- 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
Exactly. You can eliminate three answers off the bat, and the (d) value looked suspiciously low to me (did a bit of mental back-of-the-envelope calculation) and I settled on (a).
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
Yes! If this question is really a multiple choice, we don't even need to solve the equation. Just simple elimination can figure out which one is the real answer. That makes our life, may be an Oxford's life, more easy! 😂
@anigami01 Yes! you are correct that you can't spot the answer immediately with the example you provided. However, as what the title said -- this was an oxford admission paper. The question was set to those whose candidate can use quick method to eliminate those wrong answer in a flash, or, you would be eliminated. 😜
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 did something else. Drew a regular pentagon with side length 1. Then marked its outer circle radius as "r" and its diagonal length as "b" (they are unknown for now). From that picture, we know that a*sin(72°) = b / 2 and a*sin(36°) = 1/2. Dividing, we get sin(72°)/sin(36)° = b. Using the half angle formula, this simplifies to cos(36)° = b / 2. Now, we can get b by using Ptolemy's theorem on the pentagon: we get b^2 = b + 1 and b is equal to the golden ratio. Now, we know b, and by extension cos(36°), sin(36°) and (after using the double angle formula) sin(72°). Done.
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
Aurum ratio? I've not heard it called that before, so initially I was puzzled till I recognised the quantity you gave. Then I realised: Aurum = pertaining to gold, as in the chemical element. Yes, this is just a posher name for what I was taught as the golden ratio. Thanks for your comment: which turned out to teach me something about language as well as maths :)
Thanks for the clarification. In my native tongue it's called "razão áurea", as in spanish "proporción áurea", in french "nombre d'or" or in latim "aurea ratio"... I forgot that in english it doesn't follow the same pattern. All those words come from the hebrew "aur", which means divine light, hence the creationists idea that this ratio is present in all creation (which of course is just a point of view).
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
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.
Of course you can. If the 5 solutions were reasonably close, your have to it the hard way. But if 3 options eliminate themselves right away and you have a rough estimate of the ball park and one solution is way off, tadaaa. @@KaiserBob99
@@hazhup51 You should leave square roots intact. "Approximately 0.9" doesn't really mean anything. It has an actual value and for that value to mean more than just some random digits you need to determine it.
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
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.
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.
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.
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 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
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.
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
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.
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]
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.
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) )
Yes, you can be sure, as it needs to be quite close to 1, and (a) is the only one close to one. Or, since sin(60) is sqrt(3/2), it's clear out cannot be any other value.
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.
let f be the function defined by f(x)=cosx-sinx harmonic addition yields f(x)=sqrt(2)cos(x+pi/4) let I be the interval [-pi,pi] we can easily study the sign of f in I by putting t=x+pi/4 f(t)>=0 cos(t)>=0 pi/2+2kpi>=t>=-pi/2+2kpi (k an integer ) pi/2+2kpi>=x+pi/4>=-pi/2+2kpi pi/4+2kpi>=x>=-3pi/4+2kpi in the interval [0,pi] we get : pi/4>=x>=0 we have 72 degrees bigger than 45 degrees so f must be negative so sin of theta is bigger than cos of theta if we let sintheta be the square root with the negative inside cos will be bigger so by contradiction theta is 72
First way, just use intuiton, second way, draw it and guess (also intuition), third one: taylor series up to the fifth term have great convergence, so you cold look for something similar in the options. Less acurrate but okay too, linealize at sin (60). Finally 2 iter with newton-rhapson method would also destroy it
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
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.
Some simpler solution, please. sin72 = 2sin36cos36; sin36 = 2sin18cos18; cos36 = cos^2(18) - sin^2(18). Substitute all this to sin72 = cos18, and we obtain 8x^3 - 4x + 1 = 0 for x = sin18, or y^3 - 2y + 1 = 0 for y = 2x. This factorizes to y = 1 (no go) and y^2 + y +1 = 0, giving positive y = (√5 - 1)/2. Now going back to x^2 we get cos^2(18) = 1 - x^2 = (5 + √5)/8 and thus your answer for sin72 = cos18. Merry Xmas and Happy 2025!
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 am a 12th grade student in India and I alerady knew the value of sin18 so I just found out the value cos 18 since sin72=cos18 and got the answer within a minute. But it was nice learning how to solve a 5 degree polynomial.
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.
One place that real life problems come in the form of multiple choice, is with product selection. You may not need an exact answer, but you just need a best-enough answer. You have a finite number of options to choose from, since manufacturers only make certain sizes. As an example, consider the sizing of a water pipe. Suppose you are given the flow rate, the allowable pressure drop, the length and type of the pipe, and need to determine the diameter of the pipe. This is a difficult problem to solve directly, because turbulent flow (which water usually has) doesn't have an analytic solution, and the standard equations require you to know diameter and velocity in advance. It also isn't possible to algebraically invert the combination of equations to solve for these values. But you don't start with diameter and velocity; you start with flow rate and pressure drop. So, what you can do in practice, is use the options for diameter as inputs to the calculation, and find the corresponding flow velocity. From then on, the problem is ready for you to directly solve for pressure drop, and match it to your given data. The smallest pipe that produces a pressure drop less than the allowable pressure drop you were given, will validate selecting that pipe diameter.
To put specific numbers on my example: Flow rate m_dot = 8 kg/s Max allowed pressure drop deltaP = 60 kPa Length: 20 meters PVC pipe sizes considered: 25 mm, 35 mm, 50 mm, and 75 mm, internal diameter. Corresponding speeds at each diameter, using m_dot = rho*v*A, solved for v: 25 mm: v = 16.3 m/s 35 mm: v = 8.31 m/s 50 mm: v = 4.07 m/s 75 mm: v = 1.81 m/s Reynolds number at each size: 25 mm: Re = 407436 35 mm: Re = 291026 50 mm: Re = 203718 75 mm: Re = 135812 Friction factors at each size, from the Colebrook equation: 25 mm: f = 0.013 35 mm: f = 0.014 50 mm: f = 0.015 75 mm: f = 0.016 Pressure drops, from deltaP = 1/2*rho*f*(L/d)*v^2: 25 mm: deltaP = 1373 kPa 35 mm: deltaP = 273 kPa 50 mm: deltaP = 49.1 kPa 75 mm: deltaP = 6.995 kPa Of these options, the 50 mm is the smallest choice meets the criteria. 0.479 mm would be closer, but we only care about which of the options given we'd recommend.
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)
I know what they asked you to but they are really begging the question. So without begging the question is it possible to know the sin 72° a priori information that a circle has 360°. So we start with 360°/5 is 72°. The 72° interval between radii create 5 chords The chord then has a bisector and half-chord which are the sine of 36° and cos 36° So what we really need is the chord 144° to arrive at the sine 72°. Now we can use a marcov chain to ascertain the value of 72° starting from reasonable guess. But there is an esoteric property of the pentagon inscribed with a pentagram. If we remove a single point and all associated sides we get ptolemies quadrilateral 1 + d = d^2 and this alters to d^2 - d - 1 = 0 which is 1/2 + SQRT(5)/2 (golden ratio) This turns out to be the chord 108° of a unit circle. To understand why, assume all the sides on the pentagon are one, then by removing one point and its attached segments we expose 1 side that equals the two diagonals of the quadrilateral. Thus we simplify the solution. To understand why it’s 108°, each chord create an isosceles triangle. Each triangle has 72° on the inside. 180-72° = 108° which is 54° x 2 but each radii is a bisecting the angle, so the outside angle is 108°. d is the value of the chord on the unit circle. It’s halfchord is (1 + SQRT(5))/4 and it’s bisector is 1/4 SQRT(16- 1 - 2 SQRT(5) - 5) = SQRT(10-2*SQRT(5))/4 = SQRT(1/8)*SQRT(5-SQRT(5)) Ok so then we can deduce the halfchord and bisector of 72°. These are SQRT (1/8) SQRT(5-SQRT(5)) and (1 + SQRT(5))/4, respectively. So now we have sin of 36° but we really need the sin 72° which is halfchord 144° Chord 2t = 2 Chord t * bisector t = 4 halfchord t * bisector T. The 4 term and 1/4 term in the bisector cancel Chord 2t = SQRT(1/8) * SQRT(5-SQRT(5))*(1+SQRT(5)) Chord^2 2t = 8 * (5 - SQRT(5))*(1+2SQRT(5) + 5) = 1/8 * (5-SQRT(5))*(6+2SQRT(5)) =1/8 * (30 + 10 SQRT(5) - 6 SQRT(5)- 10) = 1/8 * (20 + 4 SQRT(5)) We want the halfchord^2 so we divide these by (1/2)^2 Halfchord^2 144° = sin^2 72° = 1/8 *(5 + SQRT(5) = (5+SQRT(5))/8 Thus sin 72° = SQRT ((5 + SQRT(5))/8)) The answer here is definite.
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
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.
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.
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.
Actually it is sin 72° = ¼√(10 + 2√5)) sin 36° = ¼√(10 − 2√5)) and you also have sin 18° = ¼(√5 − 1) sin 54° = ¼(√5 + 1) I also know all this by heart and no, I'm not from India.
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.
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?
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.
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.👍👍
∆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
I solved it before you showed the solution, I picked 'a' by logic, and not by proof because I'm lazy like that. 72 is close to 90, sin(90) is one, therefore, sin(72) must be close to sin(90), it must also mean it's in the first quadrant. I then looked at the choices, I don't know what root(5) is, I do know root(4) which is 2, therefore root(5) must be in between 2 and 3. I approximated root(5) and made it just 2, to make things easier and then solve. 'a' yield root(7/8) while 'd' yield root(3/8), approximately ofc. Therefore, since root(7/8) > root(3/8), it must be 'a'.
Most schools do that, that's why they specifically asked you to find it given the equation at hand. "I remember sin(72) from memory" would not be a valid answer
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...
Maybe 18 would indeed be better, lol...I mean...90=18x5, 72=18x4...we know what sin(90) is, lol, that would give us an equation involving sin(18), but a quintic again...hmm...it's the quintic I'm not certain about...
Ok, maybe one can reduce the 5th of the equation to something else, using some square angle formulas or something...I think there was a formula somewhere that involved sin(2theta) = some function of [sin^2 (theta)]...that might reduce sin^5(theat)=sin^2(2theta) * sin(theta), etc, we can perhaps use some identity to get rid of the 5th, and even the third of the polynomial...in other words, it would probably be manageable if I remembered more trigonometry, lol...I don't remember any identities...
Thank you for the upload! This method can be extended for proving constructability of polygons, and I have a nice visualization of proving with just ruler and compass! ruclips.net/video/U-MIXnOca7Y/видео.html
@@Gordy-io8sb Degrees are more common in secondary education due to being considered more intuitive. The test is for highschoolers, so it makes sense to use the notation that the kids would be more familiar with. Radians are awesome for uni math, but here it literally doesn't matter
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لو انني كنت افهم الإنجليزية لكان الأمر سهلا معك
I can take this course
I want to say that
obviously , 8>√5>2
so [5-√(5)]/8
Hello blackpenredpen!
I have found a pattern for a prime numbers.
- 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
Oops! You got bprp this time 😆
Same line of thought. Had the proposed answers had more diversity, twould be much much tougher to evaluate
We actually have to remember angles like 18,36,54,72,7½, and other angles🥲🥲
process of elimination is always simpler 😂
Exactly. You can eliminate three answers off the bat, and the (d) value looked suspiciously low to me (did a bit of mental back-of-the-envelope calculation) and I settled on (a).
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
True
Yes! If this question is really a multiple choice, we don't even need to solve the equation. Just simple elimination can figure out which one is the real answer. That makes our life, may be an Oxford's life, more easy! 😂
My thoughts, as well.
Does not necessarily mean we must choose the bigger of the remaining two.
However, note that 5 - sqrt(5)
@anigami01 Yes! you are correct that you can't spot the answer immediately with the example you provided. However, as what the title said -- this was an oxford admission paper. The question was set to those whose candidate can use quick method to eliminate those wrong answer in a flash, or, you would be eliminated. 😜
Nah, ain't no way. I just did this EXACT question at school today. I come home and, boom, there it is
U from India??
Just asking cuz we were also taught this in 11th grade
@@Blank10syes. I also did in 11
We all study this in 11th not just indians@@Blank10s
@@BroomieHERE I studied in 10th in india...
@@4fgaming925i m studying in 9th... basic maths in physics and vectors.
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 did something else. Drew a regular pentagon with side length 1. Then marked its outer circle radius as "r" and its diagonal length as "b" (they are unknown for now). From that picture, we know that a*sin(72°) = b / 2 and a*sin(36°) = 1/2. Dividing, we get sin(72°)/sin(36)° = b. Using the half angle formula, this simplifies to cos(36)° = b / 2. Now, we can get b by using Ptolemy's theorem on the pentagon: we get b^2 = b + 1 and b is equal to the golden ratio. Now, we know b, and by extension cos(36°), sin(36°) and (after using the double angle formula) sin(72°). Done.
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
Aurum ratio? I've not heard it called that before, so initially I was puzzled till I recognised the quantity you gave.
Then I realised: Aurum = pertaining to gold, as in the chemical element. Yes, this is just a posher name for what I was taught as the golden ratio.
Thanks for your comment: which turned out to teach me something about language as well as maths :)
Thanks for the clarification. In my native tongue it's called "razão áurea", as in spanish "proporción áurea", in french "nombre d'or" or in latim "aurea ratio"... I forgot that in english it doesn't follow the same pattern. All those words come from the hebrew "aur", which means divine light, hence the creationists idea that this ratio is present in all creation (which of course is just a point of view).
6:55
37 appears everywhere, even in your jokes :)
Fellow veritasium follower
veritasium was here
Funfact : 37 is the only number in the entire universe which is spelt as Thirty Seven
@@Aj3.14 wow... i can't believe it. do you have a source for that??
@@Aj3.14therry seven
I love these kinds of admission test math problems. Would love to see more of your videos on this
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
It should be (90-18)
Same in bd
Same brro. But we are required to learn the value of cos18°
for the fortunate rich
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.
You can't compute the actual value of the sin. You have to figure it out yourself. You can't say it's 0.95 therefore it's that thing
Of course you can. If the 5 solutions were reasonably close, your have to it the hard way. But if 3 options eliminate themselves right away and you have a rough estimate of the ball park and one solution is way off, tadaaa.
@@KaiserBob99
@@KaiserBob99 At least guessing is ok.
@@KaiserBob99 bro is upset the question is so easy that you can just approximate and eyeball sin to get to the correct answer
@@hazhup51 You should leave square roots intact. "Approximately 0.9" doesn't really mean anything.
It has an actual value and for that value to mean more than just some random digits you need to determine it.
Nice - I sorta got stuck on the last step, but your explanation made perfect sense.
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
I compared the values with sin 45, to get to the answer.
But ur argument is pure genious
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.
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.
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.
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
This man will certainly pass this question but will run out of time for the remaining MAT
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
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.
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
Those are for sure numbers
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.
Triangles have 3 sides, that’s all I know
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]
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.
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) )
Yes, you can be sure, as it needs to be quite close to 1, and (a) is the only one close to one.
Or, since sin(60) is sqrt(3/2), it's clear out cannot be any other value.
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.
This is the Best content on RUclips in the world 🙏🏿🌍!
let f be the function defined by f(x)=cosx-sinx
harmonic addition yields f(x)=sqrt(2)cos(x+pi/4)
let I be the interval [-pi,pi]
we can easily study the sign of f in I by putting t=x+pi/4
f(t)>=0
cos(t)>=0
pi/2+2kpi>=t>=-pi/2+2kpi (k an integer )
pi/2+2kpi>=x+pi/4>=-pi/2+2kpi
pi/4+2kpi>=x>=-3pi/4+2kpi
in the interval [0,pi]
we get : pi/4>=x>=0
we have 72 degrees bigger than 45 degrees so f must be negative so sin of theta is bigger than cos of theta
if we let sintheta be the square root with the negative inside
cos will be bigger so by contradiction theta is 72
First way, just use intuiton, second way, draw it and guess (also intuition), third one: taylor series up to the fifth term have great convergence, so you cold look for something similar in the options. Less acurrate but okay too, linealize at sin (60). Finally 2 iter with newton-rhapson method would also destroy it
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
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 had a similar question where we were supposed to find sin(72) on a complex analysis final.
Some simpler solution, please. sin72 = 2sin36cos36; sin36 = 2sin18cos18; cos36 = cos^2(18) - sin^2(18). Substitute all this to sin72 = cos18, and we obtain 8x^3 - 4x + 1 = 0 for x = sin18, or y^3 - 2y + 1 = 0 for y = 2x. This factorizes to y = 1 (no go) and y^2 + y +1 = 0, giving positive y = (√5 - 1)/2. Now going back to x^2 we get cos^2(18) = 1 - x^2 = (5 + √5)/8 and thus your answer for sin72 = cos18. Merry Xmas and Happy 2025!
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 :)
Yes yes this is smart and pretty cool but when you approximated sin3 😢😢😢 no
Bro .
@@wadieoyt723 I mastered the forbidden dark magic (Engineering) 😈
Sir videos help me a lot to make my concept very strong in mathematics love from India❤❤❤❤
I am a 12th grade student in India and I alerady knew the value of sin18 so I just found out the value cos 18 since sin72=cos18 and got the answer within a minute.
But it was nice learning how to solve a 5 degree polynomial.
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.
One place that real life problems come in the form of multiple choice, is with product selection. You may not need an exact answer, but you just need a best-enough answer. You have a finite number of options to choose from, since manufacturers only make certain sizes.
As an example, consider the sizing of a water pipe. Suppose you are given the flow rate, the allowable pressure drop, the length and type of the pipe, and need to determine the diameter of the pipe.
This is a difficult problem to solve directly, because turbulent flow (which water usually has) doesn't have an analytic solution, and the standard equations require you to know diameter and velocity in advance. It also isn't possible to algebraically invert the combination of equations to solve for these values. But you don't start with diameter and velocity; you start with flow rate and pressure drop.
So, what you can do in practice, is use the options for diameter as inputs to the calculation, and find the corresponding flow velocity. From then on, the problem is ready for you to directly solve for pressure drop, and match it to your given data. The smallest pipe that produces a pressure drop less than the allowable pressure drop you were given, will validate selecting that pipe diameter.
To put specific numbers on my example:
Flow rate m_dot = 8 kg/s
Max allowed pressure drop deltaP = 60 kPa
Length: 20 meters
PVC pipe sizes considered: 25 mm, 35 mm, 50 mm, and 75 mm, internal diameter.
Corresponding speeds at each diameter, using m_dot = rho*v*A, solved for v:
25 mm: v = 16.3 m/s
35 mm: v = 8.31 m/s
50 mm: v = 4.07 m/s
75 mm: v = 1.81 m/s
Reynolds number at each size:
25 mm: Re = 407436
35 mm: Re = 291026
50 mm: Re = 203718
75 mm: Re = 135812
Friction factors at each size, from the Colebrook equation:
25 mm: f = 0.013
35 mm: f = 0.014
50 mm: f = 0.015
75 mm: f = 0.016
Pressure drops, from deltaP = 1/2*rho*f*(L/d)*v^2:
25 mm: deltaP = 1373 kPa
35 mm: deltaP = 273 kPa
50 mm: deltaP = 49.1 kPa
75 mm: deltaP = 6.995 kPa
Of these options, the 50 mm is the smallest choice meets the criteria. 0.479 mm would be closer, but we only care about which of the options given we'd recommend.
@@carultch Great example!
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)
do more calculus 3 question on you channel, i really like them
Very nice. I would never solve this myself but love seeing the magic happen haha
Can you do integral of 1/(x^i)?
Beautiful math excercise. ❤
In order to choose the right value between the + vs -sqrt5, you can compare to the value of sin45
I know what they asked you to but they are really begging the question.
So without begging the question is it possible to know the sin 72° a priori information that a circle has 360°.
So we start with 360°/5 is 72°. The 72° interval between radii create 5 chords
The chord then has a bisector and half-chord which are the sine of 36° and cos 36°
So what we really need is the chord 144° to arrive at the sine 72°. Now we can use a marcov chain to ascertain the value of 72° starting from reasonable guess. But there is an esoteric property of the pentagon inscribed with a pentagram. If we remove a single point and all associated sides we get ptolemies quadrilateral
1 + d = d^2 and this alters to d^2 - d - 1 = 0 which is 1/2 + SQRT(5)/2 (golden ratio) This turns out to be the chord 108° of a unit circle. To understand why, assume all the sides on the pentagon are one, then by removing one point and its attached segments we expose 1 side that equals the two diagonals of the quadrilateral. Thus we simplify the solution. To understand why it’s 108°, each chord create an isosceles triangle. Each triangle has 72° on the inside. 180-72° = 108° which is 54° x 2 but each radii is a bisecting the angle, so the outside angle is 108°. d is the value of the chord on the unit circle.
It’s halfchord is (1 + SQRT(5))/4 and it’s bisector is 1/4 SQRT(16- 1 - 2 SQRT(5) - 5) = SQRT(10-2*SQRT(5))/4 = SQRT(1/8)*SQRT(5-SQRT(5))
Ok so then we can deduce the halfchord and bisector of 72°. These are SQRT (1/8) SQRT(5-SQRT(5)) and (1 + SQRT(5))/4, respectively.
So now we have sin of 36° but we really need the sin 72° which is halfchord 144°
Chord 2t = 2 Chord t * bisector t = 4 halfchord t * bisector T. The 4 term and 1/4 term in the bisector cancel
Chord 2t = SQRT(1/8) * SQRT(5-SQRT(5))*(1+SQRT(5))
Chord^2 2t = 8 * (5 - SQRT(5))*(1+2SQRT(5) + 5)
= 1/8 * (5-SQRT(5))*(6+2SQRT(5))
=1/8 * (30 + 10 SQRT(5) - 6 SQRT(5)- 10)
= 1/8 * (20 + 4 SQRT(5))
We want the halfchord^2 so we divide these by (1/2)^2
Halfchord^2 144° = sin^2 72° = 1/8 *(5 + SQRT(5) = (5+SQRT(5))/8
Thus sin 72° = SQRT ((5 + SQRT(5))/8))
The answer here is definite.
Tried and also done 👍🏻
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
yh its literally Q1 F, these are questions you are supposed to run thought in less than 3 minutes each. The actually harder questions come up later
@@UmairMalik-rj3in ooh i didnt know that, thanks
@@UmairMalik-rj3in Wait, you are saying, you have to prove that "The answer is 'A'"?
This MAT question was surprisingly easy🤓
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
You have to go with the greater positive answer, simply because 72 is approaching 90
Me trying to solve sin(72°) from the thumbnail without reading the complete problem🤡🤡🤡🤡
Me too😂😂😂
Me too😂😂😂
I was hoping for a geometric solution involving a regular pentagon inscribed in a circle.
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.
So it doesn't change the fact that I don't like minus and I remove it, and I am still correct
I actually remember doing this question 2 years ago for the MAT
There are a lot of ways to approach this that I took in a vastly different way and still got the same results :3
Please do integral of e^x/x dx, without Taylor or MacLaurin Series expansion. Thank you very muchA
Thanks for improving my "trigonometry" skills
Omg u just took me back to trigonometric algebra thnx 💀
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.
very nice....i understand it, only i wouldn't know where to start
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.
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.
simpler explanation ,we can write 5*72= 360 sin both side we get sin(3*72+2*72)=0
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
Wait WHAT
@@andromeda16384 yes and more than these it's not necessary to remember all of them but we need to remember multiples of 36
Actually it is
sin 72° = ¼√(10 + 2√5))
sin 36° = ¼√(10 − 2√5))
and you also have
sin 18° = ¼(√5 − 1)
sin 54° = ¼(√5 + 1)
I also know all this by heart and no, I'm not from India.
I don’t know what any of that stuff means :)
@@NadiehFan oh yeah it's ¼ for all and damn nice bro
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 ❤
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.
Those are for sure a bunch of words, do I know what they mean, no
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?
That's nice but the proper angle doesn't drawable so can only approximate the value .
@@Chess_for_fun_77 u dont draw to look at it u draw just to visualize the relations for angles and side lengths lol
@@Chess_for_fun_77 the proper angle is drawable: you just need more than a ruler and compass...
can you do Singapore H3 A level math? been suffering through those although the threshold for distinction isn't that high
I love your teaching so much ❤ form Cambodia teacher
Very insightful
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.
It would be so cool if you actually did a maths Oxford Admission. Other RUclipsrs have done it with Tom Rocks Maths.
"Oxford asks sin(72°)"
-Me in India having to learn the value of Sin(72°) for my high school exams💀
You should try the 2011 IMO question number 2 it is very fun
Exelente Ejercicio. 😃
sir please make video for calculus 3 for multivariable case
Teaching skill
Love it
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.👍👍
It must turn into the Gaussian integral after replacing x with e^t
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
Watching a video about sin(72°) with 72% battery
What a coincidence
Why Theta is not equal to 72/5?
I solved it before you showed the solution, I picked 'a' by logic, and not by proof because I'm lazy like that. 72 is close to 90, sin(90) is one, therefore, sin(72) must be close to sin(90), it must also mean it's in the first quadrant. I then looked at the choices, I don't know what root(5) is, I do know root(4) which is 2, therefore root(5) must be in between 2 and 3. I approximated root(5) and made it just 2, to make things easier and then solve. 'a' yield root(7/8) while 'd' yield root(3/8), approximately ofc. Therefore, since root(7/8) > root(3/8), it must be 'a'.
(Root3 +1)/2root 2
Facinating question
And we have to remember it as result for mains 😂
I finally got something right in one of these videos :D
I can calculate value of trigo ratios degree... 5,105,18,20,22.5,5,,.25........
I saw the blue pen 😂
You shoudl try the joint enterance exam advanced 2016 paper. Its a tough pill!
Agreed
You mean a tough suppository?
Please send the paper
jeeadv.ac.in/past_qps/2016_2.pdf
This ones for shift 2
In India we were asked to memorise special angles (18,72,36,54)..
Most schools do that, that's why they specifically asked you to find it given the equation at hand. "I remember sin(72) from memory" would not be a valid answer
@@asd-wd5bj When there are multiple choices use memories. But not in the description.
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...
Maybe 18 would indeed be better, lol...I mean...90=18x5, 72=18x4...we know what sin(90) is, lol, that would give us an equation involving sin(18), but a quintic again...hmm...it's the quintic I'm not certain about...
Ok, maybe one can reduce the 5th of the equation to something else, using some square angle formulas or something...I think there was a formula somewhere that involved sin(2theta) = some function of [sin^2 (theta)]...that might reduce sin^5(theat)=sin^2(2theta) * sin(theta), etc, we can perhaps use some identity to get rid of the 5th, and even the third of the polynomial...in other words, it would probably be manageable if I remembered more trigonometry, lol...I don't remember any identities...
Master, try some ITA and IME from Brazil!
Pls make the proof for sqrt of a+sqrt of a - sqrt of a so on
It is another way of asking about pi/5. We will have to solve a 5th degree poĺynomial.
Wow I'm speechless you are legendry so MAT for who ?
Thank you for the upload! This method can be extended for proving constructability of polygons, and I have a nice visualization of proving with just ruler and compass!
ruclips.net/video/U-MIXnOca7Y/видео.html
No radians?
I thought one of the most prestigious universities in the US or otherwise would know to use radians lmao
@@Gordy-io8sb Degrees are more common in secondary education due to being considered more intuitive. The test is for highschoolers, so it makes sense to use the notation that the kids would be more familiar with. Radians are awesome for uni math, but here it literally doesn't matter
@@Gordy-io8sb Also Oxford is in the UK
В Оксфорде нет таблиц Брадиса, что ли?