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
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
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.
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!!!
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.
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 ua-cam.com/video/wjJKhOZ9_K8/v-deo.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
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.
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?
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.
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.
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.
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
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
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
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.
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.
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.
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.
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
@@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.
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
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.
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
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 .
@@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 ua-cam.com/video/wjJKhOZ9_K8/v-deo.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.
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.
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.
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
@@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
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
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
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
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
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
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).
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.
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
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.
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
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).
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
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
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 ua-cam.com/video/wjJKhOZ9_K8/v-deo.html
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!
"Hello everyone" :)
Seeing this comment again after 2 months!
😁
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!
21:35 "If you divide both sides by 0" is the best thing in the derivation of the answers. 😂
😁
The good point about your way is that we can find the imaginary answers easier.
I liked it
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! 😊
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
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
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.
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 ua-cam.com/video/wjJKhOZ9_K8/v-deo.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
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.
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
21:36 "And If you divide both sides by 0"😅
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
Solving this with inequality x^2>x^5 when x
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
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.
I enjoy your math telent. Thank you !
You are welcome! Thank you!
from which book this question had been taken?
I don't think it was from a book. I thought of it myself
The quintic can also be factored using synthetic division. May be easier.
where you have got u3 from? (this complicated value with cubic roots)
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.
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!
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
This follows almost immediately from the Cauchy Schwarz inequaliity (if we restrict attention to real x, which seems to have been implicit).
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
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
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
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.
if you allow complex valued theta then you can have those solutions outside of the regular bounds
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.
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
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.
Beautiful method, dear friend.
Best regards from Serbia
Many thanks! 🧡
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.
Thx teacher very good question please dowland trigonometry questions like this really beautiful question
Np. Thank you!
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! 😁
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
Very good!!
Thank you!
6:00 sin^3x + cos^3x = (3u - u)/2
(Sin^3x +cos^3x)(sin^2x + cos^2x) = (3u - u)/2
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 ua-cam.com/video/wjJKhOZ9_K8/v-deo.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.
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.
If it can be "seen" that sin^5(x)+cos^5(x)
Very nice! I'll probably redo this and use your idea 🤩
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
Nice mathematics. Kindly tell me the app or software i need to have to also enjoy calculating that way at home
Thanks! Notability
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!
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!
Instead of raising (Sinx + Cosx) to power 5 sin3x + Cos3x can be multiple by 1 e.g sin2x + cos2x
From relations 1=sin(x)^5+cos(x)^5
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
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 .
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 ua-cam.com/video/wjJKhOZ9_K8/v-deo.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 !
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.
Phương trình về các hàm số lượng giác. Giải bởi đặt ẩn phu. Cảm ơn.
Không vấn đề gì!
Thanks for your nice work.
Thank you too!
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.
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
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.
How did u know to introduce u in the first place to aid solving?
Substitution is a good method and with trig expressions like this one, it works real well
@@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
21:37 💀witnessed a very scary statement.
Division by 0 should be allowed!!! 😜🤣
@@SyberMath 😂
Avant de répondre a la question /
Où est la formule ?
On the one part, he said "Divide by 0", but meant "Divide by 2."
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 ua-cam.com/video/wjJKhOZ9_K8/v-deo.html
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.
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.
x=2nPi or x=(2n+1)Pi where n is any integer.
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!
This, for some reason, is actually entertaining to watch.
Glad to hear that!
sinx^2+cox^2=1 then one must be 1. x=0 or pi()/2.
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 🤣
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
What is the cubic formula used in 19:50
I used Wolfram Alpha 😜😂
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
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
I had this question for my test and I am surprised how utube recommended this to me
We are being tracked! 😜😂
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).
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.
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
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.
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
.....
That's cool!
What value of x, which satisfy this equation ??
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.
Np. Thanks
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
21:37 i believe if you divide both sides by zero you have a problem
oh yeah
shouldn't the solutions be 0 + 2n*pi and pi/2 + 2n*pi? because sin5(x) = sin5(x + 2pi)
When u are obsessed with mathematics for 20mins u can even divide equations by zero ... 21:37
😜
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).
nice
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
This formal way is so complicated
Sin^5
sin^5x+cos^5x
Muy buen trabajo. Saludos desde Peru
Thank you! Greetings from the United States!
5:22 You forgot the negative sign in front of 3u
So It will be (-3u-u³)/2 not (3u-u³)/2