules out option D, and allows us to conclude option A is correct.

I think we ought to just compare the answers to sin(60) BUT to perform it differently i just did: sin(72)=sin(45+27) with addition formulae, and then again i splitted (27) as (30-3) and used same formulae. Then i approximated sin(3)~0 and cos(3)~1, last thing i did few simple divisions to compare and A is the answer :)

I did this mentally in 30 seconds using elimination strategy (100% accuracy). sin72 cant be negative or 0. That eliminates options b,c,e Comparing a and d now sin72> sin60 sin72> root(3)/2 sin72> root(3/4) sin72> root(6/8) a is apx root(6.5) d is apx root(3.5) so d is eliminated and hence a is correct answer

You could also use a simple complex numbers strategy. Let z = exp(2πi/5) which is a fifth root of unity, therefore it satisfies z^4 + z^3 + z^2 + z + 1 = 0. We know that z^4 = 1/z^2 and z^3 = 1/z so this equation becomes (z^2 + 1/z^2) + (z+1/z) + 1 = 0. We can turn this into [(z + 1/z)^2 - 2] + (z+1/z) + 1 = 0. Setting x = z+1/z gives x^2 + x - 1 = 0, which has solutions x = (-1± √5)/2. We reject the negative solution, giving us that z + 1/z = (-1+√5)/2. We know that z + 1/z = 2cos(2π/5), so cos(2π/5) = (-1+√5)/4. We use the Pythagorean identity to find that sin(2π/5) = √[(5 + √5)/8]

This MAT question was surprisingly easy🤓

It would be so cool if you actually did a maths Oxford Admission. Other KZfaqrs have done it with Tom Rocks Maths.

one with the greater valve as sin is an increasing function

I finally got something right in one of these videos :D

The 37 joke was cute 😂

This is so useful

For the finaly answer, can I say that: Since Sin(45)=root(2)/2, and sin(72)>sin(45), so the answer should be lager than root(2)/2. And since root(5) > root(4) > 2, and 5-root(5)

Before you showed the answer, I squeezed the answer to be greater than root 3 / 4, and used inequalities to prove that +-root 5 cannot be negative 😂 What a way of solving the issue - understand the properties of the value you are dealing with - in this instance sin 72 was simply greater than sin 60 ...

Wow I'm speechless you are legendry so MAT for who ?

U know already that the sinx function is increasing in the interval (0,π/2) so since 72° = 2π/5 > π/3 we have sin π/5 > sin π/3 = √3/2. And you can check that √ (5 -√5)/8 < √3/2 by calculation. So it is automatically dropped.

You can see you're going to get a quadratic formula for s², so take the positive square root. Then 72° is in a positive interval for sin, so take the positive square root again. That gives A.

How often does real life come in multiple choice? As an engineer, I'll say sometimes it does, and it really saves time and effort to eliminate the answers that are clearly incorrect and go from there. In this case, one can find the right answer without doing any of the work.

My idea to solve this even without the given equation would be to draw a unique 72/18/90 deg right triangle. The only thing we need to do now is to scale it to the proper size. It is obvious the answer is either A or D so we know side/hypotenuse ratio and can pick them to fit the answer. At this point i dont have a pen and paper to actually go forward. Any thoughts?

Master, try some ITA and IME from Brazil!

I was hoping for a geometric solution involving a regular pentagon inscribed in a circle.

I got right until the last step rigorously, but I didn’t spot the sin(36) trick, so I had to approximate some square roots in my head which led to me having a good guess that the larger one is greater than sqrt(3)/2 and the lesser one is less than sqrt(3)/2. Since sqrt(3)/2 = sin(60), the answer follows.

All 5 solutions of the original equation (0, two positive and two negative) are valid. For angles: { 0, 36, 72, 108, 144 } + k * 180 for integer k including zero. For an even k they are 0 or positive and for an odd k they are 0 or negative.

Pretty sure there is an exact value for sin(36°), could use the sin(2x) formula

You have to go with the greater positive answer, simply because 72 is approaching 90

I saw the blue pen 😂

Hi bprp, I have "found" an interesting thing, if you compute the integral of the function x^(ln(x)/ln(1/2)) from 0 to infinity i dont' know why but you will get the square root of pi in the result. I will be really happy if you can make a video about this, thanks.👍👍

without solving: zero is obv not the answer it cant be negative (value is in the range [0,1]) and the one which is bigger is the answer (since it is close to sin 90, i can infer that the answer is closer to 1, so, a, yup) did this in literal 1 minute lol

Somewhere around 24/25

Nobody ever remembers that 5x angle formula and it takes a while to derive. I’d stop at 3x for 54 and 2x at 36 and know that 54 + 36 = 90 so sin of one is cos of another. You’ll find sin18 in two minutes.

I can calculate value of trigo ratios degree... 5,105,18,20,22.5,5,,.25........

Can we just eliminate the minus because we dont like it? I laughed so hard 🤣

∆ABC, AB=AC=1 and ∠A=36° Let D is on AC such that AD=BD then ∠BDC=∠C=72° and ∆ABC~∆BCD (AA) Let AD=BC=a ( cos72°=√[1-(a/2)²] ) then 1:a=a:(1-a) => a²=1-a => a=(-1+√5)/2 Thus cos72°=√(a+3)/2 =√[2(5+√5)]/4

Sin 18 is( root 5 - 1)/ 4... expanding sin 4theta in sin 2 theta form and then expanding sin 2 theta in sin theta form will get us the answe

Pls make the proof for sqrt of a+sqrt of a - sqrt of a so on

Why Theta is not equal to 72/5?

For me in India we were taught sin,cos,tan of 18,72,7½ so on angles and i actually remember sin72 as =√10+2√5÷2√2

In India we were asked to memorise special angles (18,72,36,54)..

Can you solve this high school integral ∫(1+xcosx)dx/x(1-x^(2)*e^(2sinx))

It is another way of asking about pi/5. We will have to solve a 5th degree poĺynomial.

please solve this limit for me, I am trying to figure out this from long time lim x - 0 ((1+x)^1/x-r+ex/2)/x²

You shoudl try the joint enterance exam advanced 2016 paper. Its a tough pill!

When the question is way easier if you just use logic and common sense instead of full math lmao

Come to India these are the basics

Looking at the problem, lol, I'm guessing you can take theta = 12...we know sin(60), and that is sin(5 theta)...replace all the theta's with 12's, etc, lol...then, of course, one has to solve an equation of degree 5, etc, not sure about that...the moment one has sin(12), knowing sin(60) (I don't actually remember what sin(60) was, lol, it was either 1/2 or sqrt(3)/2, but it's something that is usually known), one can easily find sin(5theta + theta) using the sum formula, something like sin(x+y)=sin(x)cos(y) + sin(y) cos(x) or something like that, don't quote me on that one, lol...that is, we would know sin(72)=sin(5x12 + 12), etc...perhaps that's the trick, the only part I don't see is how to solve the degree 5 equation, maybe the fact that there are no 4th nor 2nd degrees will help, lol...if it weren't for that part, I would consider I basically solved it, lol...I mean, of course, it's not like I actually know the sum formula or the sin(60), etc...72 might also be a multiple of 18?...Ah, yeah, lol...perhaps that would be an easier way?...But it's not like we know what the sin(18) is, traditionally, etc...we could use sin(90) = sin(18 + 72), etc, using the sum formula and the formula provided in the video, etc...at any rate, maybe that would avoid solving a quintic...at any rate, lol, my first thoughts on how one could try to solve this problem...

В Оксфорде нет таблиц Брадиса, что ли?

Plzzz Solve a^a=a

👍 job

Plzzz Solve the eqn a^a=a

But what is sin though

i got this question for 11th grade