Yeah. It makes perfect sense once you think about it graphically: sin(x) starts at 0 and only goes up to 1, meaning cos(sin(x)) will start at 1 but never go down to zero; while cos(x) starts at 1 and goes down to -1, meaning sin(cos(x)) will start at sin(1) and go down to sin(-1), and continue to meander between those two values, which makes it strictly smaller than cos(sin(x)). But it's not exactly obvious.
At 7:45 the inequality is only correct if (4n-1) > 0, i.e. n ≥ 1. For n ≤ 0, the inequality flips: from 2√2 < 3 and 𝜋/3 > 1, we get: sin(x-𝜋/4) = (4n-1)𝜋/√8 < 4n-1. As 4n-1 ≤ 1 for all n ≤ 0, again the equation has no solutions.
I have a simpler method. We can write cos(sinx)=sin(π/2-sinx) since both π/2-sinx and cosx are in interval (0,π/2) so we need to show π/2-sinx>cosx. Now we can write sinx+cosx as √2sin(x+π/4) which is less than or equal to √2
I might have missed something but why would pi/2-sin(x)>cos(x) be sufficient to prove that sin(pi/2-sin(x))=/=sin(cos(x)) ? sin is not injective over [-1,pi/2+1]
@@hach1koko If a,b€[0,π/2] and a>b then obviously sin(a)>sin(b). But now I understood that π/2-sinx is not always ≤π/2. The prove will need some amendment when sinx
For the case sinx< 0, We can write sin(cosx)=cos(cosx-π/2) since both sinx and cosx-π/2 are in (-π, 0). If a,b€(-π, 0) and a>b then cos(a)>cos(b). so we need to show sinx>cosx-π/2. we can write sinx-cosx as √2sin(x-π/4) which is ≥ -√2 > -π/2, so sinx-cosx > -π/2. So sinx>cosx-π/2 !!!!!
To me, it all came down to |sin x| < |x| for x =/= 0. As cos(x) has a maximum from the origin to +/-pi, we find for x =/= 0 that cos(x) < cos(sin(x)) Using the same inequality, we likewise have for from -pi/2 to pi/2 exclusive sin(cos(x)) < cos(x) The rest is explained with period and checking at x = 0.
Good video, made better for me because the inequality is, to me, so unintuitive. I am so used to treating sin and cos as basically the same thing (just displaced) it's amazing that such a symmetrical set up isn't bigger sometimes and smaller at others
It may be able to show only to use more easy method. cos(sin(x)) > sin(cos(x))‥① is equivalent to cos( | sin(x) | ) > cos( π/2 - cos(x)) ‥② and is equivalent to | sin(x) | < π/2 - cos(x) ‥③ because both of | sin(x) | and π/2 - cos(x) are in [0,π], and it holds if | sin(x) ± cos(x) | < π/2‥④ holds, and one can easily check that the maximal value of the LHS is √2.
After 4:31, the inequalities and cases and subcases makes things needlessly complicated, imho. Using the fact, that |sin(x)+- cos(x)| = sqrt(2) < pi/2 (One would need to prove that, to be fair) The trinangle inequality yields for both arguments that they are strictly less than pi, so we have n=0. Then using the fact that |sin(x)+- cos(x)| = sqrt(2) < pi/2, we obtain by the reverse triangle inequality, that the argument is never zero (for Reals), showing no solution.
I feel I have found a most elegant geometric proof of this inequality. Consider the function on the plane f(x , y) = cos y - sin x . The original problem is then equivalent to the statement: cos y - sin x > 0 on the unit circle. We first solve the the equation cos y = sin x on the plane. We get an infinite number of intersecting diagonal straight lines. Then we show the distance of each of these lines from the unit circle is positive. Lastly, since f is continuous, all we need do is find one value of f on the unit circle that's positive.
Awesome approach. I have a bit more of a "brute force" approach, but to me it made more sense. First note that the inequality can be re-arranged to form cos(sinx) - sin(cosx) > 0. We will also need the equality siny = cos(pi/2 - y) First note that we have the following interval for the sin function: -pi/2 < -1 0 b/c cosine is always positive on that interval ( see above) and sin (0
Heres a neat question which I got from my prof they you could maybe do a vid on: is there a group G such that Aut(G) is isomorphic to (Q,+)? Spoiler: Because of the isomophism, Aut(G) must be locally cyclic, which means that Inn(G) is locally cyclic, which means that G is abelian, which means that the map x-> x^{-1} is an automorphism. If Its not the identity function, then we've found an order two element in Aut(G) so there's our contradiction. But if instead this map is the identity, then that means all elements of G have order 2, so then you can construct another order two automorphism using that (one way might be to show that its isomorphic to the countable direct sum of Z_2 with itself and then construct an order two automorphism from that). Therefore there is no such group such that Aut(G) is isomorphic to (Q,+)
If you put the equality in Wolfram Alpha, the graph of these two equations together is quite awesome! Also, wondering if using the alternative form with ½ie^-½i(…) etc would’ve been easier to see the inequality. Granted, I didn’t try it….
Hey, Michael! What I found out for the n>=0 subcases of the case where the second "big argument" equals 0, is that the proof is essentially the same as the proof for the n>=0 subcases of the case where the first "big argument" equals 0, but with the plus and minus signs switched. Furthermore, the n=0 subcase for both "big argument equals 0" cases results in sine value being less than -1, which is impossible.
The crazy part about cos(sin (x + k)) or sin(cos (x + k)) or cos(cos(...)) or sin(sin(...)) is that, for the graph's general shape, what's inside of the parenthesis doesn't matter, only the outside function. Because no matter what, the inside part just oscillates between -1 and 1. Whereas cos(something in [-1, 1]) and sin(something in [-1, 1]) have very different ranges. Looking at them naively you'd think they have similar graphs, just displaced. Not so. Composition throws your naive math instincts out of whack.
A plot of the two functions is amusing. sin(cos(x) has one minimum in [0,2pi) while cos(sin(x)) has two - basically has one-half the period of sin(cos(x)).
My idea was this. We can take only x from , there Is cos(sinx)>0; sin(cosx)>=0 So we can proof,that (cos(sinx))^2>(sin(cosx))^2 (cos(cosx))^2>1-(cos(sinx))^2=(cos0+cos sinx)(cos0-cos sinx)=(-4)(cos(sinx/2))^2)*(sin(sinx/2))^2 (Its not positive, cos(cosx)0) If we have cos(sinx)>sin(cosx) for x in (0;π/2) , for other intervals from (π/2;2π) Is left side the same, right side can be also Positive,0, negative.... For x0.
Great video! I have a question. Why is it necessary to do the proof after 11:27? If you have f(x) continuous and f(x)!=0 for all x real, isn't enough to show that f is positive just at a single point,? For example x=0, as you did
Well, yes. If f(x) is continuous and f(x) ≠ 0 for all x ∈ ℝ then you can visualise that the graph of f(x) must lie entirely above or entirely below the x-axis, and so finding the value of f(x) at any convenient point will resolve which it is. Michael Penn's proof after 11:27 is simply demonstrating rigorously that our conclusion from the visualisation of the graph is sound.
*Simpler proof* -1 pi/2 > sinx + cos x = √2 sin(pi/4+x) = √2 cos(pi/4-x) => pi/2 > √2 x (max=1) => pi > 2 √2 We know pi > 3.14 is indeed greater than 2 √2 = 2 x 1.41421 < 2 x 1.5 = 3. Hence, proved.
A bit of confusion: if x is pi/2, then plugging into the inequality gives cos(1)>sin(0), which I feel is incorrect. Yet we proved that should not be. What am I doing wrong?
@@TimmmmartinThanks. Actually I realized I am mixing up the arguments for the sin and cos before substitution but writing it correctly here. Big brain fart...
Hey! This inequality is the method I used for SyberMath's video on "solving cos(sinx) = sin(cosx)". He uses a different method to show no solutions exist. Very cool
It works for cos(sin(ax)) > sin(cos(ax)) where a is any real scale factor. On simple way to illustrate this is if you were to visualise the graphs of each side, then scaling x by any value simply stretches the graph in the x-direction while preserving the fact that the value of the LHS remains greater than the value of the RHS for all x. Converting radians to degrees is just a scale factor of 180/π.
Great exercise for my college students! When treating the "algebraic" proofs of goniometric formulae in a pre-calculus course such kind of articuated application in an elementary context represents a really precious didactic hint. Thanks a Lot ;)
