You can also use arcsin(t) + arccos(t) = pi/2 for all t, and take arcsin of both sides to get arcsin(sin(cos(x))) = cos(x) = arcsin(cos(sin(x))) = pi/2 - arccos(cos(sin(x))) = pi/2 - sin(x). Hence cos(x) + sin(x) = pi/2 and use your technique from there.
@@SyberMath why did you choose sqr root of 2 over 2 without explaining?? Seems arbitrary unless because its when sine equals cosine at 45 degrees? But surely no one would think lf thst..?
@@leif1075 in general if you have a•sinx+b•cosx you can always rewrite it as A•sin(x+α) , multiplying and dividing by sqrt(a^2+b^2). In this case one finds out that A=sqrt(a^2+b^2) , cosα = a/A, sinα=b/A. With this trick, if a=b, you have A=sqrt(2) and α=π/4 (easy peasy)
Taking into account that in order to have sin a= cos b the angles must add to pi/2 we can write cos x = t sin x = pi/2 - t So squaring both sides of each equation and adding them 1 = t^2+pi^2/4-pi t+ t^2 Which leads to a 2n degree eq that has no real solutions. t^2-pi/2 t +pi^2/8-1/2=0 Discriminant 2-pi^2/4 < 0
For the method used at 1:29, I feel like you should have generalised what we call the R-alpha method (where you rewrite an expression of sinx + cosx as Rsin(x + alpha) since i feel that’s clearer, good video otherwise!
You can also use the sum/product formula to rewrite this as sin(-sin(x)/2 - cos(x)/2 + π/4) sin(sin(x)/2 - cos(x)/2 + π/4) = 0 Which gives the two branches: 1. sin(x) + cos(x) - π/2 = 0 2. sin(x) - cos(x) + π/2 = 0 And then you can use the same min/max argument to see these can never be satisfied.
Ok that's just something that popped in my mind at the start of the video but I don't have the basis to solve it... f ( f' ( x ) ) = f' ( f ( x ) ) Naturally e^x satisfies it but are there other functions? How should I proceed?
It is a quite incomplete solution, for example at: sin(x)+cos(x)=Pi/2+2*n*Pi the Pi/2>sqrt(2) shows only that it has no solution if n>=0, in the negative side you need that Pi/2-2*Pi=-3/2*Pi
sin(x)=π/2 - cos(x) + nπ for integral n or sin(x) + cos(x) = π/2 + nπ or √2 * cos(x- π/4) = π/2 + nπ Now absolute value ( (π/2 + nπ )/√2) being > unity, no feasible sollution is there for x
To start the arguments inside must both be equal to 45 degrees. That implies cosx=sinx= 45 degrees but the maximum range is 1. Therefore there is no real solution.
Notice that both equations produce a oscillating waves. The wave won't intersect because both waves use x as the component that creates fist a sine and then a cosine wave and visa versa. So if you add to x you merely move the y axis in the graph. If you just multiple x than you move the values on the x axis. To get an intersection you need to make one or the other graph to either change in frequency, shift in phase or change in amplitude. We see this kind of thing on dual input oscilloscopes. Your looking at two waveforms one at twice the frequency of the other at different amplitudes with the top having a DC offset. To answer your question on the oscilloscope a slight change in the DC offset would get the lines to touch. However mathematically you need to find the point x where the two functions sin (cos x) and cos (sin x) are the closest. So you might want to use the double angle trig identity, maybe square both sides, maybe the inverse trig functions or insert a/c and b/c with a^2 + b^2 = c^2 for cos x and sin x and after getting a solution changing back to the trig functions. Maybe play with complex variables and use e and ln instead of trig functions. If you want just a numerical solution maybe write a program the inputs x and x plus looks at the differences picks a new number in between and tries that.
Let sin x = a, cos x = b Since sin a = cos b, then a+b=π/2 From this, we get sin x + cos x = π/2 If we square this we get 1+2 sin x cos x = π²/4 Since 2 sin x cos x = sin 2x, we get sin 2x = (π²-4)/4 Since both π²-4 and 4 are bigger than 0, this exists iff π²-4≤4, however, π²-4>3²-4=9-4=5>4 Thus, there is no real answer I checked for a complex one and there are 2 per log branch, pretty cool imo
It's clear that the second approach - comparison of the graphs of the two functions - that there is no solution to the equation. It's easy to dream up what looks like an equality relation as shown here, but using the geometric approach it is clear that no such equality necessarily, exists - unless of course we want to invent a new branch of mathematics. This leaves us with the realization that to treat trigonometry as pure algebra without reference to geometry can very easily lead us astray.
You made an interesting comment. I'm aware of ring/field extensions, but suppose we extend by a supposed solution S to a trig equation like this one? Will R[S] be interesting or will it just be isomorphic to a transcendental extension?
@@sk8erJG95 Your conjecture goes way beyond my mathematical knowledge to comment intelligently. That said, it seems similar to what I had in mind when I suggested in my comment that some new math could be invoked to give sense to the equation. A simple analogy might be that a polynomial with no real roots would also demonstrate a supposed " no solution " since in such a case there is no obvious Euclidean geometry involving intersection of curve and line. Perhaps my comment would have been more to the point had I said that treating trigonometry like algebra, without being hampered by Euclidean geometry leads us into unfamiliar territory. Thanks for lifting the comments to a new higher level.
As a further matter of interest we can create parametric relations which are analogous to cosine and sine : c(u) = ( u^2 -1 )/( u^2 +1 ). and s(u)= ( 2*u )/( u^2 +1 ) where u is a rational number, and it is easy to show that c (u)^2 + s(u)^2 =1 for all values of u. If we look at c(s(u)) = s(c(u)) we obtain solutions for u { 0,1,-1,i*sqrt3),-i*sqrt(3) } and this can be confirmed with a plot of the two functions, since the curves intersect at the values of u indicated.
You can't solve in standard ways. One solution is x=0. Graph them together on desmos and you'll see the intersection points. (There are infinitely many because cosine is periodic and will keep oscillating while e^x approaches 0 as x approaches negative infinity.) No positive solutions.
Method 2 (please check my work?): Identities: sin x = cos ( x - π/2) sin x = cos (-x + π/2) Substitute: cos ( cos x - π/2) = cos (sin x) cos (-cos x + π/2) = cos (sin x) Equals are equal: cos (cos x - π/2) = cos (-cos x + π/2) Arccos: cos x - π/2 = -cos x + π/2 + 2kπ Rearrange, divide by 2: cos x = kπ + π/2 Then: -1 < cos x < 1 is not true for any k ∈ ℤ Does anybody know whether arccos (π/2 + kπ) provides any of the complex solutions?
Another way is to finish up the algebra to solve explicitly for x: Just transform the sin as you did to a cos, then set cos(x) = y Which means sin(x) = sqrt(1-y^2) So you get pi/2-y = sqrt(1-y^2) Whose solution is y = pi/4+-sqrt(1/2-pi^2/16) So x = arccos(pi/4+-sqrt(1/2-pi^2/16)) Since 1/2
ton équation n'admet aucune solution ; tu peux écrire cos(pi/2 - cosx) = cos(sinx) soit donc pi/2 - cosx = + ou - sinx (inversion du cosinus) et tu considères deux fonctions périodiques de période 2pi soient f(x) = cosx + sinx - pi/2 et g(x) = cosx - sinx - pi/2 vont-elles s'annuler ? tu fais un tableau des variations (pour x variant de 0 à 2pi) de chacune des deux fonctions : pour f sa dérivée s'annule pour pi/4 et 5pi/4 et toutes les images par f pour 0 < x < 2pi sont négatives pour g sa dérivée s'annule pour 3pi/4 et 7pi/4 et là aussi toutes les images sont négatives donc f et g ne s'annulent jamais et ton équation ne comporte aucune solution Cordialement
For this Trig. Equation to be true Sin(x) and Cos(x) must be equal to π/4, Because; Sin(Cos(x)) = Cos(Sin(x)) Sin(π/4) = Cos(π/4) = 1/Sqrt(2). But to find x we must take the inverse function of Sin and/or Cos of π/4, But Csc(π/4) and Sec(π/4) isnt lie in the interval of Negative Infinity to Positive Infinity. Thus, Sin(Cos(x)) = Cos(Sin(x)) has no solutions for real value of x
Secant and cosecant are not the inverse functions of sine and cosine; they are their reciprocals. The inverse functions are arcsine and arccosine - which confusingly (and IMHO incorrectly) are often represented as sinˉ¹ and cosˉ¹
You can also use arcsin(t) + arccos(t) = pi/2 for all t, and take arcsin of both sides to get arcsin(sin(cos(x))) = cos(x) = arcsin(cos(sin(x))) = pi/2 - arccos(cos(sin(x))) = pi/2 - sin(x). Hence cos(x) + sin(x) = pi/2 and use your technique from there.
Nice!
@@SyberMath That's exaclty what I did..this has NO REAL.solutions righy??
@@SyberMath why did you choose sqr root of 2 over 2 without explaining?? Seems arbitrary unless because its when sine equals cosine at 45 degrees? But surely no one would think lf thst..?
@@leif1075 in general if you have a•sinx+b•cosx you can always rewrite it as A•sin(x+α) , multiplying and dividing by sqrt(a^2+b^2).
In this case one finds out that A=sqrt(a^2+b^2) , cosα = a/A, sinα=b/A. With this trick, if a=b, you have A=sqrt(2) and α=π/4 (easy peasy)
The graph reveals another method of doing this: Show that cos(sin(x)) > sin(cos(x)) for all x.
Taking into account that in order to have sin a= cos b the angles must add to pi/2 we can write
cos x = t
sin x = pi/2 - t
So squaring both sides of each equation and adding them
1 = t^2+pi^2/4-pi t+ t^2
Which leads to a 2n degree eq that has no real solutions.
t^2-pi/2 t +pi^2/8-1/2=0
Discriminant 2-pi^2/4 < 0
Nice !!
Nice!
For the method used at 1:29, I feel like you should have generalised what we call the R-alpha method (where you rewrite an expression of sinx + cosx as Rsin(x + alpha) since i feel that’s clearer, good video otherwise!
You can also use the sum/product formula to rewrite this as
sin(-sin(x)/2 - cos(x)/2 + π/4) sin(sin(x)/2 - cos(x)/2 + π/4) = 0
Which gives the two branches:
1. sin(x) + cos(x) - π/2 = 0
2. sin(x) - cos(x) + π/2 = 0
And then you can use the same min/max argument to see these can never be satisfied.
Ok that's just something that popped in my mind at the start of the video but I don't have the basis to solve it...
f ( f' ( x ) ) = f' ( f ( x ) )
Naturally e^x satisfies it but are there other functions? How should I proceed?
Would it also be the case for imaginary numbers ??
In this case there are no pure imaginary solutions. There are complex solutions.
that equation focused on real numbers as most others do. most people don't go as far to include the complex solutions unless they have no other choice
It is a quite incomplete solution, for example at:
sin(x)+cos(x)=Pi/2+2*n*Pi the Pi/2>sqrt(2) shows only that it has no solution if n>=0, in the negative side you need that Pi/2-2*Pi=-3/2*Pi
It's obvious
@@SyberMath Did you remark all these comments from MathsJokar
Awesome , trig are so important !
Absolutely!
sin(x)=π/2 - cos(x) + nπ for integral n
or sin(x) + cos(x) = π/2 + nπ
or √2 * cos(x- π/4) = π/2 + nπ
Now
absolute value ( (π/2 + nπ )/√2) being > unity, no feasible sollution is there for x
Are there complex solutions?
The complex solutions are pretty cool.
How does Cos(pi/2-B)=cosB at 5.17?
I was wondering tha same...
To start the arguments inside must both be equal to 45 degrees. That implies cosx=sinx= 45 degrees but the maximum range is 1. Therefore there is no real solution.
Great question and solution. Can I request you take it one step further? What do you add to sin (cos x) so that they do meet without crossing over?
Notice that both equations produce a oscillating waves. The wave won't intersect because both waves use x as the component that creates fist a sine and then a cosine wave and visa versa. So if you add to x you merely move the y axis in the graph. If you just multiple x than you move the values on the x axis. To get an intersection you need to make one or the other graph to either change in frequency, shift in phase or change in amplitude. We see this kind of thing on dual input oscilloscopes. Your looking at two waveforms one at twice the frequency of the other at different amplitudes with the top having a DC offset. To answer your question on the oscilloscope a slight change in the DC offset would get the lines to touch. However mathematically you need to find the point x where the two functions sin (cos x) and cos (sin x) are the closest. So you might want to use the double angle trig identity, maybe square both sides, maybe the inverse trig functions or insert a/c and b/c with a^2 + b^2 = c^2 for cos x and sin x and after getting a solution changing back to the trig functions. Maybe play with complex variables and use e and ln instead of trig functions. If you want just a numerical solution maybe write a program the inputs x and x plus looks at the differences picks a new number in between and tries that.
Let sin x = a, cos x = b
Since sin a = cos b, then a+b=π/2
From this, we get
sin x + cos x = π/2
If we square this we get
1+2 sin x cos x = π²/4
Since 2 sin x cos x = sin 2x, we get
sin 2x = (π²-4)/4
Since both π²-4 and 4 are bigger than 0, this exists iff π²-4≤4, however, π²-4>3²-4=9-4=5>4
Thus, there is no real answer
I checked for a complex one and there are 2 per log branch, pretty cool imo
I did this instead of my analytical mechanics homework, procrastination ftw
😁
It's clear that the second approach - comparison of the graphs of the two functions - that there is no solution to the equation. It's easy to dream up what looks like an equality relation as shown here, but using the geometric approach it is clear that no such equality necessarily, exists - unless of course we want to invent a new branch of mathematics. This leaves us with the realization that to treat trigonometry as pure algebra without reference to geometry can very easily lead us astray.
You made an interesting comment. I'm aware of ring/field extensions, but suppose we extend by a supposed solution S to a trig equation like this one? Will R[S] be interesting or will it just be isomorphic to a transcendental extension?
@@sk8erJG95 Your conjecture goes way beyond my mathematical knowledge to comment intelligently. That said, it seems similar to what I had in mind when I suggested in my comment that some new math could be invoked to give sense to the equation. A simple analogy might be that a polynomial with no real roots would also demonstrate a supposed " no solution " since in such a case there is no obvious Euclidean geometry involving intersection of curve and line. Perhaps my comment would have been more to the point had I said that treating trigonometry like algebra, without being hampered by Euclidean geometry leads us into unfamiliar territory. Thanks for lifting the comments to a new higher level.
They have that new branch of math you're looking for--complex analysis! In other words there are complex solutions
As a further matter of interest we can create parametric relations which are analogous to cosine and sine :
c(u) = ( u^2 -1 )/( u^2 +1 ). and s(u)= ( 2*u )/( u^2 +1 ) where u is a rational number, and it is easy to show that
c (u)^2 + s(u)^2 =1 for all values of u. If we look at c(s(u)) = s(c(u)) we obtain solutions for u { 0,1,-1,i*sqrt3),-i*sqrt(3) } and this can be confirmed with a plot of the two functions, since the curves intersect at the values of u indicated.
I checked on desmos it has zero sol
@@unonovezero won't work
Where are you from ? :D
Thank you
You're welcome
Can you do e^x = cos(x)?
I've been wondering how to possibly solve this but didn't get anywhere
You can't solve in standard ways. One solution is x=0. Graph them together on desmos and you'll see the intersection points. (There are infinitely many because cosine is periodic and will keep oscillating while e^x approaches 0 as x approaches negative infinity.) No positive solutions.
Can you solve this for me sir, prove tan72=tan66+tan36+tan6.
No solution problem,favorite problem for teachers,and most hated for students😃
Thank you, I like trigonometry!
You're welcome 😊
The trig equation was amazing!
I liked it a lot 🙂💖
wow!
@@tariqislam4915 OMG!
Glad to hear that! 💕
@@jimmykitty
Salve, Soror,
Quid agis?
@@tariqislam4915 Ego frater valeo, Quid de te?
I am wondering how this equation might come up at all.
Of course the line y=x crossing the unit circle immediately came to mind but there is always more to the story
Can't there be complex solutions to this
Ynot
Method 2 (please check my work?):
Identities:
sin x = cos ( x - π/2)
sin x = cos (-x + π/2)
Substitute:
cos ( cos x - π/2) = cos (sin x)
cos (-cos x + π/2) = cos (sin x)
Equals are equal:
cos (cos x - π/2) = cos (-cos x + π/2)
Arccos:
cos x - π/2 = -cos x + π/2 + 2kπ
Rearrange, divide by 2:
cos x = kπ + π/2
Then:
-1 < cos x < 1 is not true for any k ∈ ℤ
Does anybody know whether arccos (π/2 + kπ) provides any of the complex solutions?
One way to look for complex solutions is to asume they exists and substitute a+bi for the real variable.
Thanks
No problem
sin(cos("no solution")) = cos(sin("no solution"))😀
Hey sybermath. Please teach Trignometry from basics. Some of your young viewers (like me) need it very bad!!
How young are you? As in which class
There's literally thousands of lessons of trigonometry on UA-cam. You don't need it very bad
@@two697 lol ikr
9th grade. I just like his teaching 🙂
@@cube7353 chill.
For 9th 10th you don't need UA-cam for studies tbh. don't overdo without reason
Phương trình lượng giác sin lồng cos.
We can prove that cos(sinx) > sin(cosx) as following:
cos(sinx) - sin(cosx) = sin(pi/2-sinx) - sin(cosx) = 2* cos((pi/2-sinx+cox)/2) * sin((pi/2-sinx+cox)/2) > 0, because abs(sinx+-cosx) < pi/2
Nice!
No solutions is also considered a solution🥇
Thanks plz this limit limite (tan(sin(x) )- sin(tan(x) )) /(2x cos(tan(x) )-2x cos(sin(x) ) +x^5) x---->0
Nice!
Thanks!
@@SyberMath My intuition was surely there's a real solution. Nope!
Güzel bir kanıttı teşekkürler
After guessing and checking, I didn't think there were any solutions that satisfied both sides.
Trig functions aren't so simple. Usually for answers in trig equations you have smth like pi/6, pi/19, pi/something something something
X=45deg
Another way is to finish up the algebra to solve explicitly for x:
Just transform the sin as you did to a cos, then set
cos(x) = y
Which means
sin(x) = sqrt(1-y^2)
So you get
pi/2-y = sqrt(1-y^2)
Whose solution is
y = pi/4+-sqrt(1/2-pi^2/16)
So
x = arccos(pi/4+-sqrt(1/2-pi^2/16))
Since 1/2
Good!
x = pi/4
Really?
ton équation n'admet aucune solution ; tu peux écrire cos(pi/2 - cosx) = cos(sinx) soit donc pi/2 - cosx = + ou - sinx (inversion du cosinus)
et tu considères deux fonctions périodiques de période 2pi soient f(x) = cosx + sinx - pi/2 et g(x) = cosx - sinx - pi/2 vont-elles s'annuler ?
tu fais un tableau des variations (pour x variant de 0 à 2pi) de chacune des deux fonctions :
pour f sa dérivée s'annule pour pi/4 et 5pi/4 et toutes les images par f pour 0 < x < 2pi sont négatives
pour g sa dérivée s'annule pour 3pi/4 et 7pi/4 et là aussi toutes les images sont négatives
donc f et g ne s'annulent jamais et ton équation ne comporte aucune solution
Cordialement
nice
Thanks
aw man :(
👍
For this Trig. Equation to be true Sin(x) and Cos(x) must be equal to π/4, Because;
Sin(Cos(x)) = Cos(Sin(x))
Sin(π/4) = Cos(π/4) = 1/Sqrt(2).
But to find x we must take the inverse function of Sin and/or Cos of π/4, But Csc(π/4) and Sec(π/4) isnt lie in the interval of Negative Infinity to Positive Infinity.
Thus, Sin(Cos(x)) = Cos(Sin(x)) has no solutions for real value of x
Secant and cosecant are not the inverse functions of sine and cosine; they are their reciprocals. The inverse functions are arcsine and arccosine - which confusingly (and IMHO incorrectly) are often represented as sinˉ¹ and cosˉ¹
the first sentence is not correct .. sin (A) = cos (B) does not require that A = B = pi/4
However you have a good idea, which *could* produce one solution if it worked: try to solve cos x = pi/4 and sin x = pi/4. Since -1