@@ajbonmg I oversimplified my explanation. In more details, not only I should convert to radians, but limit the domain too, because f(x) is not continuous in every point. Limiting the domain to ( -1, pi -1) fix it, since the required values are inside this interval.
@@jacob.invertase That's true, and you are right - for the purposes of Braian's method, it makes no overall difference. However, he said 'use calculus' - which only really works in radians. The derivative of sin is only cos if it is in radians.
I did it in a very different way. The derivative of Sin(x) is Cos(x). Since the Cos(0)=1 and decreases when the angle increases (at least for the first 90 degrees this is true in the positive direction), that means that the difference between Sin(1 degree) and sin(2 degrees) is larger than sin(3 degrees) and sin(4 degrees), because, while the difference is both 1 degree, 1 and 2 are closer to 0. And since the Sin(x) function is increasing for the first 90 degrees, we can say that Sin(2 degrees) > Sin(1 degrees) and Sin(4 degrees) > Sin(3 degrees). If the difference between sin(3 degrees) and sin(4 degrees) is smaller (which it is as shown before) and the values are larger (which they are), than the relative difference is even smaller and the devision must be closer to 1, therefore it is a larger value.
Yes, looking only at 0 to 90 degrees, sine is an increasing function, and the derivative of sine is a decreasing function so clearly sine 2* / sine 1* would be bigger than sine 4* / sine 3* and if you invert both sides you get the answer to this question
Likewise. I tend to just formulae these kinds of problems out, but for once I tried to think visually since I'm always teaching my students to think visually for trig questions. The small angle approximation is very nice too though!
I love how many are using different methods. I also did it in a different way that doesn’t require any calculation whatsoever, I looked at the rate of change of sin(x) it’s faster the closer you’re to zero, based on this you can know which is greater, it’s the part with more angle or the slower rate for a better description.
We could solve this with basic trigonometry Multiply and divide sin 1/sin 2 with 2 cos 2 the ratio will then become (sin(1+2)+sin(1-2))/sin 4 upon simplifying it will be (sin 3 - sin 1)/sin4 which is clearly less than second ratio
The 1 second solution, possible just from seeing the video thumbnail is memorizing that for small values of a, sin(a) ~= a. So: sin(1)/sin(2) ~= 1/2 and: sin(3)/sin(4) ~= 3/4. This is shown later in the video with a formal derivation, but the sin(a) ~= a trick is great knowledge to have in your toolkit.
Thanks for spreading the true etymology of sine. In BHARAT, in some books written in hindi,sine is still called "jya". Cosine - "kojya" Tangent- "sparshjya" The chord of a circle is called "jiva".
The way i thought of it was. Sin0 ->Sin 30 we go from 0 to 1/2 which is a 0.5 change. From Sin30->Sin 60 we go from 1/2 to √3/2 which is a 0.35 change. This observation can also be made from the sine graph, which starts going flatter the closer we are to 90. Therefore the difference at a later angle is lesser than the difference at an earlier angle. Therefore the difference in sin 2 and sin 1 is greater than the difference in sin 4 and sin 3 Therefore the rhs would be larger.
This was very formal and strict, for those who like that. Those who like to be more intuitive, will notice that close to (0,0), the sine function is very close to linear. For those who are able to imagine walking 1, 2, 3 and 4 steps along the something that is very close to a straight ascending path, they will realise that the problem comes down to comparing approximately 1/2 to approximately 3/4. Which is just an intuitive version of the second method presented by Presh. This took me 7 seconds, while the video is 7 minutes long :-)
My reasoning was pretty similar, but started with "sine is an increasing fuction over the first 90°, therefore we don't actually have to worry about the sine". Sometimes solving the problem isn't the hard part, it's proving it.
I sure appreciate Presh's work with interesting math problems. The slope of the sine wave is steepest at theta = zero. This means that sin2/sin1 is a large ratio bested only by sin1/sin0 using natural numbers. Thus sin1/sin2 is immediately the smaller of the two ratios in this problem.
My quick cheat is to use the small angle approximation for sin(x). That is sin(x) approximates to (x). Of course in radians, but as the pi/180 bit conversion cancels out that's easy. Of course for 3 & 4 degrees it's pushing things a bit to use that approximation, but if it's not too close a result it will work. The first one will approximate to 1/2 and the second to 3/4, so the sin(3)/sin(4) is > than sin(1)/sin(2). At least I hope so. Very quick and dirty of course and it's not going to work for larger angles. Written before watching the video I should add.
Actually, you can use the small angle approximation for a lot more. How much do you think the difference between sin(1°) rounded to 3 digits after the decimal point and 1° (π/180) rounded to three digits after the decimal point? The difference is actually 0 The same goes for sin(2°), sin(3°), all the way until sin(8°), that's when the 3 digit approximation differs. By how much? Only by 0.001 (i.e sin(8°) ≈ 0.139 while 8° ≈ 0.140) Do you know when does the difference reach 0.010? At sin(23°) (i.e. sin(23°) ≈ 0.391 while 23° ≈ 0.401) So I think you can push it all the way until 23°. You may say though: "but we're dividing two approximations! Won't the error compound?" Well, here are the results from comparing sinx/sin(x+1°) and x/(x+1°): The difference between them at x = 1° all the way till x = 4° is 0. The difference grows pretty slowly though, at x = 23°, the difference is only 0.003, and by the time x = 90°, the difference has only reached 0.011 So actually, the error was _less_ than before. In conclusion The small angle approximation works better than you think for sinx, and it works wayyyy better than I thought for sinx/sin(x+1°)
*_Analyising the problem graphically_* As in a decreasing function ∆y per ∆x decreases as well as we go forward in x Sin x < Sin x+1 hence both are proper fraction....so the *one having small ∆y in fixed ∆x will be bigger* As we discussed ∆y will decrease on going forward in ∆x, hence ∆y = smaller for x = 3° to 4° (∆x is same for both = 1).....this way we can say Sin 3°/Sin 4° have smaller change in numerator and denominator for a proper fraction so Sin3°/Sin4° is bigger 🙌
you can also do it using differential calculus let f(x)= sinx/sin(x+1) now on differentiating we will get f'(x)= {cos(x)sin(x+1)-sin(x)cos(x+1)}/sin^2(x+1) = sin(1)/sin^2(x+1) which is clearly positive so we can say that f(x) is an increasing function so f(1)
Maybe with a Taylor series but then you need to convert angles to radians, and you need to prove how many terms are enough, which - while being feasible - is more complex than using either geometry or trigonometric formulas. In any case, for any approximation, you need to prove that it won't change the inequality.
I guessed correctly based on the approximation, but I don't have a good sense about how small of an angle is "small enough" for that approximation to "be close enough." I was concerned that 4 degrees was maybe not close enough to "small".
@@markday3145 well but since the derivative of the sine is cosine, its slope decreases between 0 and π/4. Therefore at any number x in between of 0 and π/4 sin(x)
You don't need to think about whether the gradient is increasing or decreasing, as long as you realise that the slope is close to constant for small angles.
you could also use small angle approximation, where sin(x) ≈ x (when x is in radians) and convert 1°, 2°, 3° and 4° to π/180, π/90, π/60 and π/45 rad respectively. comparing the ratios, you get 1/2 and 3/4, and we know that 3/4 > 1/2
When I did my private pilot's license, I was taught to use sin x = x for 'moderate' angles (of the wind direction relative to the track of the aircraft) to mentally estimate a course to set if you needed to divert to an unplanned airfield. Therefore I immediately used that trick here, and of course 3/4 > 1/2.
It took me awhile to get it, but I just simply used the fact that the slope of the sine curve is gradually decreasing up to x = π/2. The sin itself is increasing from x= 0 to x = π/2. So then y-difference from sin 1⁰ to sin 2⁰ is more than that from sin 3⁰ to sin 4⁰. (sin 2⁰ - sin 1⁰ > sin 3⁰ - sin 4⁰). So then (sin 1⁰ / sin 2⁰) < (sin 3⁰ / sin 4⁰). sin 88⁰ almost equals sin 89⁰, So sin 88⁰/sin 89⁰ is probably like .99. So then for a given x, sin x⁰ / sin (x+1)⁰ are more nearly equal, the closer x is to π/2., So yes, sin x⁰ / sin (x+1)⁰ keeps increasing as x increases, up to x = π/2.
I found it just by looking at it, knowing that, in Trigonometry circle, vertical bar of sinus, when the angle increase, the projected value is increasing logarithmically. Which means that the projected difference between sin2 and sin1 on the sinus bar is more than the difference between sin4 and sin3, making sin1/sin2 is closer to zero than sin3/sin4, and therefore sin1/sin2 is less than sin3/sin4.
By using the approximation sin(x) = x you can assume that sin(1)/sin(2) ~= 1/2, and sin(3)/sin(4) ~= 3/4. However, the differences between the actual values and the approximate values goes up as x increases. If we consider things in terms of ratios, we only need to consider the amount at which the numerator changes, so we can change the problem to sin(1)/sin(2) = (1+e1)/2, and sin(3)/sin(4) = (3+e2)/4. Since e2 > e1, (3+e2)/4 > (1+e1)/4, therefore sin(3)/sin(4) > sin(1)/sin(2).
Sine=Opp/Hypotenuse. The change is Opp is greatest per change in the smallest degree. Can also think of it as the gradient of the sine curve (max at zero and decreasing to 90)
Similarly I approximated that hypotenuse is not going to change significantly, so can be written as (1/h)/(2/h) and (3/h)/(4/h) the h are going to cancel out so you end up with 1/2 and 3/4.
I did this in my head, knowing that for small angles sin(a) ~= a if a is in radians. Converting degrees to radians is unnecessary in this case because it won't change the ratios. Therefore the left side is approximately 0.5 and the right side is approximately 0.75. Thus the left side is smaller. Checking with calculator: sin(1)/sin(2) = .500076... and sin(3)/sin(4) = .7502666... Amazingly close!
There's another solution to this problem: we know that , 0 < sin 1< sin 2< sin 3 < sin 4 ..(i) we can simplify as 1 , 2 ,3 & 4 wrt to their order in (i) now since , 1/2 < 3/4 therefore, we can easily say : sin 1/sin2 < sin 3/sin4
0 < S2 < S4 < 1 in the 1st quadrant. So, S2/S4 < S3/S4 Now S4 = 2*S2*C2 So, S2/S4 = 1/(2*C2) Now S2 = 2*S1*C1 S1/S2 = 1/(2*C1) Now 1 > C1 > C2 > 0 in the 1st quadrant. So, 1/(2*C1) < 1/(2*C2) So, S1/S2 < S2/S4 So, S1/S2 < S2/S4 < S3/S4 So, S1/S2 < S3/S4
I just visualised a sine wave and it is know that it approaches a stationary point at pi/2 - that means the gradient decreases gradually due to its concave down nature. Therefore the interval between sin 1 and 2 is greater than the interval between sin 3 and 4, so sin3/sin4 must be a larger fraction.
Hi I just figured out another way to prove this: Say you have a function that is continuously increasing on an open interval (p,q) not including endpoints. In addition, the slope of this function is continuously decreasing. So then if a f(a), and f(b + w) > f(b), so f(a) / f(a + w), and f(b) / f (b + w) are both positive fractions less than 1. This is assuming a, b, a+w, b+w are all in the interval. And since the slope is continuously decreasing, f(a + w) - f(a) > f(b + w) - f(b) So then we know f(a) / f(a + w) < f(b) / f (b + w). This is because not only is the absolute difference between numerator and denominator more on the left, but also the denominator on the right is greater. The sine function is just such a function from x=0 to x=π/2, and the selection of numerator and denominator on the right side fits how it was done on the left side (w = π/180). So then sin (1⁰)/sin(2⁰) < sin(3⁰)/sin(4⁰).
An old engineer's view: I'm old enough to have used a slide rule throughout my engineering training. I recall the facts that (1) a standard slide rule uses a common scale (ST) for both sin and tan below ~5.7º (0.1 radian), and that (2) the ST scale is linear within well below 1% in that range. So, sin 1º/sin 2 = 1/2 ± 2%, and sin 3º/sin 4º = 3/4 ± 2%. Therefore, sin 3º/sin 4º is larger.
I thought of the graph of y = sin x and how the gradient of the curve decreases steadily from infinity at x = 0° to 0 at 90°. So the difference between the y values at x = 1° and x = 2° is larger than the difference between the y values at x = 3° and x = 4°. Therefore the ratio of sin 1°:sin 2° is smaller than the ratio of sin 3°:sin 4°. Similarly, sin 5°/sin 6° > sin 3°/sin 4°, and so on until sin 89°/sin 90°.
The derivative of y=sin(x) = y=cos(x), and y=cos(x) decreases from y=1 at x=0, to y=0 at 90 degrees, so in turn the gradient of sin(x) decreases as x increases (to 90deg). So the value difference between sin(1) and sin(2) will be greater than sin(3) and sin(4), and since sin(0) = 0, the proportional difference between that gradient's effect on the value itself will also be greater, therefore the fraction sin(3)/sin(4) will be closer to parity than sin(1)/sin(2).
As someone who was taught by my teacher during basic math class that sinx ≈ x in radians. And That's how I know that sin3°/sin4° > sin1°/sin2° Sin(π/60)/sin(π/45)> Sin(π/180)/sin(π/90)
The slope of the function sin() at the point x=0 is maximum and decreases to zero at the point 90⁰. So, the difference between sin(1⁰) and sin(2⁰) is greater than between the values of sin(3⁰) and sin(4⁰), so sin(1⁰) / sin(2⁰) < sin(3⁰) / sin(4⁰)
f= sinx/sin(x+π/180) = 1/(cos(π/180)+cotx·sin(π/180)). Cotx is decreasing in the proximity of zero (the opposite of tanx) -> f is increasing, so the second ratio is greater.
Just looking at the graph of sine and its derivative cosine, it is obvious to me that the steps between succeeding degrees decreases with each degree between 0° and 90°. Therefore the higher degree in the ratio is on the bottom, the ratios of successive degrees increases between 0° and 90°. sin(1°)/sin(2°) must be smaller than sin(3°)/sin(4°).
To approximate sin x near 0, use the Maclaurin series expansion of sin x. The Maclaurin series expansion of sin x is given by: sin x = x - (x^3/3!) + (x^5/5!) - ... So, for small x just use the first term approximation where sin x = x and x is in radians. I believe this is the same as your Method 2
And the ratio of two angles in degrees is the same as the ratio of two angles in radians, so no need to convert to radians. (Radians and degrees differ by a constant scale factor.)
@@charleslivingston2256 The Maclaurin series expansion for sin(x) = x - (x^3/3!) + (x^5/5!) "MUST" use radians to approximate the true value of sin(x). If you plugged in degrees you would generate the wrong answers.
sin for very small angles is roughly equal to the angle itself (in radians, but it doesn't matter since we're dividing and the units cancel). 1/2 < 3/4. The more interesting challenge would be cosine.
I went with the idea that sin(x) increases quicker as x increases up until 45 degrees so the difference between sin(3) and sin(4) would intuitively result in a bigger fraction than sin(1) and sin(2). Not exactly rigorous but it works intuitively.
other way round.... we can see 1/2=.5 2/3=.67 ..... ... .. and further 999/1000=.999 we can see for small x/(x+1) value is less as compared to next ones.. therefore if we check for sine it means same as sine is inc. function between 0 to pie/2
Not only is the history of sine and cosine interesting, it is also interesting that in degrees, you can use the small angle approximation in degrees as long as you are using ratios. Makes sense since I guess since radians are a unit-less ratio, I think?
I see it as that the reason is that it works here because converting between degrees & radians is done by multiplying by a constant: You multiply a constnat into the sign on both top and bottom of the fraction, then drop the sign using the approximation, and then you're left with your original numerator & denominator, each multiplied by that constant, which you can then eliminate. If that's what you meant by "unit-less ratio" (= multiplying by a constant), sorry for rewording your thoughts. :)
There is another way to solve it. Sine is an increasing function between (0, π/2), but it plateaus as it nears π/2. So sin2° - sin1° > sin4° - sin3° sin3° is greater than sin1° and since the difference between the denominator and numerator in the first term ISS larger, that term is the smaller term.
There is no need to switch from degree to radian. in the neighborhood of 0 we have sin(x)~ x hence sin(1°)/sin(2°) ~ 1/2 and sin(3°)/sin(4°) ~ 3/4. Thank u 👍
there is yet another way, if u look at sine curve, its rate of increase is faster at lower angles than at higher angles (between 0-90 deg), therefore |sin2 - sin1| > |sin4 - sin3| so in fraction when numerator is more smaller than denominator, fraction is lesser (sin1/sin2) and smaller the gap greater the fraction (sin3/sin4) therefore conclusion - sin1/sin2 < sin3/sin4 (note: angles are in degrees)
Thanks for this awesome video! I’ll pass this around to math teacher friends. Not sure why you converted to radians. sin x° is approximately x° when x° is small, and even 4° is quite small. (Up to about 22°, this approximation is within 1/100 of the correct value: I computed this by solving x^3/3! < 1/100.)
A related fun question: which is larger, sin(1°)/sin(2°) or sin(3°)/sin(6°)? The two methods presented in the video break down in this case (at least when keeping only the first term in the Madhava-Gregory series). Instead you might think about the sign of the second derivative for small positive angles, similar to what some of the others have posted in the comments. By the way, thanks for all the great videos! I've watched a lot of them :)
Use calculus! Define f(x)=sin(x)/sin(2x). Take the derivative, which is f'(x)=2sin³(x)/sin²(2x). This expression is greater than zero from zero to pi/2, so it's increasing in this interval. 1° and 3° are in this interval, so f(1°)
We can solve this problem more easily by taking sine approximation that is if sine theta is very small angle ( less than 10degrees) we can take take its value as theta in radian so sine 1 degree equals pi /180 * 1 similarly sin 2 degree equals pi/180 * 2 in this way we can find the values and compare
From 0° to 90°, sine is an increasing function, but the rate of increase is decreasing. Therefore, the ratio between sin(1°) and sin(2°) is farther than the ratio between sin(3°) and sin(4°).
I did it in my head by looking at the slope of a sine wave. from 0 deg to 90 deg the slope decreases. eg. the slope at 5deg is greater than the slope at 10 deg. the ratios are the inverse of the slope so the one further along the line is greater. another way to look at it is : the difference between sin3 and sin4 MUST be less than that of 1 & 2 so the fraction MUST be closer to ONE.
Approximations, graphs, derivatives... I chose instead to use only the most primitive trig identities sin(A+B) = sinAcosB+cosAsinB and sin(A-B) = sinAcosB - cosBsinA. So... sin2cos1 - sin1cos2 = sin1 > 0 Add 0 in the form sin1cos2-sin1cos2 sin2cos1 + sin1cos2 - 2sin1cos2 > 0 So sin3 - 2sin1cos2 > 0 Multiply by sin2 , which is > 0 sin3sin2 - 2sin1cos2sin2 > 0. Now using sin4 = 2sin2cos2 sin3sin2 - sin1sin4 > 0. The result follows on division by sin4, which is also > 0
because a circle from pi/2 to 0 has a decreasing slope the small differences between the sines or y values are going to be larger when the angles are close to 0 and smaller differences as the angle goes up to pi/2
Wow, you really overdid that explanation as I instantly used the small angle approximation where the sine = tangent = angle in radians up to about 15degres. The degree to radian ratio is a constant. OK you formally presented the answer.
i immediately jumped to the small angle approximation method, except noticed immediately that the conversion through to radians doesn't matter, since we only care about the ratios of the values. with sin(x) / sin(2x) and sin(3x)/sin(4x) for a sufficiently small x will just be in the ratios of 1/2 and 3/4, regardless of what x actually is.
Using calculus in its essence, if you determine that the rate of change between 1 degree and 2 is, more than the rate of change between 3 and 4, then (sin 1°)/(sin 2°) < (sin 3°)/(sin 4°).
cos x is the rate at which sin x increases at any value of x because cos x is the derivative of sin x with respect to x. Since cos x peaks to 1 at x=0 and then falls to 0 at x=pi/2, the rate at which sin x increases between x=0 and x=pi/2 falls from 1 to 0 in a monotonic manner. So sin 2 degrees/ sin 1 degree > sin 4 degrees/sin 3 degrees =>.sin 1 degree/ sin 2 degrees < sin 3 degrees/sin 4 degrees.
Should be sin 3°/sin 4°. The sine function has both its highest rate of increase and lowest absolute value at 0°, so the further away from 0°, the closer sin x°/sin (x+1)° should be to 1. In fact, iirc, the derivative of sin x at 0° should be cos 0°, or 1, which should make sin 1°/sin 2° and sin 3°/sin 4° fairly close to 1/2 and 3/4 respectively.
d/dx of sinx is cosx which is decreasing function in (0,pie) hence we can say that rate of increase of sinx decreases as x increses so sin2/sin1>sin4/sin3
I used the approx. method, but instead of calculating the answer I used the prooerty that assuming 2 numbers have a constant sum, the closer they are to each other the greater their product will be. If you hand wave your wau to an answer, hand wave all the way 😅
The steepness of the curve decreases from 0-90 deg. So sin(3) is closer to sin(4) than sin(1) is to sin(2). So the left ratio must be less than the right one.
Sin1/sin2 =sin (pi/180)/(pi/90) Using small angle aprox (Pi/180)/(pi/90) Pi cancels =90/180=1/2 Sin1/sin2=1/2 Sin3/sin4 =sin(3pi/180)/sin(pi/45) Small angle aprox (Pi/60)/(pi/45) Pi cancels 45/60=3/4 Sin3/sin4=3/4 3/4 is greater than 1/2 so sin3/sin4 is greater than sin1/sin2
At the same inputs, sin(x) is approximate by f(x)=x, so if we were to use f(x), we would have f(1)/f(2) and f(3)/f(4), which are 1/2 and 3/4, so the former is less than the latter.
Hmm. I just pictured the start of the sin curve, since we're only covering 1° to 4° (which are less than 90°) we know the gradient of the sin curve is decreasing as you increase the degrees, so sin 1° and sin 2° will have a greater both absolute and relative difference than sin 3° and sin 4°, therefore the ratio will be a lower number. And this will be true of any two adjacent numbers (again, up to 90°). So (sin 51°)/(sin 52°) < (sin 52°)/(sin 53°) < (sin 53°)/(sin 54°); with the result of (sin [n-1]°)/(sin n°) approaching 1 as n approaches 90°. In fact it's 0.999848 when n = 90° In fact because it's a ratio, this trend will continue until you get to 179°, maxing out at 2, then erroring at 180° (div by zero), and then repeating infinitely.
I didnt have immediate intuition, so i pictured a sine wave. One period equals 360 degrees so its not hard to imagin the sine wave is approximately a straight line of form sin=a*degrees where a is some constant. Sin(1) simplifies to a*1, sin(2) simplifies to a*2 etc etc so basicly half is less than three quarters
wow thats really cool but i just realised sin slows down over time so the difference between sin 3/sin 4 is smaller meaning its greater (like comparing 1/2 and 3/4)
when you imagine the sine wave, you will notice that 1,2,3 and 4 degrees are extremely close to each other, thus you can treat them like a line equation 1/2 < 3/4
Sine increases at a decreasing rate from 0 to 90 degrees, meaning the differences between consecutive y-values will decrease as x increases. So sin(2)-sin(1)>sin(4)-sin(3) and sin(1)/sin(2)
There's another way, by examining the comparative slope of the line between the two points, and recognizing the concavity of sine(0) to sine(90). As you get closer to the peak, the slope of the line between two points flattens. That means the two values get closer together. Ergo, sine(1) vs sine(2) is a much larger difference than sine(anything larger) vs sine(anything larger+1), at least up to 90degrees. Putting that into fraction form, the fraction with the larger angles gets closer to 1 than the fraction with the smaller angles, therefore sine(1)/sine(2) must be smaller. Knowing nothing else, including knowing no fancy equivalences or funky approximations, you can derive the answer pictorially ("geometrically"?) in a way that a 5th grader could understand.
Sir I think there are two reason for increasing of [sin(3°)/sin(4°)] from [sin(1°)/sin(2°)]. First slope is decreasing as a function of cos(x). Which causes less change in denominator in case of sine(4°) in comparison of sin(2°). And second reason is that numerator of second one is high that why it is larger. Instead if there will be same slope. We can understand this with assumption. First we have assume that sin(1°)/sin(2°)=x/y and sin(3°)/sin(4°)=a/b And other assumption . y=x+∆x(in other word sin(2°)=sin(1°)+∆sin(1°) and similarly b=a+∆a. Case 1=> If value of x and a are same then in this situation also sin(3°)/sin(4°) will be larger than sin(1°)/sin(2°) because{ x/(x+∆x) }will be smaller than {x/(x+∆y)} because ∆y is larger than ∆x as slope changes with cosine as cosine of larger number is smaller because number is in between 0° and 90°. case 2 => If slope is constant than also sin(3°)/sin(4°) will be larger than sin(1°)/sin(2°). Because according to our assumption (x/x+∆x) and (y/y+∆x) here ∆x and ∆y are same. And y is greater than x . Just like 1/2 and 3/4. 🙏🙏🙏
wich is greater can be just reasoned. the value gets bigger, so for the aproximation, you can just ignore sin completely and see that 3/4 > 1/2 and see that the left one sin(1) / sin(2) is smaller. No actual values needed or requeted in the question.
i went about it like this: sine is the vertical projection of the unit line at an angle. its 0 at 0 degress and inversely 1 and 90 degrees. the cosine is the horizontal projection. 1 at 0 degress and inversely 0 at 90 degrees. the sinus at angle 1, 2, 3 and 4 therefore is a increasing function, at each degree roughly the same amount. therefore it estimates to 1/2 < 3/4.
I compare the two fraction as follows: sin 1° / sin 2° - sin 3° / sin 4° = (sin 1° sin 4° - sin 2° sin 3°) / (sin 2° sin 4°) = (sin 1° * 2 sin 2° cos 2° - sin 2° sin 3°) / (sin 2° sin 4°) [using the double angle identity sin 4° = 2 sin 2° cos 2°] = (sin 2° (2 sin 1° cos 2° - sin 3°)) / (sin 2° sin 4°) [factor out sin 2° from the numerator] = (2 sin 1° cos 2° - sin 3°) / sin 4° = (2 sin 1° cos 2° - sin 2° cos 1° - sin 1° cos 2°) / sin 4° [using the sum formula for sine: sin(A+B) = sin A cos B + sin B cos A] = (sin 1° cos 2° - sin 2° cos 1°) / sin 4° = sin (-1°) / sin 4° [using the difference formula for sine: sin(A-B) = sin A cos B - sin B cos A] < 0 [ since sin (-1°) < 0 and sin 4° > 0] Therefore: sin 1° / sin 2° < sin 3° / sin 4°
I don't know why 'jya' (the blue line) would refer to the length of the bow-string (half of which is the green line)- when, apart from anything else, it's the angle of the radius that determines the length of 'jya'...
I solved it without any calculations. The difference on both sides between the angles is the same 1°=2-1=4-3 The fact, that the change of the sinus value by increment of 1° in the angle gets smaller the further we move away from 0° towards 90° implies that: sin(1)/sin(2)< sin(3)/sin(4)
I just visualized the graph of sin(x). After x=0, the rate of increase becomes slower until it reaches zero. So, sin(2)/sin(1) will be greater than sin(3)/sin(4) or conversely sin(1)/sin(2) will be less than sin(4)/sin(3).
I used some know facts: The function sin(x) is increasing so: Sin 1 < sin 3 The slope of the graphic is getting smaller so: Sin2 - sin 1 > sin4 - sin3 Knowing these 2 facts out is clear that the right hand side is greater.
Consider a function f(x) = sin(x)/sin(x+a), where a is any real number, but in this case it will be 1 degree. If you diff the function f you get f'(x) = (sin(x+a)cos(x)-sin(x)cos(x+a))/(sin^2(x+a)). Use the added angle identities for sin and cos for the top part. You get: sin(a)cos^2(x)+sin(x)cos(a)cos(x)-sin(x)cos(a)cos(x)+sin^2(x)sin(a) The middle parts cancels out, and the left and right parts combined are: sin(a)(cos^2(x) + sin^2(x)) = sin(a) * 1 = sin(a) Aha! So, f'(x) = (sin(a))/(sin^2(x)), and since sin(a) is a constant and the denominator is squared... The function f(x) is monotone! And its dependent on the number a we choose. In this case one degree was chosen, so f'(x) is always > 0 => f(x) is increasing! Of course be careful with the domain, but in this videos case its all where it should be. So, overall, f(1) < f(3) => sin(1)/sin(1+1=2) < sin(3)/sin(3+1=4) {in degrees}
People.. this is probably the best place to ask this question.. I work in a builders supplies yard and I sell thousands of different products at different prices ... today something happened I never saw before (20 years )in my work. A customer paid for some products and the very next customer bought completely different products and different quantities. They both paid by card and the amount was exactly the same ie £14.69 ..wow I couldn't believe it.. what are the odds this would happen?
I'm studying Physics, so my immediate thought went to sin x~ x lol
Me too 😂😂😂😂
As an Engineer, sin (x) isn't "approximately" x. Sin(x) *is exactly* x.
And pi is 3, g is 10, and whatnot.
Good old small angle approximation, never fails
But that works when it is given in radians not in degrees
@@mrsqueaksrules
sin x=tan x = x = tan(inv)x = sin(inv)x
π=3=e=2
😂
Use calculus! f(x)=sin(x)/sin(x+1) is increasing in every point in its domain (just do the derivative and check it). That way, f(1)
Good approach, but you should convert to radians first.
@@ajbonmg I oversimplified my explanation. In more details, not only I should convert to radians, but limit the domain too, because f(x) is not continuous in every point. Limiting the domain to ( -1, pi -1) fix it, since the required values are inside this interval.
@@ajbonmg I don't think you need to convert, it's just a postive scaling
@@jacob.invertase That's true, and you are right - for the purposes of Braian's method, it makes no overall difference. However, he said 'use calculus' - which only really works in radians. The derivative of sin is only cos if it is in radians.
@@ajbonmg calculus still works in degrees haha, sin_degrees(x) = sin_radians(360x/2*pi`) so it's the same derivative up to a positive multiple :)
I just drew a sketch of a circle, imagined a radius turning through 1,2,3,4 degrees - it was obvious which ratio would be much larger....
Don't understand, I see that your comment is 6 days old, and that this video is 5 minutes old. That's unfair. 😉
@@yoops66yes bro 😂..how is this possible 🧐👀
6 days ago? How???
Time traveller
Yooo is this comment FR??
I did it in a very different way. The derivative of Sin(x) is Cos(x). Since the Cos(0)=1 and decreases when the angle increases (at least for the first 90 degrees this is true in the positive direction), that means that the difference between Sin(1 degree) and sin(2 degrees) is larger than sin(3 degrees) and sin(4 degrees), because, while the difference is both 1 degree, 1 and 2 are closer to 0. And since the Sin(x) function is increasing for the first 90 degrees, we can say that Sin(2 degrees) > Sin(1 degrees) and Sin(4 degrees) > Sin(3 degrees). If the difference between sin(3 degrees) and sin(4 degrees) is smaller (which it is as shown before) and the values are larger (which they are), than the relative difference is even smaller and the devision must be closer to 1, therefore it is a larger value.
This was my solution also.
had the same idea but sinx =x is way to neat
@@lol65XD it's definitely easier, but I'm not a big fan of using the ≈ for a mathematical proof.
Yes, looking only at 0 to 90 degrees, sine is an increasing function, and the derivative of sine is a decreasing function
so clearly sine 2* / sine 1* would be bigger than sine 4* / sine 3*
and if you invert both sides you get the answer to this question
Likewise. I tend to just formulae these kinds of problems out, but for once I tried to think visually since I'm always teaching my students to think visually for trig questions. The small angle approximation is very nice too though!
I love how many are using different methods. I also did it in a different way that doesn’t require any calculation whatsoever, I looked at the rate of change of sin(x) it’s faster the closer you’re to zero, based on this you can know which is greater, it’s the part with more angle or the slower rate for a better description.
The historical etymology of 'sine' was very nice! I had never heard of it, and I learned something new.
Yeah
Much more enlightening than past videos I've seen. Well done! I love multiple perspectives. Brings balance. Keep up the good work
We could solve this with basic trigonometry
Multiply and divide sin 1/sin 2 with 2 cos 2 the ratio will then become (sin(1+2)+sin(1-2))/sin 4 upon simplifying it will be (sin 3 - sin 1)/sin4 which is clearly less than second ratio
The 1 second solution, possible just from seeing the video thumbnail is memorizing that for small values of a, sin(a) ~= a.
So: sin(1)/sin(2) ~= 1/2
and: sin(3)/sin(4) ~= 3/4.
This is shown later in the video with a formal derivation, but the sin(a) ~= a trick is great knowledge to have in your toolkit.
Thanks for spreading the true etymology of sine.
In BHARAT, in some books written in hindi,sine is still called "jya".
Cosine - "kojya"
Tangent- "sparshjya"
The chord of a circle is called "jiva".
At such a small value sin(x) is very close x, so... since 1/2
It cancels out, but it is needed for more rigor.
The way i thought of it was. Sin0 ->Sin 30 we go from 0 to 1/2 which is a 0.5 change.
From Sin30->Sin 60 we go from 1/2 to √3/2 which is a 0.35 change.
This observation can also be made from the sine graph, which starts going flatter the closer we are to 90.
Therefore the difference at a later angle is lesser than the difference at an earlier angle.
Therefore the difference in sin 2 and sin 1 is greater than the difference in sin 4 and sin 3
Therefore the rhs would be larger.
This was very formal and strict, for those who like that.
Those who like to be more intuitive, will notice that close to (0,0), the sine function is very close to linear. For those who are able to imagine walking 1, 2, 3 and 4 steps along the something that is very close to a straight ascending path, they will realise that the problem comes down to comparing approximately 1/2 to approximately 3/4. Which is just an intuitive version of the second method presented by Presh. This took me 7 seconds, while the video is 7 minutes long :-)
My reasoning was pretty similar, but started with "sine is an increasing fuction over the first 90°, therefore we don't actually have to worry about the sine". Sometimes solving the problem isn't the hard part, it's proving it.
You are comparing small values with small values, and even dividing them, so you have to prove that these approximations do work.
I sure appreciate Presh's work with interesting math problems.
The slope of the sine wave is steepest at theta = zero. This means that sin2/sin1 is a large ratio bested only by sin1/sin0 using natural numbers. Thus sin1/sin2 is immediately the smaller of the two ratios in this problem.
As an engineer, sin(x) = x is powerful tool for small angle approximation. We use it all the time.
My quick cheat is to use the small angle approximation for sin(x). That is sin(x) approximates to (x). Of course in radians, but as the pi/180 bit conversion cancels out that's easy. Of course for 3 & 4 degrees it's pushing things a bit to use that approximation, but if it's not too close a result it will work.
The first one will approximate to 1/2 and the second to 3/4, so the sin(3)/sin(4) is > than sin(1)/sin(2). At least I hope so.
Very quick and dirty of course and it's not going to work for larger angles. Written before watching the video I should add.
Actually, you can use the small angle approximation for a lot more.
How much do you think the difference between sin(1°) rounded to 3 digits after the decimal point and 1° (π/180) rounded to three digits after the decimal point? The difference is actually 0
The same goes for sin(2°), sin(3°), all the way until sin(8°), that's when the 3 digit approximation differs. By how much? Only by 0.001 (i.e sin(8°) ≈ 0.139 while 8° ≈ 0.140)
Do you know when does the difference reach 0.010? At sin(23°) (i.e. sin(23°) ≈ 0.391 while 23° ≈ 0.401)
So I think you can push it all the way until 23°.
You may say though: "but we're dividing two approximations! Won't the error compound?" Well, here are the results from comparing sinx/sin(x+1°) and x/(x+1°):
The difference between them at x = 1° all the way till x = 4° is 0.
The difference grows pretty slowly though, at x = 23°, the difference is only 0.003, and by the time x = 90°, the difference has only reached 0.011
So actually, the error was _less_ than before.
In conclusion
The small angle approximation works better than you think for sinx, and it works wayyyy better than I thought for sinx/sin(x+1°)
*_Analyising the problem graphically_*
As in a decreasing function
∆y per ∆x decreases as well as we go forward in x
Sin x < Sin x+1 hence both are proper fraction....so the *one having small ∆y in fixed ∆x will be bigger*
As we discussed ∆y will decrease on going forward in ∆x, hence ∆y = smaller for x = 3° to 4° (∆x is same for both = 1).....this way we can say Sin 3°/Sin 4° have smaller change in numerator and denominator for a proper fraction so Sin3°/Sin4° is bigger 🙌
you can also do it using differential calculus
let f(x)= sinx/sin(x+1)
now on differentiating we will get f'(x)= {cos(x)sin(x+1)-sin(x)cos(x+1)}/sin^2(x+1) = sin(1)/sin^2(x+1) which is clearly positive so we can say that f(x) is an increasing function
so f(1)
Small angle approximation wins out here, surely?
Maybe with a Taylor series but then you need to convert angles to radians, and you need to prove how many terms are enough, which - while being feasible - is more complex than using either geometry or trigonometric formulas.
In any case, for any approximation, you need to prove that it won't change the inequality.
I guessed correctly based on the approximation, but I don't have a good sense about how small of an angle is "small enough" for that approximation to "be close enough." I was concerned that 4 degrees was maybe not close enough to "small".
@@markday3145 well but since the derivative of the sine is cosine, its slope decreases between 0 and π/4. Therefore at any number x in between of 0 and π/4 sin(x)
i just thought that the gradient would be decreasing so thered be a smaller difference between 3 and 4 then 1 and 2, so the 3/4 would be greater
You don't need to think about whether the gradient is increasing or decreasing, as long as you realise that the slope is close to constant for small angles.
Same here, I had to mentally visualize the „extrem value“ sin(pi/2-1)/sin(pi/2) instead of sin3/sin4, which is almost flat, to „see it“.
you could also use small angle approximation, where sin(x) ≈ x (when x is in radians) and convert 1°, 2°, 3° and 4° to π/180, π/90, π/60 and π/45 rad respectively. comparing the ratios, you get 1/2 and 3/4, and we know that 3/4 > 1/2
Interestingly you get the exact same solution without converting to radians.
@@escuddy3244 i noticed that too!
Amazing problem. I believe the problem can be solved geometrically using sine rules for triangles as well :)
When I did my private pilot's license, I was taught to use sin x = x for 'moderate' angles (of the wind direction relative to the track of the aircraft) to mentally estimate a course to set if you needed to divert to an unplanned airfield. Therefore I immediately used that trick here, and of course 3/4 > 1/2.
It took me awhile to get it, but I just simply used the fact that the slope of the sine curve is gradually decreasing up to x = π/2. The sin itself is increasing from x= 0 to x = π/2.
So then y-difference from sin 1⁰ to sin 2⁰ is more than that from sin 3⁰ to sin 4⁰. (sin 2⁰ - sin 1⁰ > sin 3⁰ - sin 4⁰). So then (sin 1⁰ / sin 2⁰) < (sin 3⁰ / sin 4⁰).
sin 88⁰ almost equals sin 89⁰, So sin 88⁰/sin 89⁰ is probably like .99.
So then for a given x, sin x⁰ / sin (x+1)⁰ are more nearly equal, the closer x is to π/2., So yes, sin x⁰ / sin (x+1)⁰ keeps increasing as x increases, up to x = π/2.
I found it just by looking at it, knowing that, in Trigonometry circle, vertical bar of sinus, when the angle increase, the projected value is increasing logarithmically. Which means that the projected difference between sin2 and sin1 on the sinus bar is more than the difference between sin4 and sin3, making sin1/sin2 is closer to zero than sin3/sin4, and therefore sin1/sin2 is less than sin3/sin4.
By using the approximation sin(x) = x you can assume that sin(1)/sin(2) ~= 1/2, and sin(3)/sin(4) ~= 3/4. However, the differences between the actual values and the approximate values goes up as x increases. If we consider things in terms of ratios, we only need to consider the amount at which the numerator changes, so we can change the problem to sin(1)/sin(2) = (1+e1)/2, and sin(3)/sin(4) = (3+e2)/4. Since e2 > e1, (3+e2)/4 > (1+e1)/4, therefore sin(3)/sin(4) > sin(1)/sin(2).
Sine=Opp/Hypotenuse. The change is Opp is greatest per change in the smallest degree. Can also think of it as the gradient of the sine curve (max at zero and decreasing to 90)
Similarly I approximated that hypotenuse is not going to change significantly, so can be written as (1/h)/(2/h) and (3/h)/(4/h) the h are going to cancel out so you end up with 1/2 and 3/4.
I did this in my head, knowing that for small angles sin(a) ~= a if a is in radians. Converting degrees to radians is unnecessary in this case because it won't change the ratios. Therefore the left side is approximately 0.5 and the right side is approximately 0.75. Thus the left side is smaller. Checking with calculator: sin(1)/sin(2) = .500076... and sin(3)/sin(4) = .7502666... Amazingly close!
There's another solution to this problem:
we know that ,
0 < sin 1< sin 2< sin 3 < sin 4 ..(i)
we can simplify as 1 , 2 ,3 & 4 wrt to their order in (i)
now since ,
1/2 < 3/4
therefore, we can easily say :
sin 1/sin2 < sin 3/sin4
0 < S2 < S4 < 1 in the 1st quadrant.
So, S2/S4 < S3/S4
Now S4 = 2*S2*C2
So, S2/S4 = 1/(2*C2)
Now S2 = 2*S1*C1
S1/S2 = 1/(2*C1)
Now 1 > C1 > C2 > 0 in the 1st quadrant.
So, 1/(2*C1) < 1/(2*C2)
So, S1/S2 < S2/S4
So, S1/S2 < S2/S4 < S3/S4
So, S1/S2 < S3/S4
On the interval 1 degree to 4 degrees, the sin function is increasing but concave down. This implies sin(3 deg)/sin(4 deg) is larger.
I used the small angle approximation method and got the answer in about 0.5 seconds.
I just visualised a sine wave and it is know that it approaches a stationary point at pi/2 - that means the gradient decreases gradually due to its concave down nature. Therefore the interval between sin 1 and 2 is greater than the interval between sin 3 and 4, so sin3/sin4 must be a larger fraction.
Hi I just figured out another way to prove this:
Say you have a function that is continuously increasing on an open interval (p,q) not including endpoints. In addition, the slope of this function is continuously decreasing.
So then if a f(a), and f(b + w) > f(b), so f(a) / f(a + w), and f(b) / f (b + w) are both positive fractions less than 1. This is assuming a, b, a+w, b+w are all in the interval.
And since the slope is continuously decreasing, f(a + w) - f(a) > f(b + w) - f(b) So then we know
f(a) / f(a + w) < f(b) / f (b + w). This is because not only is the absolute difference between numerator and denominator more on the left, but also the denominator on the right is greater.
The sine function is just such a function from x=0 to x=π/2, and the selection of numerator and denominator on the right side fits how it was done on the left side
(w = π/180). So then
sin (1⁰)/sin(2⁰) < sin(3⁰)/sin(4⁰).
An old engineer's view: I'm old enough to have used a slide rule throughout my engineering training. I recall the facts that (1) a standard slide rule uses a common scale (ST) for both sin and tan below ~5.7º (0.1 radian), and that (2) the ST scale is linear within well below 1% in that range. So, sin 1º/sin 2 = 1/2 ± 2%, and sin 3º/sin 4º = 3/4 ± 2%. Therefore, sin 3º/sin 4º is larger.
I solved it in 10 seconds with the small angle theorem. But I like seeing your alternative approaches to this problem.
I thought of the graph of y = sin x and how the gradient of the curve decreases steadily from infinity at x = 0° to 0 at 90°. So the difference between the y values at x = 1° and x = 2° is larger than the difference between the y values at x = 3° and x = 4°. Therefore the ratio of sin 1°:sin 2° is smaller than the ratio of sin 3°:sin 4°. Similarly, sin 5°/sin 6° > sin 3°/sin 4°, and so on until sin 89°/sin 90°.
The derivative of y=sin(x) = y=cos(x), and y=cos(x) decreases from y=1 at x=0, to y=0 at 90 degrees, so in turn the gradient of sin(x) decreases as x increases (to 90deg). So the value difference between sin(1) and sin(2) will be greater than sin(3) and sin(4), and since sin(0) = 0, the proportional difference between that gradient's effect on the value itself will also be greater, therefore the fraction sin(3)/sin(4) will be closer to parity than sin(1)/sin(2).
As someone who was taught by my teacher during basic math class that sinx ≈ x in radians.
And That's how I know that sin3°/sin4° > sin1°/sin2°
Sin(π/60)/sin(π/45)>
Sin(π/180)/sin(π/90)
The slope of the function sin() at the point x=0 is maximum and decreases to zero at the point 90⁰. So, the difference between sin(1⁰) and sin(2⁰) is greater than between the values of sin(3⁰) and sin(4⁰), so sin(1⁰) / sin(2⁰) < sin(3⁰) / sin(4⁰)
f= sinx/sin(x+π/180) = 1/(cos(π/180)+cotx·sin(π/180)). Cotx is decreasing in the proximity of zero (the opposite of tanx) -> f is increasing, so the second ratio is greater.
Just looking at the graph of sine and its derivative cosine, it is obvious to me that the steps between succeeding degrees decreases with each degree between 0° and 90°. Therefore the higher degree in the ratio is on the bottom, the ratios of successive degrees increases between 0° and 90°. sin(1°)/sin(2°) must be smaller than sin(3°)/sin(4°).
To approximate sin x near 0, use the Maclaurin series expansion of sin x. The Maclaurin series expansion of sin x is given by: sin x = x - (x^3/3!) + (x^5/5!) - ... So, for small x just use the first term approximation where sin x = x and x is in radians. I believe this is the same as your Method 2
That's what I thought of, and therefore this one is easy to do in your head so then it's obvious what ratio has to be larger
And the ratio of two angles in degrees is the same as the ratio of two angles in radians, so no need to convert to radians. (Radians and degrees differ by a constant scale factor.)
@@charleslivingston2256 The Maclaurin series expansion for sin(x) = x - (x^3/3!) + (x^5/5!) "MUST" use radians to approximate the true value of sin(x). If you plugged in degrees you would generate the wrong answers.
The hard part is to prove that your approximation doesn't break the result.
@@stevenz933 It's just the ratio, so using rad or deg gives the same answer.
sin for very small angles is roughly equal to the angle itself (in radians, but it doesn't matter since we're dividing and the units cancel). 1/2 < 3/4. The more interesting challenge would be cosine.
I went with the idea that sin(x) increases quicker as x increases up until 45 degrees so the difference between sin(3) and sin(4) would intuitively result in a bigger fraction than sin(1) and sin(2). Not exactly rigorous but it works intuitively.
other way round....
we can see
1/2=.5
2/3=.67
.....
...
..
and further 999/1000=.999
we can see for small x/(x+1)
value is less as compared to next ones..
therefore if we check for sine it means same as sine is inc. function between 0 to pie/2
Not only is the history of sine and cosine interesting, it is also interesting that in degrees, you can use the small angle approximation in degrees as long as you are using ratios. Makes sense since I guess since radians are a unit-less ratio, I think?
I see it as that the reason is that it works here because converting between degrees & radians is done by multiplying by a constant: You multiply a constnat into the sign on both top and bottom of the fraction, then drop the sign using the approximation, and then you're left with your original numerator & denominator, each multiplied by that constant, which you can then eliminate.
If that's what you meant by "unit-less ratio" (= multiplying by a constant), sorry for rewording your thoughts. :)
🤠 🤠 📴 for the way you depicted the history behind without any manipulation...Keep Going 😊
You don't have to convert them to radians to use the small-angle approximation
Yep, in fact I just took the numbers and directly compared their ratios.
There is another way to solve it.
Sine is an increasing function between (0, π/2), but it plateaus as it nears π/2.
So
sin2° - sin1° > sin4° - sin3°
sin3° is greater than sin1° and since the difference between the denominator and numerator in the first term ISS larger, that term is the smaller term.
There is no need to switch from degree to radian.
in the neighborhood of 0 we have
sin(x)~ x hence
sin(1°)/sin(2°) ~ 1/2 and
sin(3°)/sin(4°) ~ 3/4.
Thank u 👍
there is yet another way,
if u look at sine curve, its rate of increase is faster at lower angles than at higher angles (between 0-90 deg), therefore |sin2 - sin1| > |sin4 - sin3| so in fraction when numerator is more smaller than denominator, fraction is lesser (sin1/sin2) and smaller the gap greater the fraction (sin3/sin4)
therefore conclusion - sin1/sin2 < sin3/sin4
(note: angles are in degrees)
Thanks for this awesome video! I’ll pass this around to math teacher friends.
Not sure why you converted to radians. sin x° is approximately x° when x° is small, and even 4° is quite small.
(Up to about 22°, this approximation is within 1/100 of the correct value: I computed this by solving x^3/3! < 1/100.)
A related fun question: which is larger, sin(1°)/sin(2°) or sin(3°)/sin(6°)? The two methods presented in the video break down in this case (at least when keeping only the first term in the Madhava-Gregory series). Instead you might think about the sign of the second derivative for small positive angles, similar to what some of the others have posted in the comments.
By the way, thanks for all the great videos! I've watched a lot of them :)
Use calculus! Define f(x)=sin(x)/sin(2x). Take the derivative, which is f'(x)=2sin³(x)/sin²(2x). This expression is greater than zero from zero to pi/2, so it's increasing in this interval. 1° and 3° are in this interval, so f(1°)
@@braianhb13 Nice approach!
We can solve this problem more easily by taking sine approximation that is if sine theta is very small angle ( less than 10degrees) we can take take its value as theta in radian so sine 1 degree equals pi /180 * 1 similarly sin 2 degree equals pi/180 * 2 in this way we can find the values and compare
From 0° to 90°, sine is an increasing function, but the rate of increase is decreasing. Therefore, the ratio between sin(1°) and sin(2°) is farther than the ratio between sin(3°) and sin(4°).
Even in india they don't teach about this history of mathematics, it's so interesting
I did it in my head by looking at the slope of a sine wave. from 0 deg to 90 deg the slope decreases.
eg. the slope at 5deg is greater than the slope at 10 deg.
the ratios are the inverse of the slope so the one further along the line is greater.
another way to look at it is : the difference between sin3 and sin4 MUST be less than that of 1 & 2 so the fraction MUST be closer to ONE.
Approximations, graphs, derivatives... I chose instead to use only the most primitive trig identities sin(A+B) = sinAcosB+cosAsinB and sin(A-B) = sinAcosB - cosBsinA. So...
sin2cos1 - sin1cos2 = sin1 > 0
Add 0 in the form sin1cos2-sin1cos2
sin2cos1 + sin1cos2 - 2sin1cos2 > 0
So sin3 - 2sin1cos2 > 0
Multiply by sin2 , which is > 0
sin3sin2 - 2sin1cos2sin2 > 0.
Now using sin4 = 2sin2cos2
sin3sin2 - sin1sin4 > 0. The result follows on division by sin4, which is also > 0
because a circle from pi/2 to 0 has a decreasing slope the small differences between the sines or y values are going to be larger when the angles are close to 0 and smaller differences as the angle goes up to pi/2
Wow, you really overdid that explanation as I instantly used the small angle approximation where the sine = tangent = angle in radians up to about 15degres. The degree to radian ratio is a constant. OK you formally presented the answer.
i immediately jumped to the small angle approximation method, except noticed immediately that the conversion through to radians doesn't matter, since we only care about the ratios of the values. with sin(x) / sin(2x) and sin(3x)/sin(4x) for a sufficiently small x will just be in the ratios of 1/2 and 3/4, regardless of what x actually is.
Using calculus in its essence, if you determine that the rate of change between 1 degree and 2 is, more than the rate of change between 3 and 4, then (sin 1°)/(sin 2°) < (sin 3°)/(sin 4°).
cos x is the rate at which sin x increases at any value of x because cos x is the derivative of sin x with respect to x. Since cos x peaks to 1 at x=0 and then falls to 0 at x=pi/2, the rate at which sin x increases between x=0 and x=pi/2 falls from 1 to 0 in a monotonic manner. So sin 2 degrees/ sin 1 degree > sin 4 degrees/sin 3 degrees =>.sin 1 degree/ sin 2 degrees < sin 3 degrees/sin 4 degrees.
Should be sin 3°/sin 4°. The sine function has both its highest rate of increase and lowest absolute value at 0°, so the further away from 0°, the closer sin x°/sin (x+1)° should be to 1.
In fact, iirc, the derivative of sin x at 0° should be cos 0°, or 1, which should make sin 1°/sin 2° and sin 3°/sin 4° fairly close to 1/2 and 3/4 respectively.
d/dx of sinx is cosx which is decreasing function in (0,pie) hence we can say that rate of increase of sinx decreases as x increses so sin2/sin1>sin4/sin3
It's beautiful! Lots of love from India.
You can solve it intuitively too. Just exageratre the slope of the sine curve. Or compare it with sin90/sin89 which is almost=1.
Thanks, Yeah! After reading the video title I considdered 2nd approach: sin x ca. x for small x.
I used the approx. method, but instead of calculating the answer I used the prooerty that assuming 2 numbers have a constant sum, the closer they are to each other the greater their product will be.
If you hand wave your wau to an answer, hand wave all the way 😅
As an engineering student, I immediately went to sin(x) = x when x smol and got the right answer :D
i just watched the rest of the video :P
The steepness of the curve decreases from 0-90 deg. So sin(3) is closer to sin(4) than sin(1) is to sin(2). So the left ratio must be less than the right one.
Sin1/sin2
=sin (pi/180)/(pi/90)
Using small angle aprox
(Pi/180)/(pi/90)
Pi cancels
=90/180=1/2
Sin1/sin2=1/2
Sin3/sin4
=sin(3pi/180)/sin(pi/45)
Small angle aprox
(Pi/60)/(pi/45)
Pi cancels
45/60=3/4
Sin3/sin4=3/4
3/4 is greater than 1/2 so sin3/sin4 is greater than sin1/sin2
Man U make each concept very clear... Keep it up!!!
At the same inputs, sin(x) is approximate by f(x)=x, so if we were to use f(x), we would have f(1)/f(2) and f(3)/f(4), which are 1/2 and 3/4, so the former is less than the latter.
Hmm. I just pictured the start of the sin curve, since we're only covering 1° to 4° (which are less than 90°) we know the gradient of the sin curve is decreasing as you increase the degrees, so sin 1° and sin 2° will have a greater both absolute and relative difference than sin 3° and sin 4°, therefore the ratio will be a lower number.
And this will be true of any two adjacent numbers (again, up to 90°). So (sin 51°)/(sin 52°) < (sin 52°)/(sin 53°) < (sin 53°)/(sin 54°); with the result of (sin [n-1]°)/(sin n°) approaching 1 as n approaches 90°. In fact it's 0.999848 when n = 90°
In fact because it's a ratio, this trend will continue until you get to 179°, maxing out at 2, then erroring at 180° (div by zero), and then repeating infinitely.
I didnt have immediate intuition, so i pictured a sine wave. One period equals 360 degrees so its not hard to imagin the sine wave is approximately a straight line of form sin=a*degrees where a is some constant. Sin(1) simplifies to a*1, sin(2) simplifies to a*2 etc etc so basicly half is less than three quarters
wow thats really cool but i just realised sin slows down over time so the difference between sin 3/sin 4 is smaller meaning its greater
(like comparing 1/2 and 3/4)
when you imagine the sine wave, you will notice that 1,2,3 and 4 degrees are extremely close to each other, thus you can treat them like a line equation
1/2 < 3/4
According to the fundamental theorem of engineering, sinx is exactly equal to x which is exactly equal to tanx not approximately.
Sine increases at a decreasing rate from 0 to 90 degrees, meaning the differences between consecutive y-values will decrease as x increases. So sin(2)-sin(1)>sin(4)-sin(3) and sin(1)/sin(2)
I actually found a way to do this
let x=1 degree
sinx/sin2x >
Linear approximation of y= sinx is simply y =x , it make sense to simply deal with y=x .
There's another way, by examining the comparative slope of the line between the two points, and recognizing the concavity of sine(0) to sine(90).
As you get closer to the peak, the slope of the line between two points flattens. That means the two values get closer together. Ergo, sine(1) vs sine(2) is a much larger difference than sine(anything larger) vs sine(anything larger+1), at least up to 90degrees. Putting that into fraction form, the fraction with the larger angles gets closer to 1 than the fraction with the smaller angles, therefore sine(1)/sine(2) must be smaller.
Knowing nothing else, including knowing no fancy equivalences or funky approximations, you can derive the answer pictorially ("geometrically"?) in a way that a 5th grader could understand.
Sir I think there are two reason for increasing of [sin(3°)/sin(4°)] from [sin(1°)/sin(2°)].
First slope is decreasing as a function of cos(x).
Which causes less change in denominator in case of sine(4°) in comparison of sin(2°). And second reason is that numerator of second one is high that why it is larger. Instead if there will be same slope.
We can understand this with assumption.
First we have assume that sin(1°)/sin(2°)=x/y and sin(3°)/sin(4°)=a/b
And other assumption .
y=x+∆x(in other word sin(2°)=sin(1°)+∆sin(1°) and similarly b=a+∆a.
Case 1=>
If value of x and a are same then in this situation also sin(3°)/sin(4°) will be larger than sin(1°)/sin(2°) because{ x/(x+∆x) }will be smaller than {x/(x+∆y)} because ∆y is larger than ∆x as slope changes with cosine as cosine of larger number is smaller because number is in between 0° and 90°.
case 2 =>
If slope is constant than also sin(3°)/sin(4°) will be larger than sin(1°)/sin(2°).
Because according to our assumption (x/x+∆x) and (y/y+∆x) here ∆x and ∆y are same.
And y is greater than x .
Just like 1/2 and 3/4. 🙏🙏🙏
Thanks for the history Presh. It was a great addition to this video.
wich is greater can be just reasoned.
the value gets bigger, so for the aproximation, you can just ignore sin completely and see that 3/4 > 1/2 and see that the left one sin(1) / sin(2) is smaller.
No actual values needed or requeted in the question.
doing differentiation also helps , it's an increasing function at 1 degree
i went about it like this: sine is the vertical projection of the unit line at an angle. its 0 at 0 degress and inversely 1 and 90 degrees. the cosine is the horizontal projection. 1 at 0 degress and inversely 0 at 90 degrees.
the sinus at angle 1, 2, 3 and 4 therefore is a increasing function, at each degree roughly the same amount. therefore it estimates to 1/2 < 3/4.
You can just cancel out the sines
I compare the two fraction as follows:
sin 1° / sin 2° - sin 3° / sin 4°
= (sin 1° sin 4° - sin 2° sin 3°) / (sin 2° sin 4°)
= (sin 1° * 2 sin 2° cos 2° - sin 2° sin 3°) / (sin 2° sin 4°) [using the double angle identity sin 4° = 2 sin 2° cos 2°]
= (sin 2° (2 sin 1° cos 2° - sin 3°)) / (sin 2° sin 4°) [factor out sin 2° from the numerator]
= (2 sin 1° cos 2° - sin 3°) / sin 4°
= (2 sin 1° cos 2° - sin 2° cos 1° - sin 1° cos 2°) / sin 4° [using the sum formula for sine: sin(A+B) = sin A cos B + sin B cos A]
= (sin 1° cos 2° - sin 2° cos 1°) / sin 4°
= sin (-1°) / sin 4° [using the difference formula for sine: sin(A-B) = sin A cos B - sin B cos A]
< 0 [ since sin (-1°) < 0 and sin 4° > 0]
Therefore:
sin 1° / sin 2° < sin 3° / sin 4°
I don't know why 'jya' (the blue line) would refer to the length of the bow-string (half of which is the green line)- when, apart from anything else, it's the angle of the radius that determines the length of 'jya'...
sin(x-1°)/sin x = (sin x cos 1° - cos x sin 1°) / sin x = cos 1° - sin 1° cot x which is strictly increasing when x is acute
I solved it without any calculations.
The difference on both sides between the angles is the same 1°=2-1=4-3
The fact, that the change of the sinus value by increment of 1° in the angle gets smaller the further we move away from 0° towards 90° implies that:
sin(1)/sin(2)< sin(3)/sin(4)
I just visualized the graph of sin(x). After x=0, the rate of increase becomes slower until it reaches zero. So, sin(2)/sin(1) will be greater than sin(3)/sin(4) or conversely sin(1)/sin(2) will be less than sin(4)/sin(3).
I thought of it physically, and the conclusion was pretty obvious.
I used some know facts:
The function sin(x) is increasing so:
Sin 1 < sin 3
The slope of the graphic is getting smaller so:
Sin2 - sin 1 > sin4 - sin3
Knowing these 2 facts out is clear that the right hand side is greater.
Great Video with knowledge, keep it up🎉❤
Consider a function f(x) = sin(x)/sin(x+a), where a is any real number, but in this case it will be 1 degree.
If you diff the function f you get f'(x) = (sin(x+a)cos(x)-sin(x)cos(x+a))/(sin^2(x+a)).
Use the added angle identities for sin and cos for the top part. You get:
sin(a)cos^2(x)+sin(x)cos(a)cos(x)-sin(x)cos(a)cos(x)+sin^2(x)sin(a)
The middle parts cancels out, and the left and right parts combined are:
sin(a)(cos^2(x) + sin^2(x)) = sin(a) * 1 = sin(a)
Aha! So, f'(x) = (sin(a))/(sin^2(x)), and since sin(a) is a constant and the denominator is squared... The function f(x) is monotone! And its dependent on the number a we choose. In this case one degree was chosen, so f'(x) is always > 0 => f(x) is increasing! Of course be careful with the domain, but in this videos case its all where it should be.
So, overall, f(1) < f(3) => sin(1)/sin(1+1=2) < sin(3)/sin(3+1=4) {in degrees}
sin(x) is a concave and increasing function for x from 0 to 90 (omitting degree symbols because I'm lazy)
sin(1)
or, 1/2sin(x) and x-sin(x) is monotonically decreasing in the range 0
People.. this is probably the best place to ask this question.. I work in a builders supplies yard and I sell thousands of different products at different prices ... today something happened I never saw before (20 years )in my work. A customer paid for some products and the very next customer bought completely different products and different quantities. They both paid by card and the amount was exactly the same ie £14.69 ..wow I couldn't believe it.. what are the odds this would happen?
Odds are 2 in 1469
Not sure what that means M... odds the digits match, I thought, was 10 x 10 x 10 x 10 ... 10000/1. Also throw in the huge range of prices etc ...🤔
@@widescreen3 just a silly joke - it would be a much bigger coincidence if true though