*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****
@@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)
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 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
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)
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!
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
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?
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)]).
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
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?
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.
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.
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
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
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.
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....?
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.
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.
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!
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
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).
K has range of [-sqrt(2),sqrt(2)], so we do not need to check the value of sqrt(2)+1, and we need to check both boundary. This will not change the final answer, but if we want to find the maximum value, we will see the difference.
@@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.
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.
I don't understand why this method not working. If we multiple all trig functions we get 1. By AM GM -> (sum)/6 >= (1)^(1/6) So min value is 6. This is not right but why?
How Chong Actually, that has nothing to do with it: if all the functions were to be nonnegative, then the sum of the greatest lower bounds of each function would be equal to the greatest lower bound of the sum of the functions. The issue is that the functions are not nonnegative.
@@angelmendez-rivera351 you are right, however, the minimum of each of the trig functions cannot be reached simultaneously so regardless it might not attempt the lowest value
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 .
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...
@@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.
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.
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
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.
I mean that you checked the root2 +1 value but since k =root2*sinx -root2 is less then equal to k and root2 is greater than equal to k so you could have just neglected the root2 +1 value for k
@@afc6994 No. The inequality will be true (the value is indeed >= 6 on the first quadrant) but it never has a value of 6. AM-GM is only equal when all terms are equal but sin x = cos x = tan x is never true. The minimum value on the first quadrant happens for x = π/4 giving value 2+3√2 (approx 6.24)
@@Quantris U look to be a well versed guy in thus field , so how could you ignore the fact that '> or =' sign contains an "or" in between so it does not mean tht it should be equal always.
*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
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
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
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
Hello sir 🙂your videos are really helpful! Love from India
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
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!
See that Christmas three here:
twitter.com/blackpenredpen/status/1267186696136736768/photo/1
Did anyone else feel a little bit of joy to see 3blue1brown at the top of the sponsors list?
Hello bprp! Can you please make a video on negative combinatorics. It would be really helpful.
Thanks for the video!! Nice mic btw...
No intro addicts me more than;:-
Hello lets do some math for fun
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
why ur integral battle videos are not there in the playlist section??
What a mind-blowing question! Thanks always
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?
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)]).
@blackpenredpen Note: the sum of the trig functions is equal to the sum of their reciprocals.
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
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?
Wow I like the amount of identities, methods, and some calculus just to find the answer.
"No Sponsor" - Drinks obviously energy drink with a branding no one ever seen before.
¡Give us some exercises like this pls!
Nice vid ! I just have a question that may seem stupid, why have you put absolute value to f(k) ?
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.
K is between -sqrt(2) and sqrt(2)
So no need to calculate f(1+sqrt(2)).
Hi blackpenredpen. Please show me how to integrate W(x) dx. Thanks.
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.
First time I see a trigonometric problem without pi in the answer
Every time it surprises me that 3Blue1Brown is a patron
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
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).
very cool video
Can we also find the range of sum of all inverse trignometric functions?
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.
You not mentioned that A is not zero so how you can cancel out for cot(¥)
16:19 are we in the 2 is bigger than 3 territory again, are we? :D
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.
Nice question and thanks for such an awesome approach....
who does the outro musix for the video and is there a copy of it somewhere?
4:46 for a quick Michael Penn homage :)
Hmmm, I was expecting bprp to snap his fingers to clean up the whiteboard.
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
this is how a man learn math in just 17 min .
I'm pretty sure differentiation is faster
Why Am greater than Gm will not do ???
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.
Can you solve the MIT OCW 18.01SC exams?
Plz tell the value of x for which the min value occurs
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!
Can we done this by AM GM inequality?
No, it cannot
the answer is just positive cause there is a absolute value
¡¡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
Why not 2nd derivative for maxima and minima
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).
Me: this looks very hard
Also me: the board is suspiciously small
K has range of [-sqrt(2),sqrt(2)], so we do not need to check the value of sqrt(2)+1, and we need to check both boundary. This will not change the final answer, but if we want to find the maximum value, we will see the difference.
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.
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.
Love yr videos man
I don't understand why this method not working.
If we multiple all trig functions we get 1. By AM GM -> (sum)/6 >= (1)^(1/6)
So min value is 6. This is not right but why?
Because AM GM is only valid for non-negative real numbers, and Trig functions can be negative. I'm not 100% sure, but i think this is the problem.
Equality must hold when all the trig functions have the same value in order for it to work
HydroStrike HD Yes, you are 100% correct. The reason the inequality is invalid is because it only holds for nonnegative terms.
How Chong Actually, that has nothing to do with it: if all the functions were to be nonnegative, then the sum of the greatest lower bounds of each function would be equal to the greatest lower bound of the sum of the functions. The issue is that the functions are not nonnegative.
@@angelmendez-rivera351 you are right, however, the minimum of each of the trig functions cannot be reached simultaneously so regardless it might not attempt the lowest value
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)
so roughly 1.8?
Yea
Since k = sin x + cos x, k must always be within -sqrt(2) and +sqrt(2), so k = sqrt(2)+1 is not a valid answer.
At X=π/4 , because final answer must have symmetry.
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.
Why did u use calculus??
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.
Please have a look on this one minimum value of y=18 secant square x + 8 cos square x
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?
Well JEE Advance questions are much more extreme than this easy question, u should do that..
Very cool Problem, thanks for showing.
put in the inverse trig functions
Actually I just realized that the sum of all the inverse trig functions will be just 3pi/2. So it would actually have been easy! lol
@@blackpenredpen nice
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
🎶🙂 "Hello, let's do some maths for fun!" 🙃🎶
How triangle behave when we do integration or differrenciated 🤗
Why we r getting wrong by am-gm inequality
When you go to a video for fun, and discover trig identities making you wonder wth were you not taught this in school?
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
this is what happens when some random math guy drink and derive a question but end up doing this instead
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.
what is min value?
Can we apply AM>or=to GM
No, because the trig functions are not nonnegative.
I like this this guy say "Well"😁😁
But isn't k between -root2 and +root2 so why we did not consider it
What do you mean?
I mean that you checked the root2 +1 value but since k =root2*sinx -root2 is less then equal to k and root2 is greater than equal to k so you could have just neglected the root2 +1 value for k
Really enjoyed this vid
16:48 the legendary mic
why is 2/sqrt(2)=sqrt(2)
Black pen and red pen...best friends.
Wait,I got notification nowwww, after 1 week..how??
TH-cam doing it's thing again
It took him 16:54 sec to finish the drink.🤣
Thank you sir you are the best 😊
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
If 0 or = to GM
yaa but that way the answer comes up to be 6...
@@prabalbaishya6179 wont tht be correct??
@@afc6994 No. The inequality will be true (the value is indeed >= 6 on the first quadrant) but it never has a value of 6. AM-GM is only equal when all terms are equal but sin x = cos x = tan x is never true.
The minimum value on the first quadrant happens for x = π/4 giving value 2+3√2 (approx 6.24)
@@Quantris U look to be a well versed guy in thus field , so how could you ignore the fact that
'> or =' sign contains an "or" in between so it does not mean tht it should be equal always.
@@Quantris also u answered it being 6.24 so isnt tht greater than 6😒
Now hiperbolic functions!
I think you should do more to prove that these are actually absolute maximum/minimum and not just local maximum/minimum.
I think so too
should know all the trig identities, 2 sin x cos x is cos 2x so the denominator is 1/2 cos 2x or 2 sec 2x.
LOVE this one!