My approach: From the sine identity we immediately derive two families of quadratic equations relating their arguments (angles): x^2 - x = 2k*pi and x^2 + x = (2l + 1)*pi for integers k, l. Then just use the solution formula for quadratic equations and restrict k,l to non-negative integers to keep the discriminant non-negative, too. Nice exercise for solving in the head. Didn’t have to write down anything.
Here's another approach to this problem sinx² - sinx = 0 By product to sum formula, we have 2sin[½(x² - x)]*cos[½(x² + x)] = 0 Therefore, sin[½(x² - x)] = 0 or cos[½(x² + x)] = 0 For sinθ = 0, θ = 0, π, 2π … nπ Hence it can be concluded that, For sinθ = 0, θ = nπ (n ∈ Z ≥ 0) Similarly, For cosθ = 0, θ = π/2, 3π/2 … nπ/2 But for cosθ, n must be an odd number. Hence it can be concluded that, For cosθ = 0, θ = ½(2n+1)π (n ∈ Z ≥ 0) Thus we arrive at the same step where, ½(x² - x) = nπ or ½(x² + x) = ½(2n+1)π x² - x - 2nπ = 0 or x² + x - (2n+1)π = 0 (where n ∈ Z ≥ 0)
Here is another possible way sin(x^2)-sin(x) = 0 We have the identity sinP - sinQ = 2cos[(P+Q)/2]sin[(P-Q)/2] So the LHS becomes 2cos[(x^2 + x)/2]sin[(x^2 - x)/2] = 0 cos[(x^2 + x)/2] = 0 or sin[(x^2 - x)/2] = 0 The first equation gives (x^2 + x)/2 = (2n+1)/2 * pi x^2 + x = (2n+1)pi x^2 + x - (2n+1)pi = 0 The second equation gives (x^2 - x)/2 = npi x^2 - x = 2npi x^2 - x - 2npi = 0 Which are the same two quadratic equations you obtained. Great video still.
sin(x²) = sin(x) One group of solutions will come from exploiting the fact that sin(x) = sin(x + 2nπ) for interger values of n. This won't be ALL of the solutions, but it will be a set of them. sin(x²) = sin(x + 2nπ) We are JUST going to explore the solutions that we get by setting x² = x + 2nπ We don't need to use arcsin because we are just looking at which values of x will work in this relationship. In case 2 we use the identity sin(x) = sin(π - x) sin(x²) = sin(π - x) We can now explore all the solutions we get by setting x² = π - x We are able to find ALL of them solutions in terms of n using these two cases.
This teaching is wrong. Because of there is an x = two possible values of 2nπ then there isn't an x^2 = x + 2nπ , which contradicts one solution when there is two (a domain of two mapping to a range of one). Ignore his bad teaching concepts. It has to be a piecewise continuous function like step functions in engineering and viewed as engineering students do in intervals of the composite function sin(x) is. NOTE: Fourier Series describes cos (x) and sin (x) by approximating the function from its higher frequencies harmonics to avoid looking at all infinite solutions.
It actually depends on the domain of the functions if it's between -pi/2 and pi/2. you can use the Inverse Trigonometric identity and cancel the sin out.
Please those 1st 30s of the video gave me heart attack... i lost at least 10 years of my life seeing YOU do it (despite full knowing you intentionally doing that crime against math to correct anyone who would dare to do that)...
@@Aryan-o4y7gYou just said only valid solutions are 0 and 1. NO we cant do that here, that causes a lost of infinitely many solutions. We can limit the domain in 3 cases 1st case: 1 or more functions in equation are undifined at a point or at an interval (we are then forced to) 2nd case: limiting the interval doesnt cause lost of solutions OR we solve rest separately 3rd case: We want solutions with a criteria. x = (1+sqrt(1+8pi))/2 is bigger than pi/2 and is still a solution. So we cant limit the interval.
because the +2npi is there just to say that the difference of left and right side is some multiple of 2pi if you wrote it on the other side you would basically say it twice
If you included the 2*n*pi on one side and 2*m*pi on the other side, you could bring them to one side and then combine them into 2*k*pi where k is another integer. So we only need to put it on one side.
The reason that those just combine into other constants is that the period is outside the squaring. It would be x^2 + 2n*pi. It's not a composition going into the input. It is a function of the range having a period.
You can kinda cancel them(as you're technically just applying the sin inverse function on both sides), but you wouldn't get all possible solutions with that. So, in case you need only one value, it's a faster way to get it.
You CAN just cancel the two sines, but then you only get two of the infinite number of results. Ignoring complex numbers, x^2=x has two solutions: x = 0 or x = 1.
@@PuneetSharma-uw1kdNote that sin(x) as a function sin: R -> [-1, 1] -- not being an injective function -- does NOT have an inverse function. The function arcsin(y) which is usually called the inverse of sin(x) in a casual way, is actually only the right-inverse of sin(x). However, what we need here for cancellation is a left-inverse, and that does not exist!
Are negative values of n really not solutions? True, you get a complex number, but does taking the sine of a complex argument (by Euler's formula) give a valid solution?
I think it does; I think those complex numbers are indeed solutions. No matter if n is positive or negative, for the first solution set we'll have x² = x + 2πn , and for the second solution set we'll have x² = (π - x) + 2πn (just replace n = -K where integer K>0 , write x as a complex number by factoring out the i outside the square root operator, then evaluate x² by squaring the formula; takes quite a bit of work, but a good exercise). Now, _if I'm not mistaken_ , if z = (A + iB) is a complex number (with real numbers A and B), then sin(z) = sin(A+iB) = sin(A)*cosh(B) + i*cos(A)*sinh(B) [identity #1] This identity can be derived by either starting from sin(A+iB) = sin(A)cos(iB) + sin(iB)cos(A) , or starting from sin(A+iB) = (e^[i(A+iB)] - e^[-i(A+iB)])/(2i) . (again takes quite a bit of work, but a good exercise!) Now let x be a complex-valued solution by having a negative n ; then x has the form of x = a+ib , for some real values of a and b . So, using the identity #1 above, sin(x) = sin(a+ib) = sin(a)*cosh(b) + i*cos(a)*sinh(b) [equation #2] If x is from the first solution set, we'll then have x² = x + 2πn = a+ib + 2πn = (a+2πn) + ib and hence sin(x²) = sin(a+2πn + ib) = ... apply identity #1 ... = sin(a+2πn)*cosh(b) + i*cos(a+2πn)*sinh(b) ... note: for any integer n , sin(a+2πn) = sin(a) , cos(a+2πn) = cos(a) ... = sin(a)*cosh(b) + i*cos(a)*sinh(b) ... see equation #2 ... = sin(x) If x is from the second solution set, we'll then have x² = (π - x) + 2πn = (π - (a+ib)) + 2πn = (π-a + 2πn) - ib and hence sin(x²) = sin((π-a + 2πn) - ib) = ... apply identity #1 ... = sin(π-a + 2πn)*cosh(-b) + i*cos(π-a + 2πn)*sinh(-b) ... note: cosh(-b) = cosh(b) , sinh(-b) = -sinh(b) ... = sin(π-a + 2πn)*cosh(b) - i*cos(π-a + 2πn)*sinh(b) ... note: for any integer n , sin(X+2πn) = sin(X) , cos(X+2πn) = cos(X) ... = sin(π-a)*cosh(b) - i*cos(π-a)*sinh(b) ... note: sin(π-a) = sin(a) , cos(π-a) = -cos(a) ... = sin(a)*cosh(b) + i*cos(a)*sinh(b) ... see equation #2 ... = sin(x) So in both solution sets, even when n is negative, the original equation sin(x²) = sin(x) is satisfied. Therefore, these formulas with negative values of n give complex values of x that are also solutions to the original equation. (Please, feel free to comment if there are any mistakes.) EDIT: Correcting "cosh(-b) = cos(b)", which misses a letter h due to a typo.
Yes, I think those complex values when n is negative are also solutions. In short: I think the key is proving that the trigonometry identities sin(z+2πn) = sin(z) and sin(π-z) = sin(z) are also valid when z is a complex number. No matter if n is positive or negative, for the first solution set we'll have x² = x + 2πn , and for the second solution set we'll have x² = (π-x) + 2πn . (Verify this by squaring the found formulas while n is negative.) So if the above trig identities hold for complex-valued z , then in the first solution set we'll have sin(x²) = sin(x + 2πn) = sin(x) , and in the second solution set we'll have sin(x²) = sin((π-x) + 2πn) = sin(π-x) = sin(x) and hence the equation sin(x²) = sin(x) is also satisfied by the complex values of x from the found solution formulas.
@@yurenchu Quote: "(Please, feel free to comment if there are any mistakes.)" As you wish 😉 : the hyperbolic cosine of negative b isn't equal to the cosine of b, but the hyperbolic cosine of b instead. (I guess you just forgot to type a necessary h, that's all.)
I was thinking of an approach based on replacing the sines by their expressions in terms of exponentials (sin (x) = (e^ix - e^-ix)/2i = (y - 1/y)/2i, sin(2x) = (y^2 - 1/y^2)/2i etc) but this was a bad idea, leading to a nasty cubic equation that doesn't easily handle the multivalued nature of sine.
Ahhhk. Here's a better explanation. The reason you can "cross out the sin's" is because you would then be dividing both sides by function notation. "Sin(x)" is no different than F(x) except the function notation uses 3 letters. This would be akin to solving F(x) =2x+5 by dividing both sides by the letter F.
I liked your point about the arcsine is not 1 to 1, here is how I would explain that. Arcsin (1) equals an infinite number of possible values --> π/2 + 2πk. and -π/2 - 2πk. But Sin (π/2) only has one range value, y = 1. Notice that if you inverse both sides of Sine(x) =1. that is different than inversing both sides of Sin(π/2) = x for the reason that Sin(x) is not 1 to 1 which means the inverse function is problematic
You can "cross out the sin's", you might not get all the solutions if the function is not injective, but you will not introduce extraneous solutions: Given f(x)=f(y)... one possible solution if x=y In practice maybe you can have that... say f(x)=f(x+1)... one candidate would be x=x+1... but that's impossible. If you can have x=y... then it is a solution for f(x)=f(y) (because you can't have f giving different outputs for the same input... else f wouldn't be a function), might not be the only one(like he explained on sin(x)=sin(x²)
@@tarentinobg i haven't watched the video yet but wouldent sin(x^2)=sin(x) be sin(1^2)=sin(1) by definition if we put in 1 as x and since 1^2=1 i dont see how any wierd sin stuff would stop that from being true
Not quite. The two solutions have different signs for the 1 before the +/- sign. To write write a single expression to represent all solutions you would need to write [(-1)^k +/- root(1+4kpi)]/2 so that the sign of the first 1 will alternate with even/odd values of k.
The sine values repeat every 2pi. The number of repetitions is the n. If you want to include the 5th repetition, you add 5*2*pi. To generalize, just add 2*pi*n
if it was sin^2(x) = sin(x) , it would be easy for sure, you would only have to tell when sin(x) = 0 or = 1 but my guess is that you would have to make sure to write it down as something like x^2 = x*2kπ and even then it would probably only cover half of the solutions also there's the fact that negative solutions still exist
Nice problem! I've been thinking about conditions for when sin(x) = sin(y) & cos(x) = cos(y)... I feel you're proof though is a bit handwavy since it only really justifies for angles between 0 and 90 degrees, but how would you justify for angles greater than that and even greater than 2pi? The proof of such identities and why they cover all cases is what I'm a bit confused on here. Kindly if you or anyone could help explain this, I'd be grateful!
His proof works for all angles, with no bounds. Pi - x will always reflect the angle about the y-axis on the unit circle, because pi-x is by definition the supplement to x.
@@phiefer3 Sorry kindly could you elaborate how this explains for all angles w/ no bound... don't supplementary angles add up to 180 and are both between 0 & 180?
@@bobmarley9905 Yes, supplementary angles add up to 180, but there is no restriction that they must be between 0 and 180. A 210 degree angle and a -30 degree angle are supplementary, as are 480 degrees and -300 degrees. You can find the supplement of any angle with no bounds, because x + (180-x) = 180 is always true. And of course, 180 degrees is pi radians. As for covering all possibilities, this doesn't give you all of them. You still need the +2npi for that. Each y-value occurs exactly twice per period (except for 1 and -1 which each only occur once per period). x+2npi gives you one of these angles in every period, specifically the angle that is equivalent to x. The other angle in each period is the supplement of one of the x+2npi angles, so we get all of the supplements with pi-x+2npi.
Technically, the division step wasn't "wrong", your result from it was just not correct/complete. The correct result should be "x = 1 *when x != 0* ". You can then separately check for the case of "x = 0" to see if it is actually a solution too (which it is). This still ignores all the other possible solutions, though (because cancelling the "sin"s is still wrong).
Yes, it's wrong because arcsin is only a right-inverse of sin. And what you'd need for cancellation in this context is a left-inverse. A true inverse of sin (defined as a function on all of R) cannot exist because sin is not an injective function as bprp has shown.
Most people don't consider all angles past 360 but you are right if you want to get all possible answers. Perhaps teachers should specify like all [0, 360] as the range or [0, 720] etc etc. Also, I find it strange that sin(1) = sin(1^2) because it doesn't change no matter what units we are using. 1 degree, 1 radian, 1 whatever unit we use it is still the same answer so if that is true then 1 X pi and 1^2 X pi is the same sine value. Does that mean that sin(pi) = sin(pi^2) and pi = pi^2 or pi = pi^n? If we use degrees as our units then degree^2 = degree just like it did for pi? I know this isn't true but it was something I thought of while watching this video and it reminded me how much mathematicians hate units. Does the unit of a circle's angle squared = the unit of a circle's angle? Does 30 degrees squared = 30 degrees itself when talking about the angle of a circle?
Is it strange that if a function f(x) is defined for the value 1, then the equation f(x^2)=f(x) has at least x=1 as a solution because 1^2=1 and f(1)=f(1) ? If f(a)=b then x=a implies f(x)=b And, by the way: 1*pi = (1^2)*pi =/= (1*pi)^2 If g(x)=h(ax) then 1) g(y+2) isn't h( ay+2 ) but h( a(y+2) ) instead 2) g(y^2) isn't h( (ay)^2 ) but h( a(y^2) ) instead And that is true both for a=180/pi and a=pi/180
@@shashankg1006 I know that it has to be an integer, but I thought that if n was equal to -1/8π then X was equal to 1/2, I was trying to avoid complex numbers and didn't look any further. Now I see that if I substitute x in one of the formulas above, it isn't equal. Sorry for my english, im not very good at it.
I think the 1 to 1 explanation is a bit meaningless without understanding the inner workings. The concept of doing something to both sides is the same as applying the same function to both sides. So cancelling out a function involves applying the inverse to both sides. sine fails the horizontal line test and thus has no inverse function over its full domain. Thus you can't cross it out. exp(x), also called e^x is invertible so you can cross it out, which is equivalent to applying ln(x) to both sides. I don't know if e is invertible in the complex plane. I don't know if invertible functions exist in the complex plane.
Or x=[ (-1)^n (+ or -) root(1+4npi)]/2. And this is how I got this result, for the equation sinx=sin(a) the general solution is x=npi+(-1)^n a, and then you will get a quadratic
That possibility is covered in the formulas of the found solutions. For example, x = [1 ± √(1+8kπ)]/2 is negative when k is positive and the ± sign is a minus-sign ("-") . So for example, for k = 3 , one of the solutions is x = [1 - √(1+8*3*π)]/2 = [1 - √(1+24π)]/2 ≈ −3,8703 which is negative. In that case, x² = ([1 - √(1+24π)]/2)² = = [1 - √(1+24π)]² / 2² = [1 - 2√(1+24π) + (1+24π)] / 4 = [(2+24π) - 2√(1+24π)] / 4 = [(1+12π) - √(1+24π)] / 2 = [12π + 1 - √(1+24π)] / 2 = 6π + [1 - √(1+24π)]/2 = 6π + x and therefore, sin(x²) = sin(6π + x) = sin(x)
It doesn't matter if x is negative. As long as the difference (x² - x) is a multiple of 2π , the values of sin(x²) and sin(x) will be equal (because the sine function is periodic; sin(θ+2πk) = sin(θ) for any integer k). And the same goes for when the mean value (x² + x)/2 equals π/2 {+ a multiple of π} (because sin(π-θ + 2πk) = sin(θ) for any integer k).
That's like saying it's fine to say that we can apply sqrt() on both sides on x^2=4 and get x=2. When solving an equation it is a crime to leave valid answers out.
arcsine is injective, but it doesn't matter, in fact, what matters is that it's a function (in order to use on both sides, that is). The problem is that arcsin(sin(x))≠x, but rather arcsin(sin(x))=y, where y is a number such that sin(y)=sin(x) and y∈[-π/2,+π/2], that is arcsin(x) returns a principal argument. So it's not an inverse function of sine, strictly speaking (in fact, sine does not have an inverse function since it's not bijective and only bijections have inverse functions). It does have an inverse multifunction, however, commonly(?) denoted as Arcsin(x), so Arcsin(sin(x))=(x-πk)×(-1)^k for all integer k. Proceeding with the Arcsin you will get x²=((-1)^k)x+πk for all integer k, so you will get the same equation. It's really no different from (x²)²=x², you can use √, but √(x²)≠x, same thing here, you either use a multifunction and get ±x²=±x or you use a principal value and get |x²|=|x|, either way x²=x is just wrong. P.S. If you want to see the whole arcsin(sin(x)) identity, here it is, but it's somewhat nasty. arcsin(sin(x))=(x-π[x/π])×(-1)^-[x/π] where [x] denotes "round to nearest" function, that is [x]=⌊⌈2x⌉/2⌋ or ⌈⌊2x⌋/2⌉ (one rounds .5 to 0, another one rounds 0.5 to 1) where ⌈x⌉ is a ceiling function (round to +∞), ⌊x⌋ is a floor function (round to -∞).
It doesn't matter which side he puts it, it results in the same solution (with the only difference that N = -n , and since n must be non-negative , N must be non-positive). Whether sin(x^2) is periodic or not, is irrelevant.
Couldn't 'n' be negative? Like for case 1 'n' can be equal or greater than negative 1/(8pi)? Even for case 2: 'n' can be negative (1+4pi)/8. Making the the square root still have a real solution without going to complex number. Because if 'n' is equal to these numbers it would still result to X being 1/2.
“infinitely many” is more specific it means you can keep finding solutions without limit but the term “infinite” can be a broad as there are different types of infinity (ex. countable and uncountable)
@alghihend Yes, if any function f(x) has 1 in its domain, then f(x*x)=f(x) has x=1 as one of its solutions, because 1*1=1 and f(1)=f(1). edit: And if 0 is in its domain, then x=0 is (also?) one if its because 0*0=0 and f(0)=f(0).
It solution is Little bit confusing and difficult I just thought about a new one what if we just take sininverse both side then we get x^2 =x now x^2 - x =0 now x=0 or x = 1 as simple as that 😂
Umm Value of any angle doesn’t change if you add multiples of 2π to it. Coz is one full rotation, and sine is periodic-it repeats every circle.(2π = 360° btw) In short: Adding is like walking in circles and ending up at the same spot!
@Sneeehhhh No i understand that. What i dont understand is when he concluded that he can just write x^2=x+2nπ. But since he adds all the possible solutions now it makes sense.
@NamiZu00I mean your reasoning is nonexistant (I was gonna say wrong, but you have no reasoning) but you are correct in the sense that there is really nothing to solve, it's just stating a result in math.
Oh, you're missing out the solution x=0 also in the correct proof! Again: You can't divide by x, if x can be zero. You can't divide by sin x, if sin x can be zero (for x=0). Incomplete answer.
No, the misunderstanding comes from the fact that f(a)=f(b) implies a=b only if f(x) is one-to-one. For example, log(p)=log(q) means that p=q but only because the logarithmic function is one-to-one. Nobody thinks sin(x) = variable sin times variable x.
I mean, that problem is easy to tell from a different example: sin(π) = sin(2π). π ≠ 2π, but sin is a repeating function (as in, it repeats over the period 2π), so both π and 2π give the same result when plugged into sine.
0:22 NO… Why are you dividing by a variable - which could be 0? Here’s how this part should be done: x^2=x x^2-x=x-x - Therefore: x^2-x=0 - Then factor: x(x-1)=0 - Therefore: x=0 OR x-1=0 x=0 OR x=0+1 x=0 OR x=1 Then, of course, because this is a sinusoidal equation, you have to add +2*Pi*n to each of these - Therefore: x=0+2*Pi*n OR x=1+2*Pi*n
@bprpmathbasics Sorry sir but should we not set the argument of the square roots equal to zero and find the minimal value? Because it could be that the outcome of the inequality will give n≥ -c which is also an integer right?; for example 1+8nπ≥0 8nπ≥-1 N≥ -1/(8nπ) now because n must be an integer you should put the restriction N e Z therefore n≥0 But I imagine it could be -1,-2,-3 which is an integer too and then it would be n≥-1(or -2 or -3 etc)
it's very easy, firstly, I'm going to say that sin(x)=sin(x^2) so the angle x = the angle x^2. so x^2 = x+360 because they are equal. then use the quadratic equation & you'll have 2 values you will try them and the correct value is 19,28,48.91 degree
So, pretty smart, high IQ, never had anything past algebra 2 and stat. No trig, no calc. No college. I understand the basics of sin, cos, tan in triangles. I understand if you graph them, they are waves. I am a software engineer, so when i see sin(of some value), i see that as a function which returns a value. Id never expect that you’d be able to drop the function/ cancel it.
When you showed us the wrong way to solve it you said that zero is an answer to the equation. but when you solved for both cases, neither had zero as one of the answers
@@joshuahillerup4290 If n is defined as being an (non-negative) integer, then I see no problem or obstacle with having n^n = 1 for n=0, because all counter-arguments using limits go out of the window.
Factoring confusion! Trig equation 2sin^2(x)=1: ua-cam.com/video/5ckR_5M7Pig/v-deo.html
Using difference formula of sinA - sinB = 2 sin((A-B)/2)cos((A+B)/2) ..I got all values of x ,, but Is it always be applicable ??
Thanks for it ..
My approach: From the sine identity we immediately derive two families of quadratic equations relating their arguments (angles): x^2 - x = 2k*pi and x^2 + x = (2l + 1)*pi for integers k, l. Then just use the solution formula for quadratic equations and restrict k,l to non-negative integers to keep the discriminant non-negative, too.
Nice exercise for solving in the head. Didn’t have to write down anything.
Note that the 2 cases can be merged. 8n=4(2n). Taken together with 4(2n+1), you get 4 times *any* integer (the >=0 restriction still applies).
Only if you replace the -1 in front with (-1)^n, so that it will be +1 in the even case.
Here's another approach to this problem
sinx² - sinx = 0
By product to sum formula, we have
2sin[½(x² - x)]*cos[½(x² + x)] = 0
Therefore,
sin[½(x² - x)] = 0 or cos[½(x² + x)] = 0
For sinθ = 0, θ = 0, π, 2π … nπ
Hence it can be concluded that,
For sinθ = 0, θ = nπ (n ∈ Z ≥ 0)
Similarly,
For cosθ = 0, θ = π/2, 3π/2 … nπ/2
But for cosθ, n must be an odd number.
Hence it can be concluded that,
For cosθ = 0, θ = ½(2n+1)π (n ∈ Z ≥ 0)
Thus we arrive at the same step where,
½(x² - x) = nπ or ½(x² + x) = ½(2n+1)π
x² - x - 2nπ = 0 or x² + x - (2n+1)π = 0
(where n ∈ Z ≥ 0)
Wow awesome method used. Awesome explanation given. You are really genius. Well done
at least I did the first step
sin x^2 - sin x = 0 and then I'm stuck
product to sum formula ?
@@whoff59I hate that sort of 'solution'.
0:10 just some small angles aproximation
Here is another possible way
sin(x^2)-sin(x) = 0
We have the identity
sinP - sinQ = 2cos[(P+Q)/2]sin[(P-Q)/2]
So the LHS becomes
2cos[(x^2 + x)/2]sin[(x^2 - x)/2] = 0
cos[(x^2 + x)/2] = 0 or sin[(x^2 - x)/2] = 0
The first equation gives
(x^2 + x)/2 = (2n+1)/2 * pi
x^2 + x = (2n+1)pi
x^2 + x - (2n+1)pi = 0
The second equation gives
(x^2 - x)/2 = npi
x^2 - x = 2npi
x^2 - x - 2npi = 0
Which are the same two quadratic equations you obtained.
Great video still.
Yeah that actually explains how it came to be in the video. Good work.
Was thinking of that. Combine them to form the equation
Nice work!
I didn’t think of that! Thanks.
I love calculations that are right and wrong at the same time.
Nice to visualise the solutions in Desmos
Very interesting and well explained, like all your videos. Also, I've always wondered, can't you find a clip-on microphone that works?
3:57 I am confused - with case 1, how did you go from sin(x^2)=sin(x+2*Pi*n) to x^2=x+2*Pi*n??
What happened to negating it using the arcsin??
sin(x²) = sin(x)
One group of solutions will come from exploiting the fact that sin(x) = sin(x + 2nπ) for interger values of n. This won't be ALL of the solutions, but it will be a set of them.
sin(x²) = sin(x + 2nπ)
We are JUST going to explore the solutions that we get by setting x² = x + 2nπ
We don't need to use arcsin because we are just looking at which values of x will work in this relationship.
In case 2 we use the identity
sin(x) = sin(π - x)
sin(x²) = sin(π - x)
We can now explore all the solutions we get by setting x² = π - x
We are able to find ALL of them solutions in terms of n using these two cases.
Hope that helped. It's a good question - best of luck with all your mathematical explorations!
This teaching is wrong. Because of there is an x = two possible values of 2nπ then there isn't an x^2 = x + 2nπ , which contradicts one solution when there is two (a domain of two mapping to a range of one). Ignore his bad teaching concepts. It has to be a piecewise continuous function like step functions in engineering and viewed as engineering students do in intervals of the composite function sin(x) is. NOTE: Fourier Series describes cos (x) and sin (x) by approximating the function from its higher frequencies harmonics to avoid looking at all infinite solutions.
Because sine is a periodic function. If A is a solution for Sine, so is A + 2πn.
Oh, snap! The blue pen is out in full force today!
i know you never read comments but im early so can u please start a basic physics channel?
No
Does he even teach physics? Is he qualified to do that? And even if he is: Does he have time for that?
maybe not just basic physics, but it would be cool if he did more stuff with the maths side of physics
It would be really cool but I think he is a mathematician not a physic teacher unfortunately
He does read comments wdym
It actually depends on the domain of the functions if it's between -pi/2 and pi/2. you can use the Inverse Trigonometric identity and cancel the sin out.
SIN (X) is not a “bijection,” I learned at school …
Please those 1st 30s of the video gave me heart attack... i lost at least 10 years of my life seeing YOU do it (despite full knowing you intentionally doing that crime against math to correct anyone who would dare to do that)...
Well, that method does get *a* solution, just not *all* solutions. 😁
It should be stated x belongs to (-π/2,π/2).
@@Aryan-o4y7gYou just said only valid solutions are 0 and 1. NO we cant do that here, that causes a lost of infinitely many solutions. We can limit the domain in 3 cases
1st case: 1 or more functions in equation are undifined at a point or at an interval (we are then forced to)
2nd case: limiting the interval doesnt cause lost of solutions OR we solve rest separately
3rd case: We want solutions with a criteria.
x = (1+sqrt(1+8pi))/2 is bigger than pi/2 and is still a solution. So we cant limit the interval.
Not a general formula as you did but I think differentiating implicitly gives us x = 1/2 , which fits the formula you extracted
0:18 Really? A mistake in the flawed solution? You can't just divide by x (potentially 0). Solutions are x=0 and x=1.
Watch 2:20
@nasdfigol He did the same with the final solution and missed x=0 (sin x = 0) for that reason as a solution!
But why don't I need to substitute the x^2 the same way? Why only the x-term?
Because you would end up getting the same thing due to symmetry.
because the +2npi is there just to say that the difference of left and right side is some multiple of 2pi
if you wrote it on the other side you would basically say it twice
If you included the 2*n*pi on one side and 2*m*pi on the other side, you could bring them to one side and then combine them into 2*k*pi where k is another integer. So we only need to put it on one side.
The reason that those just combine into other constants is that the period is outside the squaring. It would be x^2 + 2n*pi. It's not a composition going into the input. It is a function of the range having a period.
Because he's not making a substitution. He's saying that it is ALSO equal to this other thing sin (x + 2npi).
Thanks a lot!
P.S. Also it can be solved by graph method
Have to say, I'd do the first section for trivial answers, then the second.
damn... i never knew why we couldnt cancel them, just that we couldnt. That actually makes so much sense
You can kinda cancel them(as you're technically just applying the sin inverse function on both sides), but you wouldn't get all possible solutions with that.
So, in case you need only one value, it's a faster way to get it.
@@PuneetSharma-uw1kd that is true, but its usually all the solutions in a given range for trig functions
You CAN just cancel the two sines, but then you only get two of the infinite number of results. Ignoring complex numbers, x^2=x has two solutions: x = 0 or x = 1.
@@PuneetSharma-uw1kdNote that sin(x) as a function sin: R -> [-1, 1] -- not being an injective function -- does NOT have an inverse function. The function arcsin(y) which is usually called the inverse of sin(x) in a casual way, is actually only the right-inverse of sin(x). However, what we need here for cancellation is a left-inverse, and that does not exist!
Would it work out the same way if we used degrees instead of radians?
How about using newton-raphson rule to determine its roots
In a problem like this we seek an analytical solution, not numerical. A numerical approach will yield only a single solution.
Are negative values of n really not solutions? True, you get a complex number, but does taking the sine of a complex argument (by Euler's formula) give a valid solution?
I think it does; I think those complex numbers are indeed solutions.
No matter if n is positive or negative, for the first solution set we'll have x² = x + 2πn , and for the second solution set we'll have x² = (π - x) + 2πn (just replace n = -K where integer K>0 , write x as a complex number by factoring out the i outside the square root operator, then evaluate x² by squaring the formula; takes quite a bit of work, but a good exercise).
Now, _if I'm not mistaken_ , if z = (A + iB) is a complex number (with real numbers A and B), then
sin(z) = sin(A+iB) = sin(A)*cosh(B) + i*cos(A)*sinh(B) [identity #1]
This identity can be derived by either starting from sin(A+iB) = sin(A)cos(iB) + sin(iB)cos(A) , or starting from sin(A+iB) = (e^[i(A+iB)] - e^[-i(A+iB)])/(2i) .
(again takes quite a bit of work, but a good exercise!)
Now let x be a complex-valued solution by having a negative n ; then x has the form of x = a+ib , for some real values of a and b . So, using the identity #1 above,
sin(x) = sin(a+ib) = sin(a)*cosh(b) + i*cos(a)*sinh(b) [equation #2]
If x is from the first solution set, we'll then have
x² = x + 2πn = a+ib + 2πn = (a+2πn) + ib
and hence
sin(x²) = sin(a+2πn + ib) =
... apply identity #1 ...
= sin(a+2πn)*cosh(b) + i*cos(a+2πn)*sinh(b)
... note: for any integer n , sin(a+2πn) = sin(a) , cos(a+2πn) = cos(a) ...
= sin(a)*cosh(b) + i*cos(a)*sinh(b)
... see equation #2 ...
= sin(x)
If x is from the second solution set, we'll then have
x² = (π - x) + 2πn = (π - (a+ib)) + 2πn = (π-a + 2πn) - ib
and hence
sin(x²) = sin((π-a + 2πn) - ib) =
... apply identity #1 ...
= sin(π-a + 2πn)*cosh(-b) + i*cos(π-a + 2πn)*sinh(-b)
... note: cosh(-b) = cosh(b) , sinh(-b) = -sinh(b) ...
= sin(π-a + 2πn)*cosh(b) - i*cos(π-a + 2πn)*sinh(b)
... note: for any integer n , sin(X+2πn) = sin(X) , cos(X+2πn) = cos(X) ...
= sin(π-a)*cosh(b) - i*cos(π-a)*sinh(b)
... note: sin(π-a) = sin(a) , cos(π-a) = -cos(a) ...
= sin(a)*cosh(b) + i*cos(a)*sinh(b)
... see equation #2 ...
= sin(x)
So in both solution sets, even when n is negative, the original equation sin(x²) = sin(x) is satisfied.
Therefore, these formulas with negative values of n give complex values of x that are also solutions to the original equation.
(Please, feel free to comment if there are any mistakes.)
EDIT: Correcting "cosh(-b) = cos(b)", which misses a letter h due to a typo.
Yes, I think those complex values when n is negative are also solutions.
In short:
I think the key is proving that the trigonometry identities sin(z+2πn) = sin(z) and sin(π-z) = sin(z) are also valid when z is a complex number.
No matter if n is positive or negative, for the first solution set we'll have x² = x + 2πn , and for the second solution set we'll have x² = (π-x) + 2πn . (Verify this by squaring the found formulas while n is negative.)
So if the above trig identities hold for complex-valued z , then
in the first solution set we'll have sin(x²) = sin(x + 2πn) = sin(x) , and
in the second solution set we'll have sin(x²) = sin((π-x) + 2πn) = sin(π-x) = sin(x)
and hence the equation sin(x²) = sin(x) is also satisfied by the complex values of x from the found solution formulas.
@@yurenchu Quote: "(Please, feel free to comment if there are any mistakes.)"
As you wish 😉 : the hyperbolic cosine of negative b isn't equal to the cosine of b, but the hyperbolic cosine of b instead.
(I guess you just forgot to type a necessary h, that's all.)
@@Apollorion Thanks! That was indeed a typo of mine; luckily it didn't affect the rest of the derivation. I'll fix it.
did you make the cool poster with identities?
I did!
Do you plan of uploading it?
I was thinking of an approach based on replacing the sines by their expressions in terms of exponentials (sin (x) = (e^ix - e^-ix)/2i = (y - 1/y)/2i, sin(2x) = (y^2 - 1/y^2)/2i etc) but this was a bad idea, leading to a nasty cubic equation that doesn't easily handle the multivalued nature of sine.
2^x-e^x = ?what's deafrent between both e is a number or curve please explain
why 2^x cover 2^x so 3^x why any numbers It becomes a curve when it reaches the power of x ؟!!!
Ahhhk. Here's a better explanation.
The reason you can "cross out the sin's" is because you would then be dividing both sides by function notation. "Sin(x)" is no different than F(x) except the function notation uses 3 letters.
This would be akin to solving F(x) =2x+5 by dividing both sides by the letter F.
I liked your point about the arcsine is not 1 to 1, here is how I would explain that.
Arcsin (1) equals an infinite number of possible values --> π/2 + 2πk. and -π/2 - 2πk.
But Sin (π/2) only has one range value, y = 1. Notice that if you inverse both sides of Sine(x) =1. that is different than inversing both sides of Sin(π/2) = x for the reason that Sin(x) is not 1 to 1 which means the inverse function is problematic
You can "cross out the sin's", you might not get all the solutions if the function is not injective, but you will not introduce extraneous solutions:
Given f(x)=f(y)... one possible solution if x=y
In practice maybe you can have that... say f(x)=f(x+1)... one candidate would be x=x+1... but that's impossible.
If you can have x=y... then it is a solution for f(x)=f(y) (because you can't have f giving different outputs for the same input... else f wouldn't be a function), might not be the only one(like he explained on sin(x)=sin(x²)
@@tarentinobg i haven't watched the video yet but wouldent sin(x^2)=sin(x) be sin(1^2)=sin(1) by definition if we put in 1 as x and since 1^2=1 i dont see how any wierd sin stuff would stop that from being true
You can write under the root in the answer: 1 + 4kπ, and you have both cases included.
Not quite. The two solutions have different signs for the 1 before the +/- sign. To write write a single expression to represent all solutions you would need to write
[(-1)^k +/- root(1+4kpi)]/2 so that the sign of the first 1 will alternate with even/odd values of k.
If you look at the graphs of sin x and sin(x^2) in the interval [0, 2π] you see that the equation sin x= sin(x^2) has 14 solutions.
EȘTI UN😢 TALENT CARE COMPLICĂ,
x²=x+2kπ or x²=π-x+2kπ
Use quadratic fromula on both
3:23 I still don’t get how 2n pi fits here
The sine values repeat every 2pi. The number of repetitions is the n. If you want to include the 5th repetition, you add 5*2*pi.
To generalize, just add 2*pi*n
Because the same y value repeats every 2pi since the sin function repeats every 2pi
if it was sin^2(x) = sin(x) , it would be easy for sure, you would only have to tell when sin(x) = 0 or = 1
but my guess is that you would have to make sure to write it down as something like x^2 = x*2kπ and even then it would probably only cover half of the solutions
also there's the fact that negative solutions still exist
Actually its right we just need limit x->0 and all set 👍
First x = 1 is a solution
Nice problem! I've been thinking about conditions for when sin(x) = sin(y) & cos(x) = cos(y)... I feel you're proof though is a bit handwavy since it only really justifies for angles between 0 and 90 degrees, but how would you justify for angles greater than that and even greater than 2pi? The proof of such identities and why they cover all cases is what I'm a bit confused on here. Kindly if you or anyone could help explain this, I'd be grateful!
His proof works for all angles, with no bounds. Pi - x will always reflect the angle about the y-axis on the unit circle, because pi-x is by definition the supplement to x.
@@phiefer3 Sorry kindly could you elaborate how this explains for all angles w/ no bound... don't supplementary angles add up to 180 and are both between 0 & 180?
also how does only this identity cover all possible cases when 2 diff angles have same sin-value?
@@bobmarley9905 Yes, supplementary angles add up to 180, but there is no restriction that they must be between 0 and 180. A 210 degree angle and a -30 degree angle are supplementary, as are 480 degrees and -300 degrees. You can find the supplement of any angle with no bounds, because x + (180-x) = 180 is always true. And of course, 180 degrees is pi radians.
As for covering all possibilities, this doesn't give you all of them. You still need the +2npi for that. Each y-value occurs exactly twice per period (except for 1 and -1 which each only occur once per period). x+2npi gives you one of these angles in every period, specifically the angle that is equivalent to x. The other angle in each period is the supplement of one of the x+2npi angles, so we get all of the supplements with pi-x+2npi.
Technically, the division step wasn't "wrong", your result from it was just not correct/complete. The correct result should be "x = 1 *when x != 0* ". You can then separately check for the case of "x = 0" to see if it is actually a solution too (which it is). This still ignores all the other possible solutions, though (because cancelling the "sin"s is still wrong).
sin(a)=sin(b)
a=n*pi + (-1)^n * b
Most trig equations can be solved by graphing. Algebraically most are impossible. Only other solving method is by substitutions.
Then If I aply arcsin to both sides and cancel the sin it's wrong still?
Yes, arcsin(sin(x)) =x only if -pi/2
Yes, it's wrong because arcsin is only a right-inverse of sin. And what you'd need for cancellation in this context is a left-inverse. A true inverse of sin (defined as a function on all of R) cannot exist because sin is not an injective function as bprp has shown.
Wish I could see the conclusion.
I substituted x as 0 degrees
Sin 0 square is same as sin 0?
IS this wrong?
Most people don't consider all angles past 360 but you are right if you want to get all possible answers. Perhaps teachers should specify like all [0, 360] as the range or [0, 720] etc etc. Also, I find it strange that sin(1) = sin(1^2) because it doesn't change no matter what units we are using. 1 degree, 1 radian, 1 whatever unit we use it is still the same answer so if that is true then 1 X pi and 1^2 X pi is the same sine value. Does that mean that sin(pi) = sin(pi^2) and pi = pi^2 or pi = pi^n? If we use degrees as our units then degree^2 = degree just like it did for pi? I know this isn't true but it was something I thought of while watching this video and it reminded me how much mathematicians hate units. Does the unit of a circle's angle squared = the unit of a circle's angle? Does 30 degrees squared = 30 degrees itself when talking about the angle of a circle?
Is it strange that if a function f(x) is defined for the value 1, then the equation f(x^2)=f(x) has at least x=1 as a solution because 1^2=1 and f(1)=f(1) ?
If f(a)=b then x=a implies f(x)=b
And, by the way:
1*pi = (1^2)*pi =/= (1*pi)^2
If g(x)=h(ax) then
1) g(y+2) isn't h( ay+2 ) but h( a(y+2) ) instead
2) g(y^2) isn't h( (ay)^2 ) but h( a(y^2) ) instead
And that is true both for a=180/pi and a=pi/180
4:55 isn't -1/(8 times pi) a possible n too?
At 4:55
In 2nd line it is mentioned that n belongs to integers so -1/8π can't be n as it isn't an integer..
just put it inside and you will see you get
x = 1/2. sin 1/2 isnt equal to sin 1/4
n has to be an integer
@@shashankg1006 I know that it has to be an integer, but I thought that if n was equal to -1/8π then X was equal to 1/2, I was trying to avoid complex numbers and didn't look any further. Now I see that if I substitute x in one of the formulas above, it isn't equal. Sorry for my english, im not very good at it.
@@sowndolphin5386 you're correct, I didn't saw it
I think the 1 to 1 explanation is a bit meaningless without understanding the inner workings. The concept of doing something to both sides is the same as applying the same function to both sides. So cancelling out a function involves applying the inverse to both sides. sine fails the horizontal line test and thus has no inverse function over its full domain. Thus you can't cross it out.
exp(x), also called e^x is invertible so you can cross it out, which is equivalent to applying ln(x) to both sides. I don't know if e is invertible in the complex plane. I don't know if invertible functions exist in the complex plane.
Guess if we time i at both side and use euler formula and get the 2kπ early...?
e^(x²) is not 1-1 function right
Or x=[ (-1)^n (+ or -) root(1+4npi)]/2. And this is how I got this result, for the equation sinx=sin(a) the general solution is x=npi+(-1)^n a, and then you will get a quadratic
What if x
sin(-theta) = -sin(theta) sin is not even. cos(theta) = cos(-theta) and the second equation would be that with a 2*pi*n added.
That possibility is covered in the formulas of the found solutions.
For example, x = [1 ± √(1+8kπ)]/2 is negative when k is positive and the ± sign is a minus-sign ("-") .
So for example, for k = 3 , one of the solutions is
x = [1 - √(1+8*3*π)]/2 = [1 - √(1+24π)]/2 ≈ −3,8703
which is negative.
In that case,
x² = ([1 - √(1+24π)]/2)² =
= [1 - √(1+24π)]² / 2²
= [1 - 2√(1+24π) + (1+24π)] / 4
= [(2+24π) - 2√(1+24π)] / 4
= [(1+12π) - √(1+24π)] / 2
= [12π + 1 - √(1+24π)] / 2
= 6π + [1 - √(1+24π)]/2
= 6π + x
and therefore,
sin(x²) = sin(6π + x) = sin(x)
It doesn't matter if x is negative. As long as the difference (x² - x) is a multiple of 2π , the values of sin(x²) and sin(x) will be equal (because the sine function is periodic; sin(θ+2πk) = sin(θ) for any integer k).
And the same goes for when the mean value (x² + x)/2 equals π/2 {+ a multiple of π} (because sin(π-θ + 2πk) = sin(θ) for any integer k).
is 2*n*pi right or 2*pi*n right?😅
Have you checked these solutions?
Still, isnt arcsine an injective function?
It would mean that we can arcsine() on both sides and actually get x²=x
So its not really a crime
That's like saying it's fine to say that we can apply sqrt() on both sides on x^2=4 and get x=2. When solving an equation it is a crime to leave valid answers out.
arcsine is injective, but it doesn't matter, in fact, what matters is that it's a function (in order to use on both sides, that is). The problem is that arcsin(sin(x))≠x, but rather arcsin(sin(x))=y, where y is a number such that sin(y)=sin(x) and y∈[-π/2,+π/2], that is arcsin(x) returns a principal argument. So it's not an inverse function of sine, strictly speaking (in fact, sine does not have an inverse function since it's not bijective and only bijections have inverse functions). It does have an inverse multifunction, however, commonly(?) denoted as Arcsin(x), so Arcsin(sin(x))=(x-πk)×(-1)^k for all integer k. Proceeding with the Arcsin you will get x²=((-1)^k)x+πk for all integer k, so you will get the same equation.
It's really no different from (x²)²=x², you can use √, but √(x²)≠x, same thing here, you either use a multifunction and get ±x²=±x or you use a principal value and get |x²|=|x|, either way x²=x is just wrong.
P.S. If you want to see the whole arcsin(sin(x)) identity, here it is, but it's somewhat nasty. arcsin(sin(x))=(x-π[x/π])×(-1)^-[x/π] where [x] denotes "round to nearest" function, that is [x]=⌊⌈2x⌉/2⌋ or ⌈⌊2x⌋/2⌉ (one rounds .5 to 0, another one rounds 0.5 to 1) where ⌈x⌉ is a ceiling function (round to +∞), ⌊x⌋ is a floor function (round to -∞).
However, the “crime” was when he divided x^2 by x - dividing by a variable, which could be 0 - which DID happen in this case!
Arcsin(sin(x)) does not equal x if x is not between -pi/2 and pi/2.
Why don't you put the 2Nπ on the other side? (sin(x^2)=sin(x^2 + 2Nπ))
Are there any complex solutions?
It’s because sin(x^2) isn’t periodic
It's because it yields the same result and what the guy above from me said.
It doesn't matter which side he puts it, it results in the same solution (with the only difference that N = -n , and since n must be non-negative , N must be non-positive).
Whether sin(x^2) is periodic or not, is irrelevant.
If I'm not mistaken, the formulas found in the video also give the complex solutions when n is negative.
does this always give you all solutions with any two functions inside of sin
Sure. You always can substitute any function... For example,
sin(e^2x) = sin(e^x)
e^x = t
=>
sin(t^2) = sin (t)
Couldn't 'n' be negative? Like for case 1 'n' can be equal or greater than negative 1/(8pi)? Even for case 2: 'n' can be negative (1+4pi)/8. Making the the square root still have a real solution without going to complex number. Because if 'n' is equal to these numbers it would still result to X being 1/2.
I have a question sir
Why we say infinitely many insted of infinite????
“infinitely many” is more specific it means you can keep finding solutions without limit but the term “infinite” can be a broad as there are different types of infinity (ex. countable and uncountable)
@@funrider28thanks bro 👍
e^x is 1-to-1... Only of x is real, right?
sin(1^2) does equal to sin (1)?
Yes, because the square of one is one, and that result of squaring is put into the sine function.
@ thus, 1 is a solution?
@alghihend Yes, if any function f(x) has 1 in its domain, then f(x*x)=f(x) has x=1 as one of its solutions, because 1*1=1 and f(1)=f(1).
edit: And if 0 is in its domain, then x=0 is (also?) one if its because 0*0=0 and f(0)=f(0).
It solution is Little bit confusing and difficult I just thought about a new one what if we just take sininverse both side then we get x^2 =x now x^2 - x =0 now x=0 or x = 1 as simple as that 😂
√(1+8nπ) >= 0 not n>=0
i didn't understand the 3:25 part
Umm Value of any angle doesn’t change if you add multiples of 2π to it. Coz is one full rotation, and sine is periodic-it repeats every circle.(2π = 360° btw)
In short: Adding is like walking in circles and ending up at the same spot!
@Sneeehhhh No i understand that. What i dont understand is when he concluded that he can just write x^2=x+2nπ. But since he adds all the possible solutions now it makes sense.
How to solve the equation on his shirt??
what do you mean
It's a function I think, so it can't be solved in that sense
@NamiZu00I mean your reasoning is nonexistant (I was gonna say wrong, but you have no reasoning) but you are correct in the sense that there is really nothing to solve, it's just stating a result in math.
This is called taylor series its not an equation :)
its the taylor series expansion :]
sin x = sin θ χ=2kπ+θ or x=(2k+1)π-θ, kεZ
Oh, you're missing out the solution x=0 also in the correct proof!
Again: You can't divide by x, if x can be zero.
You can't divide by sin x, if sin x can be zero (for x=0).
Incomplete answer.
Cancel the sins out ... 😂
0:23 - canceling sin both sides is wrong, because sin(x) IS NOT EQUAL sin * (x), so we CAN NOT DIVIDE BOTH SIDES BY SIN!
he means to cross out the sin as we have sin(x^2)=sin(x) and so arcsin(sin(x^2))=arcsin(sin(x)) and we can cancel the arcsin and sin
No, the misunderstanding comes from the fact that f(a)=f(b) implies a=b only if f(x) is one-to-one. For example, log(p)=log(q) means that p=q but only because the logarithmic function is one-to-one. Nobody thinks sin(x) = variable sin times variable x.
I mean, that problem is easy to tell from a different example: sin(π) = sin(2π). π ≠ 2π, but sin is a repeating function (as in, it repeats over the period 2π), so both π and 2π give the same result when plugged into sine.
@@jmvr I think you meant sin(0) = sin(2π)? cuz sin(π) is not equal to sin(2π).
@@desmosperson-d9wIt is though
👍👍👍
E.g., 0, 1.
WTF? Do they really think sines can be canceled out?! 😂
0:22 NO…
Why are you dividing by a variable - which could be 0?
Here’s how this part should be done:
x^2=x
x^2-x=x-x - Therefore:
x^2-x=0 - Then factor:
x(x-1)=0 - Therefore:
x=0 OR x-1=0
x=0 OR x=0+1
x=0 OR x=1
Then, of course, because this is a sinusoidal equation, you have to add +2*Pi*n to each of these - Therefore:
x=0+2*Pi*n OR x=1+2*Pi*n
Have you watched the video?
he quite literelly said this is wrong and dont do it, are you deaf and blind?
Bro didn’t watch the video.
5:15
n ≥ -1/8π
N is an integer though
@bprpmathbasics
Sorry sir but should we not set the argument of the square roots equal to zero and find the minimal value? Because it could be that the outcome of the inequality will give n≥ -c which is also an integer right?; for example
1+8nπ≥0
8nπ≥-1
N≥ -1/(8nπ) now because n must be an integer you should put the restriction
N e Z therefore n≥0
But I imagine it could be -1,-2,-3 which is an integer too and then it would be n≥-1(or -2 or -3 etc)
peak
Anti spiral
i would just do arcsin on both sides and then use the quadratic formula
it's very easy, firstly, I'm going to say that sin(x)=sin(x^2) so the angle x = the angle x^2. so x^2 = x+360 because they are equal. then use the quadratic equation & you'll have 2 values you will try them and the correct value is 19,28,48.91 degree
So, pretty smart, high IQ, never had anything past algebra 2 and stat. No trig, no calc. No college. I understand the basics of sin, cos, tan in triangles. I understand if you graph them, they are waves. I am a software engineer, so when i see sin(of some value), i see that as a function which returns a value. Id never expect that you’d be able to drop the function/ cancel it.
When you showed us the wrong way to solve it you said that zero is an answer to the equation. but when you solved for both cases, neither had zero as one of the answers
0 is one of the solutions in the first solution set, when n = 0 .
(In the first solution set, when n = 0 , then x=0 or x=1 .)
If X^2 = x. The only number where this is true is x=1
There is no real maths concepts being used in any of your explanation. It's all just procedural driven instructions, not conceptual explanations.
In the final condition could have just been n ∈ ℕ
true
But there’s a usual disagreement that “is 0 a natural number?”
@@bprpmathbasics true. You should do a video asking if n^n is defined for all n ∈ ℕ
@@joshuahillerup4290 If n is defined as being an (non-negative) integer, then I see no problem or obstacle with having n^n = 1 for n=0, because all counter-arguments using limits go out of the window.