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

This video is about a trigonometric equation that hasn't been solved before
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[@syberrus]

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

[@SyberMath]

Cool!

[@playomar5106]

Wow.

[@SanctBlack]

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

[@garvellokenxvi]

0

[@RigoVids]

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]

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]

Good thinking!

[@BCS-IshtiyakAhmadKhan]

I also did the same way

[@davidseed2939]

This method is used on youtube for C^100+S^100=1 as a competition question

[@dustinbachstein3729]

Yeah, this video could have been made 22 minutes shorter xD

[@TJStellmach]

S^5+C^5

[@SyberMath]

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]

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

[@marklevin3236]

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

[@phandinhthanh2295]

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

[@italixgaming915]

@@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]

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]

Interesting

[@AtomicMathematics]

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 kzbin.info/www/bejne/rZutfJuFj56SgZo

[@vascomanteigas9433]

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.

[@SyberMath]

Wow!

[@kingbeauregard]

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]

Nice!

[@VSN1001]

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

[@telnobynoyator_6183]

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

[@bilal42nga47]

.

[@cheesefrogsnail]

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

[@זאבגלברד]

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 ...

[@SyberMath]

That's a good one!

[@luismuller6505]

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

[@SyberMath]

😁

[@宇佐見英晴]

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.

[@bkclaud2135]

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

[@SyberMath]

Thanks! 😊

[@dneary]

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

[@hosseinmohammadi2800]

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

[@sekaibelle]

"Hello everyone" :)

[@SyberMath]

Seeing this comment again after 2 months! 😁

[@Jono98806]

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.

[@marioalejandrocaisaguanomu8005]

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.

[@jpolanco2004]

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.

[@alax155]

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

[@tonywong8677]

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]

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 kzbin.info/www/bejne/rZutfJuFj56SgZo

[@akshaybagaria4941]

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

[@omsingh7683]

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

[@eliasmazhukin2009]

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

[@AtomicMathematics]

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 kzbin.info/www/bejne/rZutfJuFj56SgZo

[@peterwan9076]

@@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.

[@think_logically_]

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.

[@benvickers8821]

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

[@SyberMath]

Glad you like them!

[@avrahambercovich667]

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

[@mhm6421]

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

[@liyuan-chuanli8468]

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.

[@dmitryweinstein315]

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

[@vijaysrini27]

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]

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

[@krisbrandenberger544]

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

[@Mosux2007]

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

[@AmooBaktash]

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

[@זאבגלברד]

I enjoy your math telent. Thank you !

[@SyberMath]

You are welcome! Thank you!

[@chaparral82]

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

[@cdiesch7000]

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.

[@SyberMath]

Wow! Pretty good!

[@godiswatching8110]

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

[@Caturiya]

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

[@momohkaddri941]

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.

[@meowmeow-yq9xt]

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

[@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

[@tmacchant]

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.

[@miloradtomic]

Beautiful method, dear friend. Best regards from Serbia

[@SyberMath]

Many thanks! 🧡

[@salloom1949]

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.

[@SyberMath]

Excellent!

[@lequangvan6934]

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

[@Kris-hz1ns]

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]

I don't know what I was thinking!!! 🤣

[@Kris-hz1ns]

@@SyberMath 🤣

[@brentsoper345]

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]

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

[@grzegorzkondracki4630]

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.

[@MrArcan10]

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

[@jensknudsen4222]

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]

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

[@rajakumar9377]

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

[@Muslim_011]

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

[@WolfgangKais2]

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

[@Muslim_011]

@@WolfgangKais2 yes

[@WolfgangKais2]

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

[@Muslim_011]

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

[@orenawaerenyeager]

21:37 💀witnessed a very scary statement.

[@SyberMath]

Division by 0 should be allowed!!! 😜🤣

[@orenawaerenyeager]

@@SyberMath 😂

[@mikesteele5935]

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

[@mcwulf25]

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]

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.

[@asfahanansari4669]

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

[@Zejoant]

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.

[@_wahahaha]

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

[@SyberMath]

😜

[@ngdluu]

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

[@albertmcchan]

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]

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

[@albertmcchan]

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]

@@albertmcchan Ok, I've got it.

[@veena7274]

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]

Are you sure? 🤔

[@veena7274]

@@SyberMath Sorry! I am such an idiot

[@surendraraju5754]

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.

[@ostdog9385]

Very good!!

[@SyberMath]

Thank you!

[@jarikosonen4079]

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

[@SyberMath]

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

[@shubhamjha6418]

from which book this question had been taken?

[@SyberMath]

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

[@ku6r1ck]

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).

[@user-xo1ty8dg9k]

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

[@זאבגלברד]

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 .

[@SyberMath]

That's right!

[@muhandiz6585]

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.

[@ealejandrochavez]

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.

[@huytrandang277]

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

[@SyberMath]

Glad to hear that!

[@NurmemetAbliz]

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 .

[@rubensantos1665]

(Sinx)^5+(cosx)^5=( (Sinx)^2)^(5/2)+(Cosx)^2)^(5/2). Then elevate the sum up to (2/5), 1^(2/5)=1 and the other two gives us sinx^2+cosx^2. Easier I guess.

[@tg12367]

A*B=C => (A*B)^2= C^2 => A^2* B^2=C^2 .Hier you have an addition A+B=C if you elevated it will be (A+B)^2= C^2 => A^2+2A*B+B^2=C^2 but you turned to A^2+B^2=C^2 which not correct.

[@danielalvear3406]

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

[@thierryvieville820]

# 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]

Wow! Amazing! 🤩

[@thierryvieville820]

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.

[@bichvanngoc3150]

Sin ^5x 《 sin ^2 x Sin ^5 x + cos^5 x 《 sin^2 x + cos^2 x =1

[@MyTHAIHUNG]

sin^5x+cos^5x

[@caoainguyenang1007]

This formal way is so complicated Sin^5

[@frankhofmann7263]

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]

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 kzbin.info/www/bejne/rZutfJuFj56SgZo

[@VerSalieri]

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.

[@SyberMath]

Thank you, my friend! 💖

[@zhongyuanchen8424]

Honestly, this could be solved very easily. First of all, the two solutions 0 and pi/2 can be observed rather easily. To prove that those two are unique within the cycle, observe that sin^5(x) + cos^5(x) are strictly smaller than sin^2(x) + cos^(x) = 1 when x aren't pi/2 or 0.

[@mathematicsbeautyclinics7687]

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

[@SyberMath]

Thanks! Notability

[@crustyoldfart]

Here is an approach which should appeal to lazy people like me who like to use math software. Consider the right triangle with sides 2*u, u^2-1 , u^2+1 : cos(x)=(u^2-1)/(u^2+1) and sin(x)=2*u/(u^2+1). Thus cos(x)^5 = (u^10-5*u^8+10*u^6-10*u^4+5*u^2-1) / (u^2+1)^5 and sin(x)^5 = (32*u^5)/(u^2+1)^5, from which it follows that cos(x)^5 + sin(x)^5 = (u^10-5*u^8+10*u^6+32*u^5-10*u^4+5*u^2-1) / (u^2+1)^5 = f(u) We can look at two cases of interest : f(u)=1 and f(u)=0.Solutions for the first are u=-1,0,1 and for the second u=-(sqrt(2)+1) , sqrt(2)-1 In the last case we have to find the translation from u to x : x = arccos( ( u^2-1)/( u^2+1) ) or 2*pi - arccos((u^1-1)/(u^2+1)) In the case where f(u)=0, x = 2.356 or 3*pi/4. and x = 5.498 or 7*pi/4

[@SyberMath]

Wow! Amazing!

[@ezzatabdo5027]

Thanks for your nice work.

[@SyberMath]

Thank you too!

[@wyboo2019]

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

[@andrewrao634]

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

[@AtomicMathematics]

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.

[@SyberMath]

Np. Thanks

[@akkyiofdolphin]

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

[@SyberMath]

ありがとう！

[@necrozmaftw158]

Sin⁵X + Cos⁵X = 1 Now , (sin²x)³+(cos²)³ = 1 [ x^mn = {(x)^m}n ] (Sin²x+cos²x)³=1 We know that, sin²x+cos²=1 (1)³ = 1.

[@prabhakarbk4322]

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

[@SyberMath]

Thanks a lot 😊

[@AtomicMathematics]

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 kzbin.info/www/bejne/rZutfJuFj56SgZo

[@danielsittner7570]

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

[@danielsittner7570]

Endothelial= another. English is not my native language

[@SyberMath]

Np

[@SyberMath]

This is cool! Thank you!

[@MayankSharma-jz9lc]

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

[@SyberMath]

We are being tracked! 😜😂

[@autf2_6]

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

[@SyberMath]

Np. Thank you!

[@srividhyamoorthy761]

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

[@josemariapaez9492]

Please don't use dark colours (magenta) for development of the exercise because it is not well distinguished. Other than that, your explanation is very good.

[@SyberMath]

Thank you. I try to pay attention to that

[@GourangaPL]

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

[@SyberMath]

oh yeah

[@inyks5415]

X=sinx,Y=cosx X^5+Y^5=(X^2+Y^2)(X^3+Y^3)-X^2*Y^2(X+Y) X^2+Y^2=1 X^3+Y^3=(X+Y)(X^2-XY+Y^2) なので X^5+Y^5=(X^3+Y^3)-X^2*Y^2(X+Y) =(X+Y)(X^2-XY+Y^2)-X^2*Y^2(X+Y) =(X+Y)(1-XY-X^2*Y^2) X+Y=t とすると XY=(t^2-1)/1 なので (X+Y)(1-XY-X^2*Y^2)=1 ⇆t^5-5t+4=0 後は動画と同じ感じです。

[@benjaminkarazi968]

Hello, Is Sin⁵x, or Cos⁵x a valid and legal trigonometry argument or expression? Sin⁵x+Cos⁵x ≠ Sin(x)⁵+Cos(x)⁵. The whole world does not comprehend this mistake! Regards,

[@peterbanh1364]

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

[@RhythmofSumit]

What application is this? I mean any software is there? Can you tell me? And yes nice elaboration

[@SyberMath]

Notability

[@piyushdaga357]

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]

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

[@Qermaq]

@@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]

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 kzbin.info/www/bejne/rZutfJuFj56SgZo

[@Mathskylive]

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

[@SyberMath]

Không vấn đề gì!