A Rare Quintic Trigonometric Equation | sin^5x+cos^5x=1

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  • Опубліковано 16 гру 2024

КОМЕНТАРІ • 488

  • @syberrus
    @syberrus 3 роки тому +188

    There is a much easier way. Given that sin^2 + cos^2 = 1, and 0

    • @SyberMath
      @SyberMath  3 роки тому +32

      Cool!

    • @playomar5106
      @playomar5106 2 роки тому +5

      Wow.

    • @SanctBlack
      @SanctBlack 2 роки тому +1

      On 10th second I though about the same solution when imagine picture of c^5+s^5

    • @garvellokenxvi
      @garvellokenxvi Рік тому +10

      0

    • @RigoVids
      @RigoVids Рік тому +1

      The only thing to note which maybe falsely concluded from this explaination is the range of cos^5(x). The range or codomain of the function is -1

  • @serands1840
    @serands1840 3 роки тому +213

    S² + C² = 1.
    S^5 = S² iff S = 0 or 1. Idem for C.
    Otherwise S^5 < S² and C^5 < C². (S and C between -1 and 1)
    Thus S^5 + C^5 always < 1.
    Thus the only solutions are for (S,C)=(0,1) or (1,0)… i.e. 0 and pi/2 mod 2pi…

  • @SyberMath
    @SyberMath  3 роки тому +52

    I don't think this problem has been solved and published on the internet before. Correct me if I'm wrong. I'm kind of surprised why it wasn't because it is not a super original idea. Anyone can come up with an equation like this. The fun part was to test if one equation would imply the other. I knew that one way implication worked but was not sure about the if-and-only-if scenario! Generally quintics are not solvable. (Isn't that sad?) Well, I think they are awesome! Any thoughts? Plz share down below...

    • @yogamulyadi9183
      @yogamulyadi9183 3 роки тому

      I ever seen this on your Twitter..,
      And thanks to that, I learned several things...

  • @kingbeauregard
    @kingbeauregard 3 роки тому +61

    I took the derivative of the function to see where the extrema were, and found out that the maxima were at x = 0 or pi/2, which also happened to be where the function equals 1.

    • @SyberMath
      @SyberMath  3 роки тому +7

      Nice!

    • @VSN1001
      @VSN1001 3 роки тому

      I did the same and found out that there are infinity many sol

    • @telnobynoyator_6183
      @telnobynoyator_6183 3 роки тому +4

      @@VSN1001 yeah bc cos(x + 2pi) = cos(x) and so on

    • @bilal42nga47
      @bilal42nga47 3 роки тому

      .

    • @cheesefrogsnail
      @cheesefrogsnail 3 роки тому

      Yes I did the same, there is another local maximum in 5PI/4 but it is negative

  • @marklevin3236
    @marklevin3236 3 роки тому +11

    Much simpler than that for any x sin^5(x)

    • @phandinhthanh2295
      @phandinhthanh2295 3 роки тому

      I don't really get it. Could u, please, elaborate?

    • @italixgaming915
      @italixgaming915 3 роки тому +2

      @@phandinhthanh2295 Easier to see in my solution:
      cos^5(x)+sin^5(x)=cos²(x).cos^3(x)+sin²(x).sin^3(x).
      1=cos²(x)+sin²(x).
      Then I can write your equation under the form:
      cos²(x).cos^3(x)+sin²(x).sin^3(x)=cos²(x)+sin²(x) cos²(x).[cos^3(x)-1]+sin²(x).[sin^3(x)-1]=0
      But cos^3(x)-1

  • @fm7490
    @fm7490 3 роки тому +20

    I solved for tan(x/2)
    For substitution:
    cosx= (1-t^2)/(1+t^2)
    Sinx=2t/(1-t^2)
    Where t=tan(x/2)
    Then find t, then x/2, then x

    • @SyberMath
      @SyberMath  3 роки тому +2

      Interesting

    • @AtomicMathematics
      @AtomicMathematics 3 роки тому

      This is an interesting problem from Math Olympiad. Here the general solution is x=2n⫪, (4n+1)⫪/2. You can check this video for the solution ua-cam.com/video/wjJKhOZ9_K8/v-deo.html

  • @vascomanteigas9433
    @vascomanteigas9433 3 роки тому +13

    If you apply the Weierstrass substitution: sin(x)=2*t/(1+t^2) and cos(x)=(1-t^2)/(1+t^2).
    It will get a 10th degree polynomial. Each Root are then derived from x=2*arctan(t)+2*pi*n.

  • @sekaibelle
    @sekaibelle 3 роки тому +42

    "Hello everyone" :)

    • @SyberMath
      @SyberMath  3 роки тому +7

      Seeing this comment again after 2 months!
      😁

  • @זאבגלברד
    @זאבגלברד 3 роки тому +5

    To solve sinx + cosx = 1 there is a known technic to write it as sinx + tg45 cosx =1 , then multiply both sides by cos45. You get sin(x+45)=cos45 ...

  • @luismuller6505
    @luismuller6505 2 роки тому +3

    21:35 "If you divide both sides by 0" is the best thing in the derivation of the answers. 😂

  • @hosseinmohammadi2800
    @hosseinmohammadi2800 3 роки тому +3

    The good point about your way is that we can find the imaginary answers easier.
    I liked it

  • @bkclaud2135
    @bkclaud2135 3 роки тому +8

    Thinking out of the box👏🏾. You literally had to come up with another equation and solve by substitution with the original problem. Wow ok

  • @alax155
    @alax155 3 роки тому +12

    21:37 "And if you divide both sides by 0". At first, I was like "Wait, whaaaaat", but then I saw you meant "by 2" xD

  • @vijaysrini27
    @vijaysrini27 Рік тому +1

    How did you get the cubic equation solved? Cubic equation direct problem hasn't been taught in either our school or college till date, and is still a mystery to most people, even in engineering!!!

    • @SyberMath
      @SyberMath  Рік тому

      There's a cubic formula. I used that method many times

  • @Jono98806
    @Jono98806 2 роки тому +1

    Regarding that mistake in the end, you can write sin x + cos x = sqrt(2)*(1/sqrt(2) sin x + 1/sqrt(2) cos x) = sqrt(2)*(cos(pi/4) sin x + sin(pi/4) cos x) = sqrt(2) sin(x + pi/4). Therefore, sin x + cos x = 1 implies sqrt(2) sin(x + pi/4) = 1, where dividing both sides sqrt(2) gives sin(x + pi/4) = 1/sqrt(2). Therefore, x + pi/4 = pi/4 or x + pi/4 = (3 pi)/4 and solving for x gives x = 0 or x = pi/2.

  • @tonywong8677
    @tonywong8677 3 роки тому +2

    Solved by plotting graph. The zeros of the function are at 0 and n pi/2. Hence, we can find out if there is any other solutions by plotting the graph of the corresponding function.

    • @AtomicMathematics
      @AtomicMathematics 3 роки тому +1

      This is an interesting problem from Math Olympiad. Here the general solution is x=2n⫪, (4n+1)⫪/2. You can check this video for the solution ua-cam.com/video/wjJKhOZ9_K8/v-deo.html

  • @omsingh7683
    @omsingh7683 Рік тому

    As sine and cosine function are the x coordinate and y coordinate of an arbitrary point on the unit circle centre at the origin therefore sin^2x>=sin^5x and cos^2x>= cos^5x now adding these two inequalities we get sin^5x+cos^5x

  • @marioalejandrocaisaguanomu8005
    @marioalejandrocaisaguanomu8005 3 роки тому +1

    A really interesting problem! We can also solve it using this identity: cos^5⁡(x) + sin^5⁡(x) = [cos⁡(x) + sin⁡(x)][1 - cos⁡(x)sin⁡(x) - cos^2⁡(x)sin^2⁡(x)]. We can use "u = cos⁡(x) + sin⁡(x)". After some substitutions we'll have this equation: u^5 - 5u + 4 = 0. If we undo the substitutions, we will have: sin⁡(2x) = u^2 - 1. From the equation in terms of "u", we'll have two real solutions. We only use the "u = 1" solution because the another value of "u" makes the expression "u^2 - 1" greater than one (max(u^2 - 1) = 1). The identity that I used can be proved if we factorize "cos^5⁡(x) + sin^5⁡(x)" as "[cos⁡(x) + sin⁡(x)][cos^4(x) - cos^3(x)sin(x) + cos^2(x)sin^2⁡(x) - cos(x)sin^3(x) + sin^4(x)]" and using another basic identities.

  • @宇佐見英晴
    @宇佐見英晴 Рік тому

    We can use the trigonometric identity:
    sin^2(x) + cos^2(x) = 1
    to rewrite the equation as:
    (sin^2(x) + cos^2(x))(sin^3(x) + cos^3(x)) = 1
    Expanding the second factor using the identity:
    a^3 + b^3 = (a + b)(a^2 - ab + b^2)
    with a = sin(x) and b = cos(x), we get:
    (sin^2(x) + cos^2(x))(sin(x) + cos(x))(sin^2(x) - sin(x)cos(x) + cos^2(x)) = 1
    Using the identity:
    sin(x)cos(x) = (1/2)sin(2x)
    we can simplify the third factor to:
    sin^2(x) - sin(x)cos(x) + cos^2(x) = sin^2(x) - (1/2)sin(2x) + cos^2(x) = 1 - (1/2)sin(2x)
    Substituting this back into the equation, we get:
    (sin(x) + cos(x))(1 - (1/2)sin(2x)) = 1
    Expanding the first factor using the identity:
    a+b)^2 = a^2 + 2ab + b^2
    with a = sin(x) and b = cos(x), we get:
    (sin(x) + cos(x))^2 = sin^2(x) + 2sin(x)cos(x) + cos^2(x) = 1 + sin(2x)
    Substituting this back into the equation, we get:
    (1 + sin(2x))(1 - (1/2)sin(2x)) = 1
    Expanding and simplifying, we get:
    1 - (1/2)sin^2(2x) = 1
    sin^2(2x) = 0
    Taking the square root of both sides, we get:
    sin(2x) = 0
    This means that either 2x = kπ (where k is an integer) or x = (kπ)/2. Therefore, the solutions to the equation are:
    x = kπ/2 or x = kπ/4, where k is an integer.

  • @benvickers8821
    @benvickers8821 3 роки тому +12

    What a fantastic explanation of the solution to this problem. Love your videos!!!

    • @SyberMath
      @SyberMath  3 роки тому +5

      Glad you like them!

    • @avrahambercovich667
      @avrahambercovich667 3 роки тому

      @@SyberMath אני חושב אני אוהב א אני את זה אבל אמא א

  • @mhm6421
    @mhm6421 Рік тому +1

    21:36 "And If you divide both sides by 0"😅

  • @meowmeow-yq9xt
    @meowmeow-yq9xt 2 роки тому +1

    Thanks I like this problem
    You guys are one of my hope that i will learn new and it needed from the beginning but my country education system doesn't fit in me. I can't even do something . But when I got my phone I realized that I shouldn't misuse of this device then step by step I found you and others who teaches free in YT at first I searched for graphical dynamic , optics , mechanics and other physic related then your one video came as recommendation I watched and I liked the way you teach and I can clearly understand your accent then it made me to subscribe your channel . Thanks for the videos

  • @akshaybagaria4941
    @akshaybagaria4941 3 роки тому +6

    Solving this with inequality x^2>x^5 when x

  • @jensknudsen4222
    @jensknudsen4222 3 роки тому +6

    So, all that work to show that the two solutions you can guess up front by plugging in a few points of the compass are in fact the only two solutions (mod 2pi). Isn't that just math in a nutshell?

    • @tylerduncan5908
      @tylerduncan5908 3 роки тому +2

      Yes but the difference is that he said "i will find out for sure" not "I'm gonna guess and hope those are the only 2 solutions

  • @jpolanco2004
    @jpolanco2004 3 роки тому +3

    Cos(x)^5-Cos(x)^2+Sin(x)^5-Sin(x)^2=0; from there, you can factor out 1-cos(x) and 1-sin(x), both of which give solutions when set to 0. The third factor can be reduced to 1+sin(x)cos(x)[sin(x)+cos(x)]=1+(1/sqrt(2))sin(2x)sin(x+pi/4), which is greater than 0. Hence x=0 and x=pi/2 are the only solutions, plus periodicity.

  • @זאבגלברד
    @זאבגלברד 3 роки тому +3

    I enjoy your math telent. Thank you !

    • @SyberMath
      @SyberMath  3 роки тому +1

      You are welcome! Thank you!

  • @shubhamjha6418
    @shubhamjha6418 2 роки тому +1

    from which book this question had been taken?

    • @SyberMath
      @SyberMath  Рік тому

      I don't think it was from a book. I thought of it myself

  • @Mosux2007
    @Mosux2007 Рік тому +1

    The quintic can also be factored using synthetic division. May be easier.

  • @MrArcan10
    @MrArcan10 2 роки тому

    where you have got u3 from? (this complicated value with cubic roots)

  • @think_logically_
    @think_logically_ 2 роки тому

    sin⁵x ≤ sin²x, cos⁵ x ≤ cos²x (this also covers negative values of sine and cosine), consequently sin⁵ x+ cos⁵ x ≤ sin²x + cos²x = 1. Therefore the equality holds if and only of both inequalities turn into equalities. Thus we get sin⁵x = sin²x, cos⁵x = cos²x , so sine or cosine must be either 0 or 1. Since sine and cosine cannot be both 0, or both 1, one of them must be 1, and another must be zero.

  • @cdiesch7000
    @cdiesch7000 3 роки тому +1

    interesting solution
    i would have done with inequalities
    by this we could demonstrate for
    sin(x)^n+cos(x)^n=1 n> 2 and natural number, there are in ]-pi,pi]2 solutions 0 and pi/2 if n is odd and 4 solutions, -pi/2,0,pi/2,pi if n is even.

  • @dmitryweinstein315
    @dmitryweinstein315 3 роки тому +1

    instead of solving the cubic equation exactly, you could have analysed the lines u^5 and 5u-4 and realised that they only have two points in common, u=1 (where the tangents of the two lines equal i.e. no other positive u) and a negative u which is less than -sqrt(2) because at u=-sqrt(2) the line u^5 is still higher than the line 5u-4

  • @mikesteele5935
    @mikesteele5935 Рік тому

    This follows almost immediately from the Cauchy Schwarz inequaliity (if we restrict attention to real x, which seems to have been implicit).

  • @AmooBaktash
    @AmooBaktash 3 роки тому

    There is an easier solution for this problem. Let f(x) = sin(x)^5 + cos(x)^5. Then, f'(x) = 5 sin(x)^5 cos(x) - 5 cos(x)^4 sin(x). After a few simplifications you get f'(x) = 5 sin(x) cos(x) ( sin(x) - cos(x) ) (1 + sin(x)cos(x) ) . Setting f'(x) = 0, you fill find that the maximum of f(x) is 1, which only happens at x = 0 , and x = pi/2 (for 0

  • @dneary
    @dneary 3 роки тому +5

    For the last bit, you can use the "trick" of multiplying acorss by 1/sqrt(2): sin x cos pi/4 + cos x sin pi/4 = sin(x+pi/4) = 1/sqrt(2) implies x+pi/4 = pi/4 or 3pi/4, x=0 or pi/2

  • @Muslim_011
    @Muslim_011 3 роки тому +4

    The final cubed expression have -1 as a solution too
    Correction I'm wrong it does not have -1 as a solution

    • @WolfgangKais2
      @WolfgangKais2 2 роки тому +1

      You mean that -1 + 2 - 3 + 4 = 0?

    • @Muslim_011
      @Muslim_011 2 роки тому

      @@WolfgangKais2 yes

    • @WolfgangKais2
      @WolfgangKais2 2 роки тому +1

      @@Muslim_011 Well, I thought it was 2:
      -1 + 2 = 1 and -3 + 4 = 1, 1 + 1 = 2.

    • @Muslim_011
      @Muslim_011 2 роки тому

      @@WolfgangKais2 yes you're right when I first time said that I quickly thought it is 1-2-3+4
      My bad. Thank you

  • @liyuan-chuanli8468
    @liyuan-chuanli8468 3 роки тому

    If |\sin x| < 1 and |\cos x| < 1, then sin^n(x) is strictly decreasing on n\ge 1.
    Similarly, sin^n(x) is strictly decreasing on n\ge 1. So,
    sin^2x + cos^2x = 1 implies that sin^5x + cos^5x < 1.
    Therefore sin^5x + cos^5x = 1 implies that either sin x = 0 or cos x = 0.

  • @wyboo2019
    @wyboo2019 Рік тому

    if you allow complex valued theta then you can have those solutions outside of the regular bounds

  • @grzegorzkondracki4630
    @grzegorzkondracki4630 3 роки тому

    Solution like an experience during climbing on K2 in one shoe, walking backward.
    I can admire, but Serands and some others solved this very smart and simply.

  • @srividhyamoorthy761
    @srividhyamoorthy761 2 роки тому

    Please explain asum with cubic formula and explain the cubic formula in detail and all the forms of the eqn I am notable to find any video helping me

  • @rajakumar9377
    @rajakumar9377 3 роки тому

    Put x equals 45° ......then this expression gives value < 1
    Hence given expression is false or invalid

  • @momohkaddri941
    @momohkaddri941 3 роки тому

    it suffices to write sin^5x + cos^5x= sin^2x + cos^2x which is eq. to this sum of potives sin^2x(1- sin^3x) + cos^2x(1- cos^3x)=0 and then, deduce , the solutions.

  • @miloradtomic
    @miloradtomic 2 роки тому

    Beautiful method, dear friend.
    Best regards from Serbia

  • @tmacchant
    @tmacchant 3 роки тому

    Squaring both side of sin(x)+cos(x)=1 gives extrareneaus solutions as mentioned in the end
    of part the video.
    Using relation of
    sin(x)+cos(x)=√2・sin(x+π/4)=1
    does not give extrareneaus solutions.

  • @autf2_6
    @autf2_6 2 роки тому

    Thx teacher very good question please dowland trigonometry questions like this really beautiful question

  • @CengTolga
    @CengTolga 3 роки тому +4

    Nice video. Can you revisit this problem, but including complex numbers? I'm particularly interested in u₃=-1.6506

    • @SyberMath
      @SyberMath  3 роки тому

      Thanks! Me no like complex numbers! 😁

  • @Caturiya
    @Caturiya 3 роки тому

    Anstelle der Kubischen Formel. Wir nennen das kubische Polynom K(u). Es har nur eine reele Nullstelle, da K(0) = 4 ist, ist diese negativ. etc

  • @ostdog9385
    @ostdog9385 3 роки тому +2

    Very good!!

  • @lequangvan6934
    @lequangvan6934 3 роки тому

    6:00 sin^3x + cos^3x = (3u - u)/2
    (Sin^3x +cos^3x)(sin^2x + cos^2x) = (3u - u)/2

  • @eliasmazhukin2009
    @eliasmazhukin2009 3 роки тому +5

    In Russia, we call "sin^2 (x) + cos^2 (x) = 1" the Fundamental Identity of Trigonometry (основное тригонометрическое тождество)

    • @AtomicMathematics
      @AtomicMathematics 3 роки тому +2

      This is an interesting problem from Math Olympiad. Here the general solution is x=2n⫪, (4n+1)⫪/2. You can check this video for the solution ua-cam.com/video/wjJKhOZ9_K8/v-deo.html

    • @peterwan9076
      @peterwan9076 2 роки тому

      @@AtomicMathematics I watch your solution. I appreciate the part where you figured out S^2(1 - S^3) + C^2(1-C^3) = 0 => S^2(1 - S^3)=0 and C^2(1-C^3) = 0. Genius. After that, it is straightforward.

  • @mcwulf25
    @mcwulf25 2 роки тому

    I factored out (s+c) and reduced what's left to sc(sc-1) using the sum of squares= 1 a couple of times. Both factors must be 1 or both -1. After that it's easy substitutions.

    • @mcwulf25
      @mcwulf25 2 роки тому

      Using this method it's clear that the only solutions are when s=0,c=1 and s=1,c=0. Two other solutions drop out but these have c or s > 1.

  • @jarikosonen4079
    @jarikosonen4079 3 роки тому

    If it can be "seen" that sin^5(x)+cos^5(x)

    • @SyberMath
      @SyberMath  Рік тому

      Very nice! I'll probably redo this and use your idea 🤩

  • @chaparral82
    @chaparral82 3 роки тому

    No calculation needed. The solution where either sinx or cosx = 1 are the only solutions.
    Because for any other values where |sinx| and |cosx| < 1 always sin²x + cos²x = 1 is valid, so because for those x values |sinx|² > |sinx|^k and |cosx|² > |cosx|^k is valid for any k> 2 (geometric prograssion for values smaller than 1 is strictly decreasing) and therefore 1 > |sinx|^k + |cosx|^k for any k> 2 q.e.d

  • @mathematicsbeautyclinics7687
    @mathematicsbeautyclinics7687 2 роки тому

    Nice mathematics. Kindly tell me the app or software i need to have to also enjoy calculating that way at home

  • @salloom1949
    @salloom1949 Рік тому

    I followed your methodology diligently and I understood every step along the way. Thoroughly enjoyed your explanation. But you left me dizzy and with headache. Thank you for a wonderful procedure.

  • @זאבגלברד
    @זאבגלברד 3 роки тому +1

    Like you did in the (sinx)^0.5 + cosx = 0 you could say right at the begining that x is greater or equal 0 and x is less or equal pi/2 .

  • @asfahanansari4669
    @asfahanansari4669 2 роки тому

    Instead of raising (Sinx + Cosx) to power 5 sin3x + Cos3x can be multiple by 1 e.g sin2x + cos2x

  • @godiswatching8110
    @godiswatching8110 Рік тому

    From relations 1=sin(x)^5+cos(x)^5

  • @danielalvear3406
    @danielalvear3406 3 роки тому +3

    Lo resolví en menos de un minuto recordando q con ciertos ángulos una función es cero y la otra es uno para el mismo valor del ángulo
    Por lo tanto obtuve como resultado 0, 90 y 360

  • @NurmemetAbliz
    @NurmemetAbliz 3 роки тому +1

    This solution is incorrect . Here is why:
    x = 0 -> sinx = 0 -> (0)^5 + 1^5 ;
    x = pi/2 -> cosx = 0 -> 1^5 + 0^5;
    And 0^5 has no meaning . It should not be permitted in the equation .

  • @piyushdaga357
    @piyushdaga357 3 роки тому +3

    Great video !
    But I, being a class 8 student, don't know how to solve for cubic and don't use radians in trigonometry...

    • @SyberMath
      @SyberMath  3 роки тому +1

      I understand. Solving for the cubic is not that important. Converting radians and degrees is fairly easy.

    • @Qermaq
      @Qermaq 3 роки тому

      @@SyberMath I never learned about radians until long after I was out of school. Had to learn them with Excel as it does all its trig with radians. I agree, they are no harder than converting temperature from C to F or vice versa. I hope anyone reading this realizes they're not hard, just different from degrees, and very often more useful.

    • @AtomicMathematics
      @AtomicMathematics 3 роки тому

      This is an interesting problem from Math Olympiad. Here the general solution is x=2n⫪, (4n+1)⫪/2. You can check this video for the solution ua-cam.com/video/wjJKhOZ9_K8/v-deo.html

  • @thierryvieville820
    @thierryvieville820 3 роки тому

    # This a hell of an intersting problem !
    # What about considering f_n(x) = sin(x)^n + cos(x)^n, and solve f_n(x) = 1 for n > 0 integer, in the [0, 2 Pi[ interval as the function is periodic ?
    f := (x, n) -> sin(x)^n+cos(x)^n:
    # If n = 1, then cos(x) + sin(x) = 2^1/2 sin(x + Pi/4), obtained by expanding sin(x+Pi/4) and we easily obtain f_1(x) = 1 for x = 0 or Pi/2
    solve(f(x, 1) = 1, x);
    # By the way, would n be a positive real we will find the same solutions, we the additional fact that f_n(x) is only entirely defined in [0, Pi/2] where both sin(x) and cos(x) are non negative
    plot( [f(x,1/16),f(x,1/8),f(x,1/4),f(x,1/2),f(x,1),f(x,3/2),f(x,4/3),f(x,7/3),f(x,11/2),f(x,13/3)], x=0..2*Pi,
    color=[blue, blue, blue, blue, black, green, green, red, red, red]);
    # If n = 2 f_2(x) = 1 for all x, which are all solutions
    # If n > 2 and odd obviously x = 0 or Pi/2 is solution since one of sin(x) or cos(x) is 0 the other being 1
    # If n > 2 and even x = - Pi/2 and x = Pi are also solution because sin(x) or cos(x) is 0 the other being -1 but because the exponent is even -1 turns to 1
    map(n -> n = map(s -> if has(s, I) or has(s, RootOf) then complex else s fi, {solve(f(x, n) = 1, x)}), [$3..7]);
    # And we are left to show that they are the only solutions
    # Track 1: study the function f_n(x) as proposed by e.g., @kingbeauregard,
    # the derivative vanishes for when sin(x) or cos(x) are zero or when sin(x) = cos(x) including in absolute value if n in even
    df := (x, n) -> n * sin(x) * cos(x) * (sin(x)^(n-2) - cos(x)^(n-2)):
    zero = expand(df(x,n) - diff(f(x, n), x));
    # lazy enough i simply plot the functions without a complete variation study
    plot( [f(x,1),f(x,2),f(x,3),f(x,4),f(x,5),f(x,6),f(x,7),f(x,8),f(x,9),f(x,10)], x=0..2*Pi,
    color=[black, black, green, blue, green, blue, green, blue, green, blue]);
    I share the plot here => app.box.com/file/857947696042
    # but this is a tracktable track, considering all extrema of the function
    # Track 2: study f_n(x) algebraically and @SyberMath couragously made the job,
    # 1/ using the fact that expanding (sin(x) + cos(x))^2 = f_1(x)^2 = 1 + 2 sin(x) cos(x) yields
    ok := sin(x) * cos(x) = simplify((f(x, 1)^2 - 1) / 2);
    # 2/ calculating f_n(x) f_1(x) = f_(n+1)(x) + cos(x) sin(x) f_(n-1)(x) leading to a recurrent relation
    # f_(n+1)(x) = f_n(x) f_1(x) + (1 - f_1(x)^2) / 2 f_(n-1)(x)
    zero := simplify(f(x, n+1) - (f(x, n) * f(x, 1) + (1 - f(x, 1)^2) / 2 * f(x, n-1)));
    f_x := n ->
    if n = 1 then f1
    elif n = 2 then 1
    else factor(f_x(n-1) * f_x(1) + (1 - f_x(1)^2) / 2 * f_x(n-2)) fi:
    # allowing to obtain f_n(x) as polynom in f_1(x)
    Fx := map(n -> fx(n) = f_x(n), [$1..10]);
    # to be solved with respect to f_1(x) eliminating solutions either complex or not in [-1, 1]
    map((n, Fx) -> n = map(s -> if has(s, I) or has(s, RootOf) then complex else s fi, {solve(subs(Fx[n], fx(n) = 1), f1)}), [$3..10], Fx);
    # through this is far beyond my manual algebaric calculation skills it is a feasible track
    # Track 3: using Fourier series
    # and writing s_k(k) = sqrt(2) * sin(k * x + Pi/4) and c_k(x) = sqrt(2) * cos(k * x + Pi/4)
    F := (x, n) ->
    simplify(
    subs(map(k -> (cos(k * x) = cos_x[k], sin(k * x) = sin_x[k]), {$1..n}),
    combine(f(x, n))),
    map(k -> (cos_x[k] + sin_x[k] = s[k](x), cos_x[k] - sin_x[k] = c[k](x)), {$1..n})):
    # allows to write f_n(x) as a sum of cos and sin for the different harmonics
    map(n -> n = F(x, n), [$1..10]);
    # and study for each extremum, when the value 1 is reached or not.
    # What an original and rich problem !

    • @SyberMath
      @SyberMath  3 роки тому +1

      Wow! Amazing! 🤩

    • @thierryvieville820
      @thierryvieville820 3 роки тому

      Thanks to you @@SyberMath, for me the really creative point is to have "invented" (or "discovered" :) ...) this generalization of the sin x + cos x = 1 problem , that was the smart idea , indeed.
      Merci beaucoup.

  • @albertmcchan
    @albertmcchan 3 роки тому +2

    sin(x)^5 + cos(x)^5 = 1
    → sin(x) ≥ 0 and cos(x) ≥ 0
    → u = sin(x) + cos(x) > 0
    → u^3 + 2*u^2 + 3*u + 4 > 4 ≠ 0
    For real x, no need to solve the cubic.

    • @phandinhthanh2295
      @phandinhthanh2295 3 роки тому

      No, sin(x) and cos(x) can both be negative on [0; 2*pi).

    • @albertmcchan
      @albertmcchan 3 роки тому +1

      To get sin(x)^5 + cos(x)^5 = 1, LHS 2 terms must be non-negative, because both terms at most reached 1. Because 5th power is odd function, this implied sin(x) and cos(x) also non-negative.

    • @phandinhthanh2295
      @phandinhthanh2295 3 роки тому

      @@albertmcchan Ok, I've got it.

  • @Mathskylive
    @Mathskylive 3 роки тому

    Phương trình về các hàm số lượng giác. Giải bởi đặt ẩn phu. Cảm ơn.

    • @SyberMath
      @SyberMath  3 роки тому

      Không vấn đề gì!

  • @ezzatabdo5027
    @ezzatabdo5027 Рік тому

    Thanks for your nice work.

  • @surendraraju5754
    @surendraraju5754 3 роки тому +1

    can't we let cos^2(x) = a. This implies sin^2(x) = 1-a. So, we have a^3+(1-a)^3 =1. Solving this gives a=0 or a=1 which implies cos^(2)x=1 or 0. So, x= 0 or pi/2.

  • @veena7274
    @veena7274 3 роки тому

    The problem can be done in a simpler manner if we consider u = sinx ^2 and the expression becomes x^3 + (1 -x)^3. It will work out to be an easy quadradic on x.

    • @SyberMath
      @SyberMath  3 роки тому

      Are you sure? 🤔

    • @veena7274
      @veena7274 3 роки тому

      @@SyberMath Sorry! I am such an idiot

  • @brentsoper345
    @brentsoper345 Рік тому

    So I solved for an expression in terms of tanx. I had to allow the two real solutions of x which are 0 and pi/2.
    This gave a nifty quadratic where tanx ^5= b and b^2-1b +1 =0
    I used the quadratic formula and de Moivres theorem to end up with two solutions 1 cis plus or minus pi/ 15
    In the range allowed I found two complex solutions. For the angles pi/15 and 29pi/15

    • @brentsoper345
      @brentsoper345 Рік тому

      tanx^5 gave two solutions from quadratic which are (1plus or minus root3)/2 nice little angles of 60 degrees on complex number diagram.

  • @ivornworrell
    @ivornworrell 3 роки тому

    How did u know to introduce u in the first place to aid solving?

    • @SyberMath
      @SyberMath  3 роки тому

      Substitution is a good method and with trig expressions like this one, it works real well

    • @ivornworrell
      @ivornworrell 3 роки тому

      @@SyberMath Forgive my seeming lack of advanced Mathematical knowledge, but it seems to me that a trig problem of this nature requires previous knowledge of Trig. Identities AND knowledge in Algebraic Manipulation in order to be able to quickly "spot" which trig expression should be represented by an arbitrary letter (u in this case), to ultimately determine the solution/s.Salaam

  • @orenawaerenyeager
    @orenawaerenyeager Рік тому

    21:37 💀witnessed a very scary statement.

  • @user-xo1ty8dg9k
    @user-xo1ty8dg9k 3 роки тому

    Avant de répondre a la question /
    Où est la formule ?

  • @krisbrandenberger544
    @krisbrandenberger544 3 роки тому +5

    On the one part, he said "Divide by 0", but meant "Divide by 2."

  • @prabhakarbk4322
    @prabhakarbk4322 3 роки тому +1

    Too good an explanation. I enjoyed the way it is solved

    • @SyberMath
      @SyberMath  3 роки тому

      Thanks a lot 😊

    • @AtomicMathematics
      @AtomicMathematics 3 роки тому

      This is an interesting problem from Math Olympiad. Here the general solution is x=2n⫪, (4n+1)⫪/2. You can check this video for the solution ua-cam.com/video/wjJKhOZ9_K8/v-deo.html

  • @ealejandrochavez
    @ealejandrochavez Рік тому

    True or false: For any positive integer n, the only real solutions of sin^n(x) + cos^n(x) = 1 are of the form pi * k / 2, where k is any integer.

  • @muhandiz6585
    @muhandiz6585 3 роки тому

    21:37 if you divide both side by 2* you said 0 which is was shocking.
    interesting solution refreshed the arithmetical part of my brain.

  • @peterbanh1364
    @peterbanh1364 3 роки тому

    x=2nPi or x=(2n+1)Pi where n is any integer.

  • @danielsittner7570
    @danielsittner7570 3 роки тому

    There is another way to solve this problem that I think is easier. First you substitute 1=sin^2(x)+cos^2(x) on to the original equation and you get
    sin^5(x)+cos^5(x)=sin^2(x)+cos^2(x).
    Then you move things around and you get (sin^3(x)-1)sin^2(x)=(1-cos^3(x))cos^2(x)
    Then you divide both sides by (sin^3(x)-1)cos^2(x) and you get
    tan^2(x)=(1-cos^3(x))/(sin^3(x)-1).
    now because cos and sin are between 0 and 1 then 1-cos^3(x) is bigger or 0 and sin^3(x)-1 is smaller or equal to it. And because anything squared is bigger than or equal to 0 tan^2(x)=0 and x=180k but we shouldn't forget that the stuff we divided by can be 0 so if we try those cases we get endothelial solution x=90+180k

  • @huytrandang277
    @huytrandang277 3 роки тому

    This, for some reason, is actually entertaining to watch.

  • @yoshinaokobayashi1557
    @yoshinaokobayashi1557 Рік тому

    sinx^2+cox^2=1 then one must be 1. x=0 or pi()/2.

  • @Kris-hz1ns
    @Kris-hz1ns 3 роки тому +5

    At 21:38, you say the most blasphemous thing that can be said in Mathematics !! 😆 Other than this, the whole explanation was quite gripping!👍

    • @SyberMath
      @SyberMath  3 роки тому +2

      I don't know what I was thinking!!!
      🤣

    • @Kris-hz1ns
      @Kris-hz1ns 3 роки тому +1

      @@SyberMath 🤣

  • @ГеоргийПлодущев-с2н

    Недавно нашёл замечательный способ ркшать такие уравниния заменяем sin x=t cosx=y
    t⁵+y⁵=1
    А вторым в систему дописываем
    t²+y²=1 по основному триганометрическому тождеству
    =>
    t⁵-t²+y⁵-y²=1-1
    t²(t³-1)+y²(y³-1)=0
    t²(t-1)(t²+t+1)+y²(y-1)(y²+y+1)=0
    Получаем
    Либо t=0 y=0
    Либоt=0 y=1
    Либо t=1 y=0
    Возвращаемся к синусам и косинусам
    Первый нам не подходит
    sinx=0 cosх =1 =>х=0⁰
    sin x =1 cos=0 =>х=90⁰

  • @fanan2644
    @fanan2644 3 роки тому

    What is the cubic formula used in 19:50

    • @SyberMath
      @SyberMath  Рік тому

      I used Wolfram Alpha 😜😂

  • @Djc99120
    @Djc99120 3 роки тому

    Can't we use calculas and solve in the following way,
    Sin^5x+cos^5x = 1
    Differentiating both side with r. To x
    => 5 sin^4x cosx - 5 cos^4x sinx = 0
    =>sinx cosx sin^3x - cosx sinx cos^3x = 0
    => sinx cosx (sin^3x - cos^3x) = 0
    Sinx cosx = 0
    2sinx cosx = 0
    Sin2x = 0 => 2x = 0
    => x = 0..........one solution
    Also (sin^3x - cos^3x) = 0
    => sin^3x - cos^3x = 0
    => sin^3x = cos^3x
    => sinx = cosx
    => x = π/4
    I don't know why the value x = π/4 is not satisfying the main equation. Can anyone pls help me figuring out the mistake

    • @SyberMath
      @SyberMath  Рік тому +1

      You can't just differentiate both sides of an equation and expect to get a solution to the original equation because the equation is only true for certain values of x. If you have an identity like x+x=2x, then differentiating will produce the same result

  • @MayankSharma-jz9lc
    @MayankSharma-jz9lc 3 роки тому +1

    I had this question for my test and I am surprised how utube recommended this to me

    • @SyberMath
      @SyberMath  Рік тому

      We are being tracked! 😜😂

  • @ku6r1ck
    @ku6r1ck 3 роки тому

    actually, all the point of (cost, sint) lies on unit circile x^2+y^2 = 1, and it implies that all solutions of equation cos^n(x) + sin^n(x) = 1 also must lie on the trivial vertexes, (1,0) and (0, 1).

  • @frankhofmann7263
    @frankhofmann7263 3 роки тому

    I solved this equation by finding the maxima of f(x) = sin^5(x) + cos^5(x) between [0;2pi[ , f'(x) = 0 --> Maximum at x1 = 0 and x2 = pi/2, f(0) = 1 and f(pi/2) = 1. To my mind this is much easier.

    • @AtomicMathematics
      @AtomicMathematics 3 роки тому

      This is an interesting problem from Math Olympiad. Here the general solution is x=2n⫪, (4n+1)⫪/2. You can check this video for the solution ua-cam.com/video/wjJKhOZ9_K8/v-deo.html

  • @Zejoant
    @Zejoant Рік тому

    Just look at it for a second and you can see x is either 0 or 90 degrees. 1 + 0 = 1. 1^5 + 0^5 = 1. What degree becomes 1 or 0 with sin and 1 or 0 with cos. 90 or 0 degrees.

  • @stratosleounakis2267
    @stratosleounakis2267 Рік тому

    Μuch less actions if we multipjy the
    sin^3 + cos^3 = ( 3u - u^3 ) / 2 whith
    sin^2 + cos^2 = 1 :
    ( sin^3 + cos^3 ) ( sin^2 + cos^2 ) = ( 3u - u^3 ) / 2
    .....

  • @rajakumar9377
    @rajakumar9377 3 роки тому

    What value of x, which satisfy this equation ??

  • @AtomicMathematics
    @AtomicMathematics 3 роки тому +3

    Thank you sharing this nice problem. There is an tricky method to solve this question. Here the general solution is x=2n⫪, (4n+1)⫪/2.

  • @andrewrao634
    @andrewrao634 3 роки тому

    Impressive, but far too involved.
    Putting s = sin(x), c = cos(x) for brevity:
    If 2x/pi is *not* an integer:
    0 < s^2 < 1 and s^3 < 1, so:
    s^5 = (s^3)(s^2) < s^2, and similarly c^5 < c^2, so:
    s^5 + c^5 < s^2 + c^2 = 1
    So 2x/pi *must* be an integer, so: 2sc = sin(2x) = 0, so:
    s = 0 or c = 0, so:
    c^5 = 1 or s^5 = 1, so:
    c = 1 or s = 1
    So x = 2n*pi or x = 2n*pi + pi/2
    Alternatively:
    s^5 + c^5 = 1 = s^2 + c^2, so:
    0 >= s^5 - s^2 = c^2 - c^5 >= 0
    So we must have:
    s^5 - s^2 = 0 = c^2 - c^5, so:
    s(1 - s) = 0 = c(1 - c), so:
    (1 - s)(1 - c) = 0, so:
    x = 0 or pi/2 etc as above

  • @GourangaPL
    @GourangaPL 3 роки тому

    21:37 i believe if you divide both sides by zero you have a problem

  • @armacham
    @armacham 3 роки тому

    shouldn't the solutions be 0 + 2n*pi and pi/2 + 2n*pi? because sin5(x) = sin5(x + 2pi)

  • @_wahahaha
    @_wahahaha 2 роки тому +1

    When u are obsessed with mathematics for 20mins u can even divide equations by zero ... 21:37

  • @phandinhthanh2295
    @phandinhthanh2295 3 роки тому

    Have anyone thought of taking derivative of f(x) = (sin x)^5 + (cos x)^5 -> f'(x) = 5(sin x)^4*cos x + 5(cos x)^4*(-sin x). f'(x) = 0
    (sin x)(cos x)[(sin x)^3 - (cos x)^3] = 0 x = 0, π/4, π/2, π, 5π/4, 3π/2 and 2π. Using a calculator to verify that on each interval (0; π/4), (π/4; π/2), (π/2; π),..., (3π/2; 2π), f '(x) does not change sign -> f(x) is monotonic. Demonstrate the behaviour of f(x) by a table(I don't know what this technique is called in English, in Vietnamese it' called 'khao sat ham so'), u can see that f(x), or the LHS of the given equation, only reaches 1 at x = 0, π/2 and stays below 1 everywhere else on [0, 2π).

  • @ngdluu
    @ngdluu 2 роки тому

    Remember that (sinx)^2+(cosx)^2=1. So [(sinx)^3+(cosx)^3]*[(sinx)^2+(cosx)^2]=(3u-u^3)/2 and then you have (sinx)^5+(cosx)^5+... It's easier than you use (sinx+cosx)^5

  • @caoainguyenang1007
    @caoainguyenang1007 3 роки тому

    This formal way is so complicated
    Sin^5

  • @MyTHAIHUNG
    @MyTHAIHUNG 3 роки тому

    sin^5x+cos^5x

  • @marcosgoldin5623
    @marcosgoldin5623 3 роки тому

    Muy buen trabajo. Saludos desde Peru

    • @SyberMath
      @SyberMath  3 роки тому

      Thank you! Greetings from the United States!

  • @MC_Transport
    @MC_Transport Рік тому

    5:22 You forgot the negative sign in front of 3u
    So It will be (-3u-u³)/2 not (3u-u³)/2