How to really solve sin(x^2)=sin(x)

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  • Опубліковано 29 січ 2025

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  • @bprpmathbasics
    @bprpmathbasics  Місяць тому +18

    Factoring confusion! Trig equation 2sin^2(x)=1: ua-cam.com/video/5ckR_5M7Pig/v-deo.html

    • @bikashkumar-ue6ve
      @bikashkumar-ue6ve 22 дні тому

      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 ..

  • @Grecks75
    @Grecks75 23 години тому +1

    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.

  • @jursamaj
    @jursamaj Місяць тому +27

    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).

    • @MagnusVII
      @MagnusVII Місяць тому +4

      Only if you replace the -1 in front with (-1)^n, so that it will be +1 in the even case.

  • @mangomochi9234
    @mangomochi9234 Місяць тому +61

    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)

    • @kausarali3292
      @kausarali3292 Місяць тому +2

      Wow awesome method used. Awesome explanation given. You are really genius. Well done

    • @Garfield_Minecraft
      @Garfield_Minecraft Місяць тому +3

      at least I did the first step
      sin x^2 - sin x = 0 and then I'm stuck

    • @whoff59
      @whoff59 Місяць тому +2

      product to sum formula ?

    • @travcollier
      @travcollier Місяць тому +1

      ​@@whoff59I hate that sort of 'solution'.

  • @manuelpalencia6506
    @manuelpalencia6506 Місяць тому +8

    0:10 just some small angles aproximation

  • @Ninja20704
    @Ninja20704 Місяць тому +201

    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.

    • @Sg190th
      @Sg190th Місяць тому +1

      Yeah that actually explains how it came to be in the video. Good work.

    • @itsphoenixingtime
      @itsphoenixingtime Місяць тому +1

      Was thinking of that. Combine them to form the equation

    • @graf_paper
      @graf_paper Місяць тому

      Nice work!

    • @bprpmathbasics
      @bprpmathbasics  Місяць тому +32

      I didn’t think of that! Thanks.

  • @Michael_Fischer
    @Michael_Fischer Місяць тому +1

    I love calculations that are right and wrong at the same time.

  • @charliebamford2807
    @charliebamford2807 Місяць тому +5

    Nice to visualise the solutions in Desmos

  • @MiNa-kv3lp
    @MiNa-kv3lp 9 днів тому

    Very interesting and well explained, like all your videos. Also, I've always wondered, can't you find a clip-on microphone that works?

  • @JJ_TheGreat
    @JJ_TheGreat Місяць тому +7

    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??

    • @graf_paper
      @graf_paper Місяць тому +3

      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.

    • @graf_paper
      @graf_paper Місяць тому +1

      Hope that helped. It's a good question - best of luck with all your mathematical explorations!

    • @lawrencejelsma8118
      @lawrencejelsma8118 Місяць тому

      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.

    • @nhitc6832
      @nhitc6832 Місяць тому

      Because sine is a periodic function. If A is a solution for Sine, so is A + 2πn.

  • @SatyaVenugopal
    @SatyaVenugopal Місяць тому +1

    Oh, snap! The blue pen is out in full force today!

  • @mrrandom.
    @mrrandom. Місяць тому +209

    i know you never read comments but im early so can u please start a basic physics channel?

    • @biscuit_6081
      @biscuit_6081 Місяць тому +27

      No

    • @bjornfeuerbacher5514
      @bjornfeuerbacher5514 Місяць тому +137

      Does he even teach physics? Is he qualified to do that? And even if he is: Does he have time for that?

    • @iestyn129
      @iestyn129 Місяць тому +16

      maybe not just basic physics, but it would be cool if he did more stuff with the maths side of physics

    • @kornelviktor6985
      @kornelviktor6985 Місяць тому +16

      It would be really cool but I think he is a mathematician not a physic teacher unfortunately

    • @ZainAhmed456
      @ZainAhmed456 Місяць тому +6

      He does read comments wdym

  • @VTX-Content
    @VTX-Content 16 днів тому

    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.

  • @paulerhard1252
    @paulerhard1252 Місяць тому +3

    SIN (X) is not a “bijection,” I learned at school …

  • @Ferginka
    @Ferginka Місяць тому +18

    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)...

    • @jursamaj
      @jursamaj Місяць тому

      Well, that method does get *a* solution, just not *all* solutions. 😁

    • @Aryan-o4y7g
      @Aryan-o4y7g Місяць тому

      It should be stated x belongs to (-π/2,π/2).

    • @Ferginka
      @Ferginka Місяць тому

      @@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.

  • @batuhansahin6757
    @batuhansahin6757 Місяць тому

    Not a general formula as you did but I think differentiating implicitly gives us x = 1/2 , which fits the formula you extracted

  • @rainerzufall42
    @rainerzufall42 Місяць тому +2

    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.

    • @nasdfigol
      @nasdfigol Місяць тому +1

      Watch 2:20

    • @rainerzufall42
      @rainerzufall42 Місяць тому +1

      @nasdfigol He did the same with the final solution and missed x=0 (sin x = 0) for that reason as a solution!

  • @simkoarl
    @simkoarl Місяць тому +11

    But why don't I need to substitute the x^2 the same way? Why only the x-term?

    • @sayankhan6418
      @sayankhan6418 Місяць тому +6

      Because you would end up getting the same thing due to symmetry.

    • @fluffyfox3481
      @fluffyfox3481 Місяць тому +5

      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

    • @Ninja20704
      @Ninja20704 Місяць тому +11

      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.

    • @ronaldking1054
      @ronaldking1054 Місяць тому +1

      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.

    • @cmmp5510
      @cmmp5510 Місяць тому

      Because he's not making a substitution. He's saying that it is ALSO equal to this other thing sin (x + 2npi).

  • @ОлександрУстінов-з1о
    @ОлександрУстінов-з1о Місяць тому +1

    Thanks a lot!
    P.S. Also it can be solved by graph method

  • @RichardManns
    @RichardManns Місяць тому +1

    Have to say, I'd do the first section for trivial answers, then the second.

  • @ellos_
    @ellos_ Місяць тому

    damn... i never knew why we couldnt cancel them, just that we couldnt. That actually makes so much sense

    • @PuneetSharma-uw1kd
      @PuneetSharma-uw1kd Місяць тому

      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.

    • @ellos_
      @ellos_ Місяць тому

      @@PuneetSharma-uw1kd that is true, but its usually all the solutions in a given range for trig functions

    • @Chris-hf2sl
      @Chris-hf2sl Місяць тому

      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.

    • @Grecks75
      @Grecks75 21 годину тому

      ​@@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!

  • @katlegotshabalala6250
    @katlegotshabalala6250 Місяць тому +1

    Would it work out the same way if we used degrees instead of radians?

  • @adarshkumar3184
    @adarshkumar3184 Місяць тому +1

    How about using newton-raphson rule to determine its roots

    • @gibbogle
      @gibbogle Місяць тому

      In a problem like this we seek an analytical solution, not numerical. A numerical approach will yield only a single solution.

  • @bradwilliams7198
    @bradwilliams7198 Місяць тому +2

    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?

    • @yurenchu
      @yurenchu Місяць тому +2

      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.

    • @yurenchu
      @yurenchu Місяць тому

      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.

    • @Apollorion
      @Apollorion Місяць тому +1

      @@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.)

    • @yurenchu
      @yurenchu Місяць тому

      @@Apollorion Thanks! That was indeed a typo of mine; luckily it didn't affect the rest of the derivation. I'll fix it.

  • @DANIELMABUSE
    @DANIELMABUSE Місяць тому +6

    did you make the cool poster with identities?

  • @gibbogle
    @gibbogle Місяць тому

    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.

  • @mehdizangiabadi-iw6tn
    @mehdizangiabadi-iw6tn Місяць тому +1

    2^x-e^x = ?what's deafrent between both e is a number or curve please explain

    • @mehdizangiabadi-iw6tn
      @mehdizangiabadi-iw6tn Місяць тому

      why 2^x cover 2^x so 3^x why any numbers It becomes a curve when it reaches the power of x ؟!!!

  • @tarentinobg
    @tarentinobg Місяць тому +1

    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.

    • @tarentinobg
      @tarentinobg Місяць тому +1

      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

    • @matheusjahnke8643
      @matheusjahnke8643 Місяць тому +1

      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)

    • @joziahelliott1976
      @joziahelliott1976 Місяць тому

      ​@@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

  • @xardasnecromancer590
    @xardasnecromancer590 Місяць тому

    You can write under the root in the answer: 1 + 4kπ, and you have both cases included.

    • @phiefer3
      @phiefer3 Місяць тому

      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.

  • @renesperb
    @renesperb Місяць тому

    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.

  • @dumitrudraghia5289
    @dumitrudraghia5289 Місяць тому

    EȘTI UN😢 TALENT CARE COMPLICĂ,

  • @Ahmed-kg2gf
    @Ahmed-kg2gf Місяць тому

    x²=x+2kπ or x²=π-x+2kπ
    Use quadratic fromula on both

  • @tvvt005
    @tvvt005 Місяць тому +9

    3:23 I still don’t get how 2n pi fits here

    • @jamesharmon4994
      @jamesharmon4994 Місяць тому +7

      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

    • @wintutorials2282
      @wintutorials2282 Місяць тому

      Because the same y value repeats every 2pi since the sin function repeats every 2pi

  • @lool8421
    @lool8421 Місяць тому

    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

  • @RohitSingh-rw8cz
    @RohitSingh-rw8cz Місяць тому

    Actually its right we just need limit x->0 and all set 👍

  • @林進生-k5l
    @林進生-k5l Місяць тому +2

    First x = 1 is a solution

  • @bobmarley9905
    @bobmarley9905 Місяць тому

    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!

    • @phiefer3
      @phiefer3 Місяць тому

      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.

    • @bobmarley9905
      @bobmarley9905 Місяць тому

      @@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
      @bobmarley9905 Місяць тому

      also how does only this identity cover all possible cases when 2 diff angles have same sin-value?

    • @phiefer3
      @phiefer3 Місяць тому

      @@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.

  • @foogod4237
    @foogod4237 Місяць тому

    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).

  • @Ki0212
    @Ki0212 Місяць тому

    sin(a)=sin(b)
    a=n*pi + (-1)^n * b

  • @matthewwhitted-tx3xf
    @matthewwhitted-tx3xf Місяць тому

    Most trig equations can be solved by graphing. Algebraically most are impossible. Only other solving method is by substitutions.

  • @framedthunder6436
    @framedthunder6436 Місяць тому

    Then If I aply arcsin to both sides and cancel the sin it's wrong still?

    • @Robber30
      @Robber30 Місяць тому

      Yes, arcsin(sin(x)) =x only if -pi/2

    • @Grecks75
      @Grecks75 21 годину тому +1

      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.

  • @brian554xx
    @brian554xx Місяць тому

    Wish I could see the conclusion.

  • @CR7inUefaCristianoLeague
    @CR7inUefaCristianoLeague Місяць тому

    I substituted x as 0 degrees
    Sin 0 square is same as sin 0?
    IS this wrong?

  • @chucksucks8640
    @chucksucks8640 Місяць тому +2

    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?

    • @Apollorion
      @Apollorion Місяць тому

      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

  • @felipev6559
    @felipev6559 Місяць тому +3

    4:55 isn't -1/(8 times pi) a possible n too?

    • @shashankg1006
      @shashankg1006 Місяць тому +4

      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..

    • @sowndolphin5386
      @sowndolphin5386 Місяць тому +1

      just put it inside and you will see you get
      x = 1/2. sin 1/2 isnt equal to sin 1/4

    • @duccline
      @duccline Місяць тому +1

      n has to be an integer

    • @felipev6559
      @felipev6559 Місяць тому

      @@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.

    • @felipev6559
      @felipev6559 Місяць тому

      @@sowndolphin5386 you're correct, I didn't saw it

  • @ethos8863
    @ethos8863 Місяць тому

    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.

  • @nonickname142
    @nonickname142 Місяць тому

    Guess if we time i at both side and use euler formula and get the 2kπ early...?

  • @SumithaC-ie7tz
    @SumithaC-ie7tz Місяць тому

    e^(x²) is not 1-1 function right

  • @Dharun-ge2fo
    @Dharun-ge2fo Місяць тому +2

    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

  • @jamesharmon4994
    @jamesharmon4994 Місяць тому +1

    What if x

    • @ronaldking1054
      @ronaldking1054 Місяць тому

      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.

    • @yurenchu
      @yurenchu Місяць тому +2

      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)

    • @yurenchu
      @yurenchu Місяць тому +1

      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).

  • @spicca4601
    @spicca4601 Місяць тому

    is 2*n*pi right or 2*pi*n right?😅

  • @carlosalbertovargasguerrer1654
    @carlosalbertovargasguerrer1654 Місяць тому

    Have you checked these solutions?

  • @Bangaudaala
    @Bangaudaala Місяць тому +1

    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

    • @KeimoKissa
      @KeimoKissa Місяць тому

      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.

    • @vladislavanikin3398
      @vladislavanikin3398 Місяць тому +2

      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 -∞).

    • @JJ_TheGreat
      @JJ_TheGreat Місяць тому

      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!

    • @Robber30
      @Robber30 Місяць тому

      Arcsin(sin(x)) does not equal x if x is not between -pi/2 and pi/2.

  • @beatpeat4961
    @beatpeat4961 Місяць тому

    Why don't you put the 2Nπ on the other side? (sin(x^2)=sin(x^2 + 2Nπ))
    Are there any complex solutions?

    • @marioluigimario64
      @marioluigimario64 Місяць тому

      It’s because sin(x^2) isn’t periodic

    • @megaldon1086
      @megaldon1086 Місяць тому

      It's because it yields the same result and what the guy above from me said.

    • @yurenchu
      @yurenchu Місяць тому

      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.

    • @yurenchu
      @yurenchu Місяць тому

      If I'm not mistaken, the formulas found in the video also give the complex solutions when n is negative.

  • @Ziaphy
    @Ziaphy Місяць тому

    does this always give you all solutions with any two functions inside of sin

  • @adriansolitario2664
    @adriansolitario2664 Місяць тому

    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.

  • @aazam1208
    @aazam1208 Місяць тому

    I have a question sir
    Why we say infinitely many insted of infinite????

    • @funrider28
      @funrider28 Місяць тому +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)

    • @aazam1208
      @aazam1208 26 днів тому

      ​@@funrider28thanks bro 👍

  • @travcollier
    @travcollier Місяць тому

    e^x is 1-to-1... Only of x is real, right?

  • @alghihend
    @alghihend Місяць тому

    sin(1^2) does equal to sin (1)?

    • @Apollorion
      @Apollorion Місяць тому

      Yes, because the square of one is one, and that result of squaring is put into the sine function.

    • @alghihend
      @alghihend Місяць тому

      @ thus, 1 is a solution?

    • @Apollorion
      @Apollorion Місяць тому

      @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).

  • @FilmFunZone
    @FilmFunZone Місяць тому

    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 😂

  • @swarnajeetpaul7743
    @swarnajeetpaul7743 26 днів тому

    √(1+8nπ) >= 0 not n>=0

  • @Anti_Electron
    @Anti_Electron Місяць тому

    i didn't understand the 3:25 part

    • @Sneeehhhh
      @Sneeehhhh Місяць тому +1

      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!

    • @Anti_Electron
      @Anti_Electron Місяць тому

      ​@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.

  • @tvvt005
    @tvvt005 Місяць тому +19

    How to solve the equation on his shirt??

    • @sowndolphin5386
      @sowndolphin5386 Місяць тому +2

      what do you mean

    • @NamiZu00
      @NamiZu00 Місяць тому +8

      It's a function I think, so it can't be solved in that sense

    • @KeimoKissa
      @KeimoKissa Місяць тому +2

      ​@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.

    • @fahadalshammari3627
      @fahadalshammari3627 Місяць тому +32

      This is called taylor series its not an equation :)

    • @positronium582
      @positronium582 Місяць тому +6

      its the taylor series expansion :]

  • @7Carryon
    @7Carryon Місяць тому

    sin x = sin θ χ=2kπ+θ or x=(2k+1)π-θ, kεZ

  • @rainerzufall42
    @rainerzufall42 Місяць тому

    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.

  • @whoff59
    @whoff59 Місяць тому

    Cancel the sins out ... 😂

  • @adamoksiuta4715
    @adamoksiuta4715 Місяць тому +10

    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!

    • @qaz2391
      @qaz2391 Місяць тому +10

      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

    • @deankaraniya7422
      @deankaraniya7422 Місяць тому +6

      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.

    • @jmvr
      @jmvr Місяць тому

      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.

    • @desmosperson-d9w
      @desmosperson-d9w Місяць тому

      @@jmvr I think you meant sin(0) = sin(2π)? cuz sin(π) is not equal to sin(2π).

    • @williamsantos9471
      @williamsantos9471 Місяць тому

      ​@@desmosperson-d9wIt is though

  • @Rajan-s4j
    @Rajan-s4j 3 дні тому

    👍👍👍

  • @Misha-g3b
    @Misha-g3b Місяць тому

    E.g., 0, 1.

  • @knutritter461
    @knutritter461 Місяць тому

    WTF? Do they really think sines can be canceled out?! 😂

  • @JJ_TheGreat
    @JJ_TheGreat Місяць тому +1

    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

    • @megaldon1086
      @megaldon1086 Місяць тому +5

      Have you watched the video?

    • @wg4112
      @wg4112 Місяць тому

      he quite literelly said this is wrong and dont do it, are you deaf and blind?

    • @CalculusIsFun1
      @CalculusIsFun1 Місяць тому +2

      Bro didn’t watch the video.

  • @CapitanBuffalo
    @CapitanBuffalo Місяць тому

    5:15
    n ≥ -1/8π

  • @Jason11306
    @Jason11306 26 днів тому

    @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)

  • @epicstar86
    @epicstar86 Місяць тому

    peak

  • @nilanjannayak3821
    @nilanjannayak3821 Місяць тому

    Anti spiral

  • @xunleqitrazer
    @xunleqitrazer Місяць тому

    i would just do arcsin on both sides and then use the quadratic formula

  • @Fahd2009x
    @Fahd2009x Місяць тому

    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

  • @philh8829
    @philh8829 Місяць тому

    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.

  • @basharhanna1114
    @basharhanna1114 Місяць тому

    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

    • @yurenchu
      @yurenchu Місяць тому +2

      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 .)

  • @stuartgrier5605
    @stuartgrier5605 Місяць тому

    If X^2 = x. The only number where this is true is x=1

  • @samye5537
    @samye5537 Місяць тому

    There is no real maths concepts being used in any of your explanation. It's all just procedural driven instructions, not conceptual explanations.

  • @joshuahillerup4290
    @joshuahillerup4290 Місяць тому +6

    In the final condition could have just been n ∈ ℕ

    • @duccline
      @duccline Місяць тому

      true

    • @bprpmathbasics
      @bprpmathbasics  Місяць тому +2

      But there’s a usual disagreement that “is 0 a natural number?”

    • @joshuahillerup4290
      @joshuahillerup4290 Місяць тому

      @@bprpmathbasics true. You should do a video asking if n^n is defined for all n ∈ ℕ

    • @yurenchu
      @yurenchu Місяць тому

      ​@@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.