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…
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...
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.
@@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
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.
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 ruclips.net/video/wjJKhOZ9_K8/видео.html
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.
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 ruclips.net/video/wjJKhOZ9_K8/видео.html
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
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
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.
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.
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.
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!!!
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
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.
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
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?
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.
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.
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.
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 ruclips.net/video/wjJKhOZ9_K8/видео.html
@@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.
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 .
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.
Bro, you got one little problem at the quintic because you are adding -4u^5 but you have a -5u^5 wich gives us -9u^5, not just u^5, in the minute 14:33, but afterall very good video, i like to much
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.
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.
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.
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
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.
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
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.
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.
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).
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
@@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.
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 ruclips.net/video/wjJKhOZ9_K8/видео.html
# 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 !
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.
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.
11:19 But you had said on the thumbnail that you are going to prove that equation. But you already put the answer to one of your equation to make the simple. But you had gone against the indeed rules of Mathematics. Actually you cannot take that solution to find the X. It is a joke. Because in the reality, it is not a solution. It's the answer you have to prove. Your business isn't to find the X. Your actual job is to prove this equation under the indeed rules of the Maths. But you had said one thing correctly which is, every answer cannot use to prove it. The numbers only we can use to prove this equation are 0°& 90°. Don't be fussy. I think your video of Sin^100 x +Cos^100 x = 1 is also a fussy joke like this. If you wanna prove those equations correctly you have to prove those with the trigonometry conditions and the trigonometry formulas. Start from a random side and prove the other side of the trigonometry equation. That is the reality. *Fornication Under The Consent of the King* Up! Syber Math>
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
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.
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 ruclips.net/video/wjJKhOZ9_K8/видео.html
There was no need to waste 22 minutes to solve such a simple problem, with just pure logic and common sense, I solved the problem in my head !! We all know that Sin^2x + cos^2x =1, therefore we can rewrite it as (sin^3x)^2 + (cos^3x)^2 = 1 and also sin^3x or sin^5x can be equal to 0 or 1 only when x is pie/2 or 0, same goes for cos, in the end u get 0 + 1 = 1 or 1 + 0 = 1, that's the only way possible !!
@@dr.er.raunaqdhingra Does it follow that sin²x+cos²x=1 implies (sin³x)²+(cos³x)²=1 without any information on solution x? No. x=0 or π/2 is a solution.
you could use the fact that sinx+cosx=radical2 * sin(x+pi/4)... that way you won’t be adding in solutions by squaring. Of course, that requires using the formula asinx + bcosx = radical(a^2+b^2) * sin(x + arctan(b/a)). Side note: I hate that I wasn’t taught this formula in school... and by the time i discovered it exists, I no longer needed it. As always, good content my friend. Keep up the good work.
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
There is a much easier way. Given that sin^2 + cos^2 = 1, and 0
Cool!
Wow.
On 10th second I though about the same solution when imagine picture of c^5+s^5
0
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
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…
Good thinking!
I also did the same way
This method is used on youtube for C^100+S^100=1 as a competition question
Yeah, this video could have been made 22 minutes shorter xD
S^5+C^5
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...
I ever seen this on your Twitter..,
And thanks to that, I learned several things...
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.
Nice!
I did the same and found out that there are infinity many sol
@@VSN1001 yeah bc cos(x + 2pi) = cos(x) and so on
.
Yes I did the same, there is another local maximum in 5PI/4 but it is negative
Much simpler than that for any x sin^5(x)
I don't really get it. Could u, please, elaborate?
@@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
The good point about your way is that we can find the imaginary answers easier.
I liked it
21:35 "If you divide both sides by 0" is the best thing in the derivation of the answers. 😂
😁
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.
Wow!
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
Thinking out of the box👏🏾. You literally had to come up with another equation and solve by substitution with the original problem. Wow ok
Thanks! 😊
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
Interesting
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 ruclips.net/video/wjJKhOZ9_K8/видео.html
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 ...
That's a good one!
21:36 "And If you divide both sides by 0"😅
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.
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 ruclips.net/video/wjJKhOZ9_K8/видео.html
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
"Hello everyone" :)
Seeing this comment again after 2 months!
😁
The quintic can also be factored using synthetic division. May be easier.
What a fantastic explanation of the solution to this problem. Love your videos!!!
Glad you like them!
@@SyberMath אני חושב אני אוהב א אני את זה אבל אמא א
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
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.
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.
Solving this with inequality x^2>x^5 when x
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.
This follows almost immediately from the Cauchy Schwarz inequaliity (if we restrict attention to real x, which seems to have been implicit).
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!!!
There's a cubic formula. I used that method many times
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
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.
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
if you allow complex valued theta then you can have those solutions outside of the regular bounds
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?
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
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.
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.
I enjoy your math telent. Thank you !
You are welcome! Thank you!
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.
6:00 sin^3x + cos^3x = (3u - u)/2
(Sin^3x +cos^3x)(sin^2x + cos^2x) = (3u - u)/2
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.
On the one part, he said "Divide by 0", but meant "Divide by 2."
Instead of raising (Sinx + Cosx) to power 5 sin3x + Cos3x can be multiple by 1 e.g sin2x + cos2x
In Russia, we call "sin^2 (x) + cos^2 (x) = 1" the Fundamental Identity of Trigonometry (основное тригонометрическое тождество)
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 ruclips.net/video/wjJKhOZ9_K8/видео.html
@@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.
21:37 i believe if you divide both sides by zero you have a problem
oh yeah
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
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 .
That's right!
When u are obsessed with mathematics for 20mins u can even divide equations by zero ... 21:37
😜
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 .
If it can be "seen" that sin^5(x)+cos^5(x)
Very nice! I'll probably redo this and use your idea 🤩
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.
Bro, you got one little problem at the quintic because you are adding -4u^5 but you have a -5u^5 wich gives us -9u^5, not just u^5, in the minute 14:33, but afterall very good video, i like to much
Oopsies
Beautiful method, dear friend.
Best regards from Serbia
Many thanks! 🧡
21:37 💀witnessed a very scary statement.
Division by 0 should be allowed!!! 😜🤣
@@SyberMath 😂
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.
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.
Wow! Pretty good!
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.
Are you sure? 🤔
@@SyberMath Sorry! I am such an idiot
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
I had this question for my test and I am surprised how utube recommended this to me
We are being tracked! 😜😂
5:22 You forgot the negative sign in front of 3u
So It will be (-3u-u³)/2 not (3u-u³)/2
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.
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.
The final cubed expression have -1 as a solution too
Correction I'm wrong it does not have -1 as a solution
You mean that -1 + 2 - 3 + 4 = 0?
@@WolfgangKais2 yes
@@Muslim_011 Well, I thought it was 2:
-1 + 2 = 1 and -3 + 4 = 1, 1 + 1 = 2.
@@WolfgangKais2 yes you're right when I first time said that I quickly thought it is 1-2-3+4
My bad. Thank you
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
At 21:38, you say the most blasphemous thing that can be said in Mathematics !! 😆 Other than this, the whole explanation was quite gripping!👍
I don't know what I was thinking!!!
🤣
@@SyberMath 🤣
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.
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.
Excellent!
from which book this question had been taken?
I don't think it was from a book. I thought of it myself
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).
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
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
Put x equals 45° ......then this expression gives value < 1
Hence given expression is false or invalid
when sin=0 cos=1 and the other way around so x=0 or x=0,5pi.it repeats every 2pi
Nice video. Can you revisit this problem, but including complex numbers? I'm particularly interested in u₃=-1.6506
Thanks! Me no like complex numbers! 😁
where you have got u3 from? (this complicated value with cubic roots)
From relations 1=sin(x)^5+cos(x)^5
Phương trình về các hàm số lượng giác. Giải bởi đặt ẩn phu. Cảm ơn.
Không vấn đề gì!
Great video !
But I, being a class 8 student, don't know how to solve for cubic and don't use radians in trigonometry...
I understand. Solving for the cubic is not that important. Converting radians and degrees is fairly easy.
@@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.
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 ruclips.net/video/wjJKhOZ9_K8/видео.html
とはいえ、日本ではこの「回りくどく大変な」方法を遂行する力は養われていないように思います
そのいみでとても教育的なビデオでした
ありがとう！
L2 or L3 or M1 or M2.... In LMD in university ?
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.
x=2nPi or x=(2n+1)Pi where n is any integer.
Thanks for your nice work.
Thank you too!
This, for some reason, is actually entertaining to watch.
Glad to hear that!
# 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 !
Wow! Amazing! 🤩
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.
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.
No, sin(x) and cos(x) can both be negative on [0; 2*pi).
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.
@@albertmcchan Ok, I've got it.
11:19 But you had said on the thumbnail that you are going to prove that equation. But you already put the answer to one of your equation to make the simple. But you had gone against the indeed rules of Mathematics. Actually you cannot take that solution to find the X. It is a joke. Because in the reality, it is not a solution. It's the answer you have to prove. Your business isn't to find the X. Your actual job is to prove this equation under the indeed rules of the Maths. But you had said one thing correctly which is, every answer cannot use to prove it. The numbers only we can use to prove this equation are 0°& 90°. Don't be fussy. I think your video of Sin^100 x +Cos^100 x = 1
is also a fussy joke like this. If you wanna prove those equations correctly you have to prove those with the trigonometry conditions and the trigonometry formulas. Start from a random side and prove the other side of the trigonometry equation. That is the reality. *Fornication Under The Consent of the King* Up! Syber Math>
*🤣Syber Math, reply to this comment if you are from the earth👍*
It is kind of strange -- to don't use () for function's arguments: cos(x)^5
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
tanx^5 gave two solutions from quadratic which are (1plus or minus root3)/2 nice little angles of 60 degrees on complex number diagram.
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.
Sin ^5x 《 sin ^2 x
Sin ^5 x + cos^5 x 《 sin^2 x + cos^2 x =1
Solution using trigonometric identities is much shorter and quite easy-half page.
Too good an explanation. I enjoyed the way it is solved
Thanks a lot 😊
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 ruclips.net/video/wjJKhOZ9_K8/видео.html
There was no need to waste 22 minutes to solve such a simple problem, with just pure logic and common sense, I solved the problem in my head !! We all know that Sin^2x + cos^2x =1, therefore we can rewrite it as (sin^3x)^2 + (cos^3x)^2 = 1 and also sin^3x or sin^5x can be equal to 0 or 1 only when x is pie/2 or 0, same goes for cos, in the end u get 0 + 1 = 1 or 1 + 0 = 1, that's the only way possible !!
We cannot rewrite sin²x+cos²x=1 as (sin³x)²+(cos³x)²=1 !!
@@ジョン永遠 We can if x is 0 or Pie/2 !!!
@@dr.er.raunaqdhingra
Does it follow that sin²x+cos²x=1 implies (sin³x)²+(cos³x)²=1 without any information on solution x? No. x=0 or π/2 is a solution.
Nice mathematics. Kindly tell me the app or software i need to have to also enjoy calculating that way at home
Thanks! Notability
Avant de répondre a la question /
Où est la formule ?
you could use the fact that sinx+cosx=radical2 * sin(x+pi/4)... that way you won’t be adding in solutions by squaring. Of course, that requires using the formula asinx + bcosx = radical(a^2+b^2) * sin(x + arctan(b/a)).
Side note: I hate that I wasn’t taught this formula in school... and by the time i discovered it exists, I no longer needed it.
As always, good content my friend. Keep up the good work.
Thank you, my friend! 💖
Very good!!
Thank you!
This earth gonna be better if all lecturer like you
Aww, thank you!!! 🧡🥰
What value of x, which satisfy this equation ??
An ap whose common diff is 2pi and 1st term is pi by 2...😊
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
Endothelial= another. English is not my native language
Np
This is cool! Thank you!
Thx teacher very good question please dowland trigonometry questions like this really beautiful question
Np. Thank you!
Also how you found out the coefficient using pythagorean triangle
What is the cubic formula used in 19:50
I used Wolfram Alpha 😜😂
sinx^2+cox^2=1 then one must be 1. x=0 or pi()/2.