Extreme Trig Question
ฝัง
- เผยแพร่เมื่อ 30 พ.ค. 2020
- Addendum notice that k=sin(x)+cos(x)=sqrt(2)*sin(x+pi/4) and we want k to be -sqrt(2)+1.
So sin(x+pi/4)=1/sqrt(2)-1, which is about -0.3, then x+pi/4=arcsin(1/sqrt(2)-1), thus x= -pi/4 +arcsin(1/sqrt(2)-1)
** At 11:42, I showed that those two functions have the same minimum by showing you guys graphs. However, I should had done it more careful by finding the x that would give you such k. Thanks for the viewers who pointed this out.
*** Example: (e^x)^2+4(e^x)+4 and k^2+4k+4 do NOT have the same min. The first one has the min value of 4 "at neg inf" and the second has the min value of 0 at -2.The problem is that e^x cannot be -2 in reals so the first function doesn't achieve the same min. as the second.
**** sin^-1(x)+cos^-1(x)+tan^-1(x)+cot^-1(x)+sec^-1(x)+csc^-1(x) is actually MUCH easier.
Because the function is just the constant 3π/2****
Find the minimum of abs(sin(x)+cos(x)+tan(x)+csc(x)+sec(x)+cot(x)), Putnam exam 2003, problem A3.
If you enjoy my videos, then you can click here to subscribe th-cam.com/users/blackpenredpe...
Shop math t-shirts: teespring.com/stores/blackpen...
Support Math: / blackpenredpen
Follow Math: / blackpenredpen
blackpenredpen
![[Full Episode] MasterChef Junior Thailand มาสเตอร์เชฟ จูเนียร์ ประเทศไทย Season 3 Episode 5](http://i.ytimg.com/vi/g5wpgeAIzMs/mqdefault.jpg)
![[4K직캠] BABYMONSTER(베이비몬스터) - FOREVER](http://i.ytimg.com/vi/PS3UAFto1z8/mqdefault.jpg)
*Addendum* notice that k=sin(x)+cos(x)=sqrt(2)*sin(x+pi/4) and we want k to be -sqrt(2)+1.
So sin(x+pi/4)=1/sqrt(2)-1, which is about -0.3, then x+pi/4=arcsin(1/sqrt(2)-1), thus x= -pi/4 +arcsin(1/sqrt(2)-1)
** At 11:42, I showed that those two functions have the same minimum by showing you guys graphs. However, I should had done it more careful by finding the x that would give you such k. Thanks for the viewers who pointed this out.
*** Example: (e^x)^2+4(e^x)+4 and k^2+4k+4 do NOT have the same min. The first one has the min value of 4 "at neg inf" and the second has the min value of 0 at -2.The problem is that e^x cannot be -2 in reals so the first function doesn't achieve the same min. as the second.
**** sin^-1(x)+cos^-1(x)+tan^-1(x)+cot^-1(x)+sec^-1(x)+csc^-1(x) is actually MUCH easier.
Because the function is just the constant 3π/2****
Why can't we use the face that Arithmetic Mean >= Geometric Mean just at the start with all the trig functions?
I have to ask you a question how can i contact with you
@@bharatsethia9243 The equality conditions for AM >= GM inequalities are not the same for all trig functions. For example, if we use sinx + cscx >= 2sqrt(sinx cscx), the equality only holds at sinx = cscx, which is (sinx)^2 = 1. However, in cosx + secx >= 2sqrt(cosx secx), the equality only holds at cosx = secx, which is (cosx)^2 = 1.
But if you want to find the value of X that minimizes the function, shouldn't you take the X derivative? (dy/dx). In that case you should have used the chain rule since you had y=f(k(x)) so dy/dx=dy/dk*dk/dx.
@@adb012 Yes, this is why it's a bad idea to write "k" instead of k(x). For this function he got lucky because the other critical points end up leading to a greater local minimum than what he found (for x = π/4) and a local maximum (x = 5π/4). Also for this function using the chain rule does not actually make the algebra significantly harder.
EDIT: For example this would come into play if the equation we were looking at was |k(x) + 6/(k(x) - 1)|. In this case differentiating w.r.t. k leads only to k(x) values that are not attainable with k(x) = sin(x) + cos(x). But the equation *does* have a minimum value (at x = 5π/4)
Sqrt(2)+1 cannot be solution though at all, because it is outside of domain of k which is between -sqrt(2) and sqrt(2)
Agree. Sqrt(2)*sin(x+pi()/4) has range of -sqrt(2) and sqrt(2). So, sqrt(2)+1 cannot be a solution
This is the ultimate test of your trig skills
This is good, but too far from the Ultimate skills.
And I ultimately failed it.
Not that hard tho
no,ive done a lot harder questipns than this.
For those of you who might did i do this question ?yes i did and with a different method.
@@manik1477 I found the minimum value if there was not any modulus, which occurs at angle π/2.4083
Whoa, extreme algebra!!! This is so neat!
I was too lazy tho. I didn't find the x that will give u the k thought this was not really completed.
Haha, it’s the analysis way; proving that it exists without finding it 😝
@@drpeyam hahaha
16:19 This made me sad..
Mathematician Physicist lol
00:25 _Pause the video and tell us how you'd do it._
I'd think about what the graphs probably looked like, then I'd take a stab at a half root two. Then I'd watch the video and be wrong.
@@samarth6312 I tried taking sin,csc and cos,sec and tan,cot together and tried with graph but got 2 as the answer sadly
When I go to Desmos and type the function, it seems to have an asymptote at 2 √2 -1. I’m confused, can someone help me?
Oh sorry, I forgot the absolute value
@@samarth6312 taking x = -π/4 would cancel sin x and cos x and Similarly for csc x and sec x. So answer would be tan x + cot x hence 2. But obviously this answer is wrong
What a mind-blowing question! Thanks always
Constantly changing markers with 1 hand also adjusting camera, drinking some juice and solving trig problem at the same time is definitely talent.
It definitely is
Did anyone else feel a little bit of joy to see 3blue1brown at the top of the sponsors list?
Thanks for the video!! Nice mic btw...
Nice question and thanks for such an awesome approach....
Hello sir 🙂your videos are really helpful! Love from India
Hello bprp! Can you please make a video on negative combinatorics. It would be really helpful.
Love yr videos man
But shouldn't you show that K can actually take the values you found assuming x is real (not stated, but you didn't use Z as a variable)? Given K = sqrt(2)sin(x+phi), we know -sqrt(2)
Yes you are correct. I only showed the graph at 11:42 but I should had also showed the algebra at the end. Thanks for pointing it out!
Find perimeter of an ellipse
@@blackpenredpen has anyone checked out my channel for the solution of the riemman hypothesis? Idk if you had.
@@olympiad3830 Matt Parker did a video on that, and it looks like there are no formulae for perimeter of an ellipse
@@Kokurorokuko an infinite sum
Very cool Problem, thanks for showing.
Awesome solution! I wouldn't have gotten a grip on that problem ever.
However, when it comes down to testing which of the two solutions for the final quadratic equation would be the real minimum, you don't have to plug in both values into the k-function. Just remember that your k is of the form k= sqrt(2)*sin(a), where a is some kind of angle. Dividing by sqrt(2) gives you sin(a)=1/sqrt(2) +- 1. However, 1/sqrt(2) + 1 is greater than 1, which is off the domain of the sine-function, while the solution with the negative sign in inside the sine-domain. Therefore, only the k-solution with the negative sign can be a valid solution of the original equation, at least in the real world.
I have a request, if you take requests. I was playing around with desmos and I found that x/ (x!(-x)! ) = sin(pi * x)/pi. I was wondering if you could make a video about this because... idk, it's interesting and doesn't make sense why it works (at least to me). I've subbed and I really enjoy your videos.
This is essentially the reciprocal of Euler’s reflection formula (en.m.wikipedia.org/wiki/Reflection_formula mentions it), but you use factorials instead of the analytic continuation, the gamma function. Also, you didn’t cancel the (x) with the (x!). You would end up with 1/(x-1)!(-x)!=sin(pi*x)/pi. Taking reciprocals, you would get (x-1)!(-x)!=pi/sin(pi*x). Using the definition of the gamma function (Γ(n)=(n-1)! where factorials are defined and a particular integral where factorials are undefined (like negatives and between positive integers)), we get Γ(x)Γ(1-x)=pi/sin(pi*x), the typical form of the Euler reflection formula.
Also, greetings from one Jesse to another!
Pls do the same, but add the hyperbolic and inverses.
That's too much! lol
@@blackpenredpen you did 10h math videos but this is too much?! xD
do it
@@charliebaker1427 are you Palpatine?
how hyper did you from 1 comment to post this
¡Give us some exercises like this pls!
Every time it surprises me that 3Blue1Brown is a patron
¡¡Que ingeniosa resolución. Bravo, buen video!!
En verdad muy ingeniosa, pensé que buscaría el valor de x, pero eso no era lo que pedía el ejercicio. Lo que pedía era el valor mínimo de toda la expresión y me gustó la forma de resolverlo
Wait! It is impossible to have k=sqrt(2)+1! As sin x+cos x
the solution is still valid but nice catch!
Maybe x is not real
Correct. It's sloppy math.
@@samarth6312 If x is complex then it can reach zero
You don't know the domain of X, could be complex as well which will give crazy results
Wow I like the amount of identities, methods, and some calculus just to find the answer.
Please talk about the definite integral of (1+(cosx)^2)^(1/2)dx from 0 to pi, which is the length of the sine curve from 0 to pi. I'm curious about it but I was not able to find out the answer. Thank you so much!
very cool video
Can we also find the range of sum of all inverse trignometric functions?
Really enjoyed this vid
LOVE this one!
No intro addicts me more than;:-
Hello lets do some math for fun
12:47 wouldn’t it be better to find the min of f^2 (same location as min of |f| and then just square root) in case the absolute value makes a new extremum?
I think before you assume that one of them is giving the minimum number, you have to calculate the limits of this function on the extremities and the boundaries of the domain to prove that these are gonna minimum and local minimum values
"No Sponsor" - Drinks obviously energy drink with a branding no one ever seen before.
Nice vid ! I just have a question that may seem stupid, why have you put absolute value to f(k) ?
Thank you sir you are the best 😊
Eccellent Work!
why ur integral battle videos are not there in the playlist section??
this is how a man learn math in just 17 min .
First time I see a trigonometric problem without pi in the answer
0:19 I tried it, it did work in the end but it was a _huge_ pain, I have like 10 pages of just simplifications going on. Eventually I found the min of 2√(2) - 1 at x = arcsin(1/2 * [1 - √(2) - √(2√[2] - 1)]).
who does the outro musix for the video and is there a copy of it somewhere?
@blackpenredpen Note: the sum of the trig functions is equal to the sum of their reciprocals.
Please have a look on this one minimum value of y=18 secant square x + 8 cos square x
When you go to a video for fun, and discover trig identities making you wonder wth were you not taught this in school?
Can you solve the MIT OCW 18.01SC exams?
Hi blackpenredpen. Please show me how to integrate W(x) dx. Thanks.
Well JEE Advance questions are much more extreme than this easy question, u should do that..
I ended up with essentially similar approach. Though I'd phrase it a bit more carefully? This function is not continuous after all.
So all the stuff about finding A and phi is just a waste of time, the nice thing is you don't need to "simplify" more than k(x) = sin(x) + cos(x).
I chose to write the original equation as |f(x)|, meaning let's ignore the absolute value for now. As you noted we later need to check for f(x) = 0 cases.
Writing k(x) = sin(x) + cos(x) we can write f(x) = k(x) + 2/(k(x) - 1). It's good to double-check that this equation has discontinuities at the same points as the original equation (it must if we did our algebra correctly; anyway that is at k(x) = 1 or whenever sin(x) = 0 or cos(x) = 0).
We can also conclude here that f(x) = 0 has no solution by expanding that into (k(x) - 1/2)^2 + 7/4 = 0. If it *did* have solutions those would be the answer.
Differentiating w.r.t. k here is wrong. We should differentiate w.r.t. x to find critical points of f(x).
So we get f'(x) = k'(x) - 2k'(x)/(k(x)-1)^2 = k'(x)[1 - 2/(k(x)-1)^2]. So in addition to the point you found, we also need to examine k'(x) = 0.
k'(x) is just cos(x) - sin(x) and k'(x) = 0 happens at x = π/4, 5π/4. These lead to k(x) = ±√2 giving f(x) = 2 ± 3√2
The 2nd factor is 0 for k(x) = 1 ± √2 . Just above we already found the extreme values for k(x) are ±√2, and noting that k(x) is continuous, we can reject k(x) = 1 + √2 as impossible and also affirm that there is a solution for k(x) = 1 - √2 (noting that this number is inside the range). This value leads to f(x) = 1 - 2√2.
So in the end we need to compare absolute values of 2 + 3√2, 2 - 3√2, and 1 - 2√2 to find the answer to the original problem. It's easy to reject the first as largest, and also easy to look at the difference between the other two to conclude that 1 - 2√2 is the one with smallest absolute value, giving the solution as 2√2 - 1
th-cam.com/video/N6T4C8CJXxs/w-d-xo.html
Best jee integration techniques
In our school admission test for science we were asked to prove that whether this equation was equal to four or not. We had just 1/3 minute per question.
I like this this guy say "Well"😁😁
You not mentioned that A is not zero so how you can cancel out for cot(¥)
Plz tell the value of x for which the min value occurs
this is what happens when some random math guy drink and derive a question but end up doing this instead
🎶🙂 "Hello, let's do some maths for fun!" 🙃🎶
the answer is just positive cause there is a absolute value
I'm unsure about the first move. Can anyone show that there exists constant A and phi such that A*sin(x+phi) = sin(x) + cos(x) for all x in Real? If so, wouldn't that lead to a general trig identity with A and phi as known amplitude and phase shift? Or is the claim being made the weaker one that for every Real x there exists A and phi such that A*sin(x+phi) = sin(x) + cos(x)? I think I agree with the latter, but still don't know how to prove it generally.
K is between -sqrt(2) and sqrt(2)
So no need to calculate f(1+sqrt(2)).
Can you make a video on Euler's identity?
If e^(i*pi) = (-1), and suppose we square both sides and take a log, it implies that somehow i has to take a value of 0. Where am I making a mistake?
a^x = 1 has several solutions I think. Especially complex exponents are much more complicated.
If I restrict mysezlf to real numbers then, (a,x) = (1,x) , (a,0) , (-1,2k) for some integer k are all possible solutions I think. But over complex numbers, there are more solutions. Since you are solving for x, then you are looking the unit circle.
16:19 are we in the 2 is bigger than 3 territory again, are we? :D
But for the inverse trig functions it is easy.
If arcsin(x) + arccos(x) = arcsec(x) + arccsc(x) = arctan(x) + arccot(x) =π/2, then, with the right domain, it's exactly equal to 3π/2 always, or else it is undefined, like at x=0 or x=±i I think.
Just an additional information: This question appeared in IMO Hong Kong Preliminary Selection Contest 2007 Question 25.
hkage.org.hk/en/download/Student/IMO/Prelim07_Q.pdf
Me: this looks very hard
Also me: the board is suspiciously small
How triangle behave when we do integration or differrenciated 🤗
You cannot use the AM-GM inequality because the AM-GM inequality requires that every summand be nonnegative to be true, and I can find counterexamples to the inequality if I use negative quantities. For example, the arithmetic mean of the set {-1, -2, -3, -4} is equal to (-1 + -2 + -3 + -4)/4 = -10/4 = -5/2, while the geometric mean of the same set is equal to [(-1)(-2)(-3)(-4)]^(1/4) = 24^(1/4). However, -5/2 < 0 and 0 < 24^(1/4), hence -5/2 < 24^(1/4), contradicting the AM-GM inequality. Hence, the inequality can only hold if the summands are nonnegative.
Why is this important? Because the trigonometric functions can all take on negative values, hence they are *not* nonnegative summands. Therefore, the inequality does not hold. In fact, if you even try to use the inequality at all, which would be incorrect, you would obtain a minimum value of 6, which is obviously false, since this video already proved the minimum value is 1 - sqrt(2).
So, the A and φ were never used, he could have started defining k=sinx+cosx. Also need to deal with k=-1 when dividing by k+1. This happens at π and 3π/2 where either secx or cscx are ±∞, so there's not a minimum there.
Thanks bro
My approach to this would be to imagine a triangle of sides "q", "sqrt(1 - q^2)", and "1". Then replace the trig functions with ratios, make a common denominator, look for zeros in the numerator, and otherwise look for when the derivative goes to zero. But I need a nap so I'll let someone else tackle it.
Why Am greater than Gm will not do ???
I mean, you showed that it is a locally minimum value. But it really didn't show if it is an extreme minimum or there's no minimum because it goes to negative infinity.... or am i missing something?
Why not 2nd derivative for maxima and minima
hey my boy Ryan
This guy puts the black T shirt to hide the black wire
Black pen and red pen...best friends.
NICELY EQUAL
K is a function of x. So, k(x), not k. When you take a derivative respect to x, should you use the chain rule?
See that Christmas three here:
twitter.com/blackpenredpen/status/1267186696136736768/photo/1
Can we done this by AM GM inequality?
No, it cannot
I'm pretty sure differentiation is faster
Plss do integration of
(1+x^2)√(1+x+x^2)
This problem came in Putnam when I was just less than a year old 👶
16:48 the legendary mic
Now hiperbolic functions!
a small question here, can i just let sinx + cosx =k, without doing it as Asin(x+d)? for my level i dont actually understand why we need to do that step...
You are actually correct.
I wanted to do that so that I could solve for x that would give you the k. But I ended up not doing that in the video. : (
I think I will redo this video sometime soon since it's really not that good.
@@blackpenredpen oh ok thanks a lot
@@blackpenredpen ... That would be great. I feel always highly satisfied with your videos. This is "the exception that confirms the rule". And since k is a function of x, don;t forget to use the chain rule when taking the derivative, since you are looking for the x (not of k) that minimizes the original function.
adb012 *since you are looking for the x that minimizes the function*
No, we're not, we're simply looking for the value f takes on after being minimized. Read the problem again.
Wait,I got notification nowwww, after 1 week..how??
TH-cam doing it's thing again
Cool!!!!!
If this comes in my jee mains...thank you so much
Why couldn't we differentiate the original equation? Because of the abs?
It would be hard setting the derivative to 0, as it contains many trig functions
Use A.M greater than equal to G.M.
The cosec sec and cot can be written as 1/sin 1/cos 1/tan
Then (sin+cos+tan+csc+sec+cot)/6 is greater than equal to 1^1/6.... (since all trigo terms will cancel out)
Which gives that whole term is greater than equal to 6
Hence minimum value is 6
Hey @BlackpenRedpen why can't we do it like this....?
Because the minimum value is not 6.
@@Quantris i have the proof
The problem is that the AM-GM inequality only holds for elements of a set that are non-negative. However, each of the trig functions can take on negative values, since for example, cos(π) = -1 < 0. As such, the inequality does not hold.
Lovely
Why we r getting wrong by am-gm inequality
11:43, the second graph is for k = x, not k = √2sin(x + π/4). the minimum y value when k = x is 2√2 - 1 at x= 1 - √2= - 0.41, but when k = the sin function, the minimum of y is 2√2 - 1 at x = - 1.08 + 2πn and x= 2.65 + 2πn, and both of these are ≠ -0.41. it was just a coincidence that they had the same min y values in this case lol
It’s not a coincidence.
@@stephenbeck7222 it is because x isn't √2 sin(x+π/4).
Why do you differentiate wrt k and not x ? Is it always guaranteed to have the same minimum value?
Yes I believe so.
@@blackpenredpen i suppose max/min values are preserved through change of variables? and the only change is the value of the input where the max/min occurs?
@@nuklearboysymbiote
Actually after a careful thought, I need to take it back.
Example: (e^x)^2+4(e^x)+4 and k^2+4k+4 do NOT have the same min. The first one has the min value of 4 "at neg inf" and the second has the min value of 0 at -2. The problem is e^x can NEVER be -2 thus the first one cannot achieve the same min value.
@@blackpenredpen oh god. so to complete the proof for the min value I guess u just need to prove that the required k can be attained with x?
@@nuklearboysymbiote Yes. I just added that in the description now. Thanks and sorry about my first response.
what is min value?
Why did u use calculus??
can someone help me with a question?
- p,r are positive integers and s is a prime number.
1/p + 1r = 1/s
i need the sum of every possible p in terms of s .
What? Did you mean 1/p + 1/r = 1/s? What does "the sum over all possible p" mean?
the only feasible values of p and r such that 1/p + 1/r =1/ s for any fixed prime no. s is
p=r=1/(2s)
3:57 I tried in a calculator sqrt(2) * sin(pi/4) and I get 0.019385... I don't get a 1 as it says on the board, which means that theres something wrong over there
You calculator calculate in degrees, not in radians
At X=π/4 , because final answer must have symmetry.
Sorry for the question if it is very basic: Where does it go sinx+cosx=Asin(x+phi)?? Thanks and sorry
Harmonic addition theorem
I should have used to do solve for x
@@gregorsamsa9762 Thank you very much. Greetings from Spain
@@blackpenredpen Thanks, and greetings from Spain
I thought of using the average inequality:
Sinx+cosx+tanx+cscx+secx+cotx/6> or = to (sinx.cosx.tanx.cscx.secx.cotx)^1/6. Therefore sinx+cosx+tanx+cscx+secx+cotx > or = to 6, the minimum must be the equal sign witch means that the minimum is 6. I'm right? Can someone check this?
AM-GM requires non-negative values but here we can have negative values for cos, tan etc. and the absolute value is also coming into play. Because of that we can have values less than 6 (the solution is actually less than 2).
Also even if this inequality did apply, it doesn't prove that the bound is tight.
Love From India ❤️
I see ninja Melk
Can't we do this by graph????
You can but the answer would display as a decimal approximation. This process gives u the actual min. value