Pretty easy, in fact. Just set x+2*π*m = π*x, where m is an integer (and π*m refers nicely to our friend Dr. Peyam) and solve for x. However, the solutions x=(2*π/(π-))*m are equidistant, which came as a surprise to me. (Okay, I missed the 2nd case!)
sin(pi*x)-sin(x)=0, then use: sin(A)-sin(B) = 2cos( [A+B]/2)sin([A-B]/2), from calculations the answers are: x= [pi(2k+1)]/[pi+1] and x = [pi(2k)]/[pi-1], for k being part of Z
Very nice problem and a cool video! I would just like to note that since we assume that x is a real number, n cannot be 0, or -1, because we have (2n+1)^2 π^2 - 16 under the square root, which is negative in these cases.
Great equation. My first instinct was to write both tangents as sin/cos, rearrange, and use compound angle formula for cosine. Your method looks neater, but you have to be careful with the tan(a)=tan(b), whereas I only needed to deal with cos(a)=0.
I got to the quadratic a little differently, and I think it's pretty cool. If you multiply through by tan(x), you end up with the equation tan(1/x)*tan(x) = 1. What THAT means is that tan(x + 1/x) must be undefined, due to the denominator of the tan compound angle identity. Therefore x+ 1/x = pi/2 +n*pi, as this is the set of values for which tan is undefined.
and then you've got so many people on the internet who argue that anything other than cos is useless. Some unironically, most ironically (by a huge margin, but the fact that there are any people who argue that in seriousness is like WHAT)
sin(1/x) = 1/sin(x) : If we're looking for real solutions only, then we can use the fact that |sin(x)| = 1, so the only way sin(1/x) could equal 1/sin(x) is if both were equal to +/- 1. But the solutions to 1/sin(x) = +/- 1 are x = n*π and the solutions to sin(1/x) = +/- 1 are x = 1/(n*π). They have no solutions in common, so sin(1/x) = 1/sin(x) has no solution. If we're looking for complex solutions, wolfram alpha gives 4 solutions, x = +/- 0.719 +/- 0.695 i. I noticed these solutions have |x| = 1 so we can set x = exp(i*t) and solve sin(exp(-i*t)) = 1/sin(exp(i*t)). This can be rewritten as cosh(2 sin t) - cos(2 cos t) = 2, which means I don't think there's any hope of expressing the solutions in closed form.
1:11 I'm always telling the students I tutor what "co" means because somehow teachers and textbooks don't often make it clear enough. And students often forget the co-function identities to boot!
Thank you for explaining the meaning of the cotx now i don't have to mug up the identity i understand the meaning why we take cot x = tan(90-x) , i have studied that a complementary angle is when two angles add up to 90* ,but my teacher never explained me the meaning of cot x is and how to connect the two concepts
I noticed a thing with a similar equation, y = x tan(pi/x), where the limit as x approaches infinity is just pi, although actually just whatever constant you replace pi with works too. doesn't seem to special, but you can also use it to solve area and perimeter of any regular polygon. All you have to do is take the equations A = pi r^2 and C = 2 pi r and replace a few values. First replace the radius with the distance from the center of the regular polygon to the midpoint of a side. Then take the number of sides, replace x in y = x tan(pi/x) with it, and solve for y. replace pi with the value of y and you are done! representing the area and perimeter of regular polygons this way isn't too useful, except it maintains the property of the circle's equations where the perimeter is the derivative of the area. another interesting effect of the 1/x in the tangent, is as x approaches infinity, you are basically taking tangent of infinitely small values, therefore, infinitely close to tangent at x=0. However at x=0, y = tan(x) has a slope of 1, meaning it basically acts like y = x when you get infinitely small. That means the function basically has it's own limit, where as x goes to infinity, y = tan(1/x) approaches the function y = 1/x completely cancelling tangent out.
Here's my take on evaluating the limit. Not as elegant as your solutions but it gets the job done. lim(x->+inf) x*tan(a/x) = lim(x->+inf) (ax/a)*tan(a/x) Let u = a/x //WLOG, assume positive a lim(x->+inf) u = 0 lim(u->0+) (a/u)*tan(u) //since x is positive u is always positive = a*lim(u->0+) tan(u)/u SQUEEZE THEOREM: For positive u: tan(u) > u > sin(u) 1/tan(u) < 1/u < 1/sin(u) tan(u)/tan(u) < tan(u)/u < tan(u)/sin(u) 1 < tan(u)/u < sec(u) lim(u->0+) 1 = 1 lim(u->0+) sec(u) = sec(0) = 1 THEREFORE: lim(u->0+) tan(u)/u = 1 lim(u->0+) a*tan(u)/u = a lim(x->+inf) x*tan(a/x) = a
@@zengakukatsu also the derivative logic in the second half is actually l'hospital I think, not quite sure that works out considering lim(x->0) sin(x)/x does not work with l'hospital due to circular reasoning and all that (bprp has a video on that)
If you write it as tan(1/x)-1/tan(x)=0 and then plot it you get a very, very… strange curve. It shoots off to infinity/minus infinity often at the beginning and then settles to almost a periodicity passing through the zero line at 4.4895 radians (257.23°) and then again at almost but not quite regular intervals for ever more. Weird!
Eulers formula gives Sinx =[ e^ix-e^(-ix)]/2i And Cosx=[e^ix+e^(-ix)]/2 Calculate tanx And then set 1/tanx=tan1/x And solve for x And you will get what I got
1/tanx = tan(1/x) cosx/sinx=sin(1/x)/cos(1/x) cos(x)cos(1/x)=sin(x)sin(1/x) Using Factor Formulae, we get: cos(x - 1/x) + cos(x + 1/x) = cos (x - 1/x) - cos (x + 1/x) Then, 2cos(x + 1/x) = 0 x + 1/x = (x^2 + 1)/x = (2n+1)π/2, where n is an integer x^2 + 1 = (2n+1)πx/2 x^2 - [(2n+1)π/2]x + 1 = 0 By solving the quadratic eqn, we get: x = (1/4)([2n+1]π±√[(2n+1)^2 x π^2 - 16]), which is the same as blackpenredpen's answer
just write tan x = sinx /cosx and tan(1/x) = sin (1/x) /cos (1/x) and cross multiply you will land up with cos(x+1/x)=0 now that gives a solution x + 1/x = 4n-1(pie/2) where n is an interger since tanx is in denominator so we have to remove inegral pie from domain and solving these conditions you will get the answer
What about the system of equations: y = x+1 and y^x = x^y +1? Just interested in finding an algebraic solution that requires no graphing, cuz no online tool could do it. Also, the solutions (real value solutions) to this system of equations are (x,y)=(0,1),(1,2) and (2,3). :D If you could do a video on this, thanks!
I did the exact same thing with inverse trignometric functions. Just didn't added the nPI. Certainly, this version of equation is more correct since there should always be infinite solns for a trignometric equation. This is how I did it: tan(1/x) = cot(x) 1/x = tan(-1)(tan(PI/2-x)) 1/x = PI/2 - x. this would only give solns in range of 0 to 180deg since this is the natural domain of Tan.
Is it possible to find a generalized form for all sin, cos and tan? Is there a geometric interpretation of the solution? Just throwing it out there for the math gods since I spent the last 5 days trying it
In trying the tan problem for myself, I'm getting pi/4+(sqrt(4-(pi^2)/4)/2)*i or pi/4-(sqrt(4-(pi^2)/4)/2)*i. At least when I restrict arccos(0) to pi/2
Another way to get to the main equation is the property that if two angles are such that one's tangent is the other's reciprocal then their sum must be π/2 + kπ.
haven't watched your video for a year, looks like you've done some things new to them. it's time for me to get back and learn some math cuz I've been slacking off for a year. there's almost nothing left inside my head now.😔
Method 2: tan(x)tan(1/x)=1 So we can say tan(x+1/x)=infinity B/w 1-tan(x)tan(1/x) on denominator will be zero In other words x+1/x=(2n+1)pi/2 And then we get the same quadratic
I don't know why my teachers never taught that cot{co tan} thing in the class, but I figured out that myself. cosine of @=sine of complementary angle of @.
Another way to get to the quadratic equation tan(1/x) = 1/tan(x) sin(1/x)/cos(1/x) = cos(x)/sin(x) sin(x)*sin(1/x) = cos(x)*cos(1/x) cos(x)*cos(1/x) - sin(x)*sin(1/x) = 0 Using the identity cos(α+β) = cos(α)cos(β)-sin(α)sin(β) cos(x+1/x) = 0 x+1/x = +-π/2 + 2πn Which can also be written as x+1/x = π/2 + πn
Hello? When you took arctan{tan(1/x)} =arctan{tan(pi/2-x)} You gotta make sure 1/x and pi/2-x both belongs to (-pi/2,pi/2) Upon solving this inequality we get x belongs to null set.. So tan(1/x) is not equal to 1/tan(x) for any value of x belongs to R(Real no.)
I did it like this... tan(x).tan(1/x)=1 1- tan(x).tan(1/x) =0 We know, tan(A+B) = (tanA+ tanB)/ 1- tanAtanB Hence, tan(x+1/x) = undefined, since the denominator is 0 x+1/x = (2n-1)π/2 Then we get the ans...
Solve for X: sqrt(sin(x))/(e^x) = sqrt(sin(x)/(e^x)) 2: From e^(x-e^a) = ln(x), come up with a function/formula to calculate all possible values of a for a given value x. 3: Solve for x, x^(1/e) = ln(x) 4: x = d/dx(sin(x)/pi) 5 **(CAREFULLY MADE)**: integral from 0 to inf of ((e^sin(x))*(cos(x)^a)/(1+sin(x)^2)) dx 6: explain integrals in a minute they're pretty hard
My approach:- tan(1/x)=1/tan(x) => tan(1/x).tan(x)=1 ----> Equation (1) Now, we know:- tan(x+1/x)= [tan(x)+tan(1/x)]/[1-tan(x).tan(1/x)] Now, using Equation (1):- tan(x+1/x)= [tan(x)+tan(1/x)]/0 ---->(Possible only if RHS is tan(pi/2) Then, tan(x+1/x)=tan(pi/2) => x+1/x = n*pi+(pi/2) where, n is an integer Now, solve further for final solution 🙃
sin(1/x) = 1/sin(x) |sin(1/x)| = 1 they could only be equal when sin(1/x) = sin(x) = 1 or sin(1/x) = sin(x) = -1 however when sin(x) = 1 (so x = pi/2 + 2kpi), sin(1/x) is not equal to 1. same thing when sin(x) is negative 1 no solutions in R
I have a question that wolframalpha couldn't solve but I know has solutions. Maybe they aren't nice closed forms idk but here goes: For x in ]0, pi/2[ tan(x)=arctan(2x) solve for x, good luck
Why cannot we do this, 1.First take tan(x+(1/x)) 2. This will give (tan(x)+tan(1/x))/(1-(tan(x))(tan(1/x))) 3. When we substitute tan(1/x)=1/tan(x) in the denominator, we will be getting infinity 4. From the above we can say that x+1/x = 90 5. Solve the quadratic equation to get the answer. Please correct me if you find any mistakes
he got to that final equation in the vid, but you have to account for the fact that there are infinitely many angles for which this equation holds (since angles can be greater than 360), so two answers from the quadratic aren't enough. what he did was sum 180 (pi in radians) multiplied by an integer, and this doesn't alter the value because tg(180+x)= tg(x). So by taking into account the infinite solutions, the final result can only be x expressed in terms of n.
I dont know if I did something wrong but cant you do this; First multiply both sides by tan(x): ___tan(1/x) x tan(x) = 1 convert tan into sin and cos:_________(sin(1/x) x sin(x)) / (cos(1/x) x cos(x)) = 1 times by (cos(1/x) x cos(x)):_________sin(1/x) x sin(x) = cos(1/x) x cos(x) subtract by (sin(1/x) x sin(x)):______(cos(1/x) x cos(x)) - (sin(1/x) x sin(x)) = 0 This is in the form cos(A+B), so:____cos((1/x) + x) = 0 Then, (1/x) + x = arccos(0) (1/x) + x = pi/2 (x^2 +1)/x = pi/2 x^2 + 1= (pi/2)x (x^2) - (pi/2)x +1 = 0 And you have a quadratic, Solving this give you 2 complex numbers.
Since we have tan(x)*tan(1/x)=1 I though of the formulas for tan(a+b) and tan(a-b), for the sum you get tan(x+1/x) = [tan(x)+tan(1/x)]/0 => x+1/x=π/2 like the solution in the video, but for the difference you’d get tan(x-1/x) = [tan(x)-tan(1/x)]/2 which apparently has no solutions but idk why
Why split 1 by infinity variable x, and sin cos or tan it. What is it exactly, because I haven't learned anything with math since 2018, and just trying it in a program like desmos seems graphically pleasing, and if I could really understand coding like I understand desmos it'd be good too, but I'm still stuck on how an abstract concept like trig functions explain curves so well, but without a smoothing of x, it becomes boxy. I know desmos is a program that allows me to see the numbers, and what they're equal to but I sometimes wish I can do a linear progression that shows from the point of selection, another 0 - tau rotation with index that explains how the next bit of the lines movement from the selected point is influenced in its momental path, with a choosable arc length, and curve strength. Basically write loops like a bezier curve, without the bezier curve.
Sir I'm from Bangladesh and in my calculas entrance exam I found a very interesting question which is integral tan(x^e)+tan(e^x)/e^2π-2 limits are from -π/2 to π/4 could you please make a video on this sir pls make a video 😭❤️
I think you need to evaluate the domain. When you add nπ ... there seems to me we would have to add some constraints like n=odd integers to satisfy 1/x ; x>0 I'm amateur, but am i wrong?
hello Mr. just wanna ask when to use certain method for solving a quadratic equation like when to use quadratic formula method or factoring method or completing the square method.
I tend to try factorising when the coefficients are integers and look like they could easily factorise, if they don't look nice or I can't factorise I use the quadratic equation. Completing the square is a nice method but it tends to be tedious and slow to do in practice (also, doing this method with variables for coefficients yields the quadratic equation anyway).
1:23 These meanings are related, though. The complementary angle goes with the original angle to make up 90°, sorry, π/2. So too a complementary breakfast or whatever goes with your hotel room to make up the complete package. Something like that.
Dumb question: If the derivative of arcsin x is *1/sqrt(1-x^2)*, and the derivative of arccos x is *-1/sqrt(1-x^2)*, then what will you obtain if you integrate *1/sqrt(1-x^2)*? Will you get arcsin x or will you get -arccos x?
For f(x) = f(a) where f(x) is a periodic function of period k, sure x = a will give you one solution to your equation. But in fact owing to the periodicity, there are actually infinite solutions to the equation f(x) = f(a) which can be written as x = a + nk (n belongs to integers). In our case f(x) is nothing but tan x and the period k is π.
Solve sin(x)=sin(πx) ua-cam.com/video/-aedH9Pusr4/v-deo.html
Pretty easy, in fact. Just set x+2*π*m = π*x, where m is an integer (and π*m refers nicely to our friend Dr. Peyam) and solve for x. However, the solutions x=(2*π/(π-))*m are equidistant, which came as a surprise to me.
(Okay, I missed the 2nd case!)
I want to send you a problem....
I want to send you a problem... ✍️
sin(pi*x)-sin(x)=0, then use: sin(A)-sin(B) = 2cos( [A+B]/2)sin([A-B]/2), from calculations the answers are: x= [pi(2k+1)]/[pi+1] and x = [pi(2k)]/[pi-1], for k being part of Z
Sir please integration of x^x dx
"I know because it's on my board, it must be true." lol Not sure if my prof would have accepted that particular proof, but loved it. :)
😆
Where I can find that board? 🥺
I literally died when he said that 😂
@@芳龄十八-u1k link perhaps please
Very nice problem and a cool video! I would just like to note that since we assume that x is a real number, n cannot be 0, or -1, because we have (2n+1)^2 π^2 - 16 under the square root, which is negative in these cases.
First, multiply by tanx
tanx * tan(1/x) = 1
then:
(sinx * sin(1/x))/(cosx * cos(1/x)) = 1
multiply by the denominator:
sinx * sin(1/x) = cosx * cos(1/x)
subtract the rhs:
sinx * sin(1/x) - cosx * cos(1/x) = 0
use the sum identity:
-cos(x + 1/x) = 0
therefore:
x + 1/x = pi/2 + -2*k*pi- k*pi
and the same from here.
_edit: corrected 2*k*pi_
Nice
I don't even know tanx*tan(1/x)=1
yes that was my solution too and in my opinion it was even more fun to figure out than using the identity with cotangent :}
actually x + 1/x = π/2 + kπ but otherwise smart solution
@@khoozu7802 It isn’t an identity, but he is cross multiplying the original equation.
Great equation. My first instinct was to write both tangents as sin/cos, rearrange, and use compound angle formula for cosine. Your method looks neater, but you have to be careful with the tan(a)=tan(b), whereas I only needed to deal with cos(a)=0.
Me too
I got to the quadratic a little differently, and I think it's pretty cool. If you multiply through by tan(x), you end up with the equation tan(1/x)*tan(x) = 1. What THAT means is that tan(x + 1/x) must be undefined, due to the denominator of the tan compound angle identity. Therefore x+ 1/x = pi/2 +n*pi, as this is the set of values for which tan is undefined.
Did the exact same thing lol
Same here
bro what
@@Proxuius I am flabbergasted by this as well!
@@geeky_001 lmaooo
The complementary identities for trigo functions are underrated imo. Most of my classmates dont even know the one with sine and cosine
hyperbolic identities are more underrated
the lemniscate elliptic functions are more underrated tho
What even is the use for those functions
@@eunkyungcho3477 idk but Gauss, Legendre, and others studied them so that makes them cool
and then you've got so many people on the internet who argue that anything other than cos is useless. Some unironically, most ironically (by a huge margin, but the fact that there are any people who argue that in seriousness is like WHAT)
I love how any equation where you apply the same function to both the input and main function/operation always has such a good answer and explanation
sin(1/x) = 1/sin(x) : If we're looking for real solutions only, then we can use the fact that |sin(x)| = 1, so the only way sin(1/x) could equal 1/sin(x) is if both were equal to +/- 1. But the solutions to 1/sin(x) = +/- 1 are x = n*π and the solutions to sin(1/x) = +/- 1 are x = 1/(n*π). They have no solutions in common, so sin(1/x) = 1/sin(x) has no solution.
If we're looking for complex solutions, wolfram alpha gives 4 solutions, x = +/- 0.719 +/- 0.695 i. I noticed these solutions have |x| = 1 so we can set x = exp(i*t) and solve sin(exp(-i*t)) = 1/sin(exp(i*t)). This can be rewritten as cosh(2 sin t) - cos(2 cos t) = 2, which means I don't think there's any hope of expressing the solutions in closed form.
I think its nπ/2 not nπ
@@acuriousmind6217 Oops ... you're right it should be (n + 1/2)*π
1:11 I'm always telling the students I tutor what "co" means because somehow teachers and textbooks don't often make it clear enough. And students often forget the co-function identities to boot!
Freshmen's dream, but level 15 septillion.
bro got ∞ social credits after solving that
Idea: prove Σ(n = 1, ∞) of n^(2k) is always 0 for k is an integer. Tip: use ramanujan summation f(0)/2 + Σ f(n) = i ∫ (f(it)-f(-it))/(e^(2πt)-1) dt
Dude ur 10 and can do very advanced calculus tf
@@KirinSD bro it's just a name don't be fooled by name on the internet
@@JasClaren dude look at his videos
Those series diverges.
bro what the.. 10 years old?!!!?!?!?!?!?!
Thank you for explaining the meaning of the cotx now i don't have to mug up the identity i understand the meaning why we take cot x = tan(90-x) , i have studied that a complementary angle is when two angles add up to 90* ,but my teacher never explained me the meaning of cot x is and how to connect the two concepts
I didn't even know you made a video of this. Good work!!! 🤩🤩🤩
4:32 *PI not pi but 0
tauism for the win #tauismftw
@@nutronstar45 what do you mean?
@@SuperYoonHo join the tauism
@@nutronstar45 why and how?
@@nutronstar45 plz explain wht tauism is
I noticed a thing with a similar equation, y = x tan(pi/x), where the limit as x approaches infinity is just pi, although actually just whatever constant you replace pi with works too.
doesn't seem to special, but you can also use it to solve area and perimeter of any regular polygon. All you have to do is take the equations A = pi r^2 and C = 2 pi r and replace a few values. First replace the radius with the distance from the center of the regular polygon to the midpoint of a side. Then take the number of sides, replace x in y = x tan(pi/x) with it, and solve for y. replace pi with the value of y and you are done! representing the area and perimeter of regular polygons this way isn't too useful, except it maintains the property of the circle's equations where the perimeter is the derivative of the area.
another interesting effect of the 1/x in the tangent, is as x approaches infinity, you are basically taking tangent of infinitely small values, therefore, infinitely close to tangent at x=0. However at x=0, y = tan(x) has a slope of 1, meaning it basically acts like y = x when you get infinitely small. That means the function basically has it's own limit, where as x goes to infinity, y = tan(1/x) approaches the function y = 1/x completely cancelling tangent out.
Here's my take on evaluating the limit. Not as elegant as your solutions but it gets the job done.
lim(x->+inf) x*tan(a/x)
= lim(x->+inf) (ax/a)*tan(a/x)
Let u = a/x //WLOG, assume positive a
lim(x->+inf) u = 0
lim(u->0+) (a/u)*tan(u) //since x is positive u is always positive
= a*lim(u->0+) tan(u)/u
SQUEEZE THEOREM:
For positive u:
tan(u) > u > sin(u)
1/tan(u) < 1/u < 1/sin(u)
tan(u)/tan(u) < tan(u)/u < tan(u)/sin(u)
1 < tan(u)/u < sec(u)
lim(u->0+) 1 = 1
lim(u->0+) sec(u) = sec(0) = 1
THEREFORE:
lim(u->0+) tan(u)/u = 1
lim(u->0+) a*tan(u)/u = a
lim(x->+inf) x*tan(a/x) = a
@@nanamacapagal8342 I completely forgot about the squeeze theorem, I have never actually had to use it before.
@@zengakukatsu also the derivative logic in the second half is actually l'hospital I think, not quite sure that works out considering lim(x->0) sin(x)/x does not work with l'hospital due to circular reasoning and all that (bprp has a video on that)
If you write it as tan(1/x)-1/tan(x)=0 and then plot it you get a very, very… strange curve. It shoots off to infinity/minus infinity often at the beginning and then settles to almost a periodicity passing through the zero line at 4.4895 radians (257.23°) and then again at almost but not quite regular intervals for ever more. Weird!
I used the euler's formula to calculate tanx as sinx/cosx. Then I applied simple algebra and I ended up with x=-[π+-sqrt(π^2+16)]/4
Intentionally vague
Seeing as it's simple algebra, I presume the minor details are left to the reader.
tauism for the win #tauismftw
Eulers formula gives
Sinx =[ e^ix-e^(-ix)]/2i
And
Cosx=[e^ix+e^(-ix)]/2
Calculate tanx
And then set 1/tanx=tan1/x
And solve for x
And you will get what I got
@@georget8008 it's just like saying a lot of things but actually saying nothing
1/tanx = tan(1/x)
cosx/sinx=sin(1/x)/cos(1/x)
cos(x)cos(1/x)=sin(x)sin(1/x)
Using Factor Formulae, we get:
cos(x - 1/x) + cos(x + 1/x) = cos (x - 1/x) - cos (x + 1/x)
Then, 2cos(x + 1/x) = 0
x + 1/x = (x^2 + 1)/x = (2n+1)π/2, where n is an integer
x^2 + 1 = (2n+1)πx/2
x^2 - [(2n+1)π/2]x + 1 = 0
By solving the quadratic eqn, we get:
x = (1/4)([2n+1]π±√[(2n+1)^2 x π^2 - 16]), which is the same as blackpenredpen's answer
I did exactly the same way as you do. But I only solved for the principle solutions.
@@professorpoke nice 👍
tauism for the win #tauismftw
just write tan x = sinx /cosx and tan(1/x) = sin (1/x) /cos (1/x) and cross multiply you will land up with cos(x+1/x)=0 now that gives a solution x + 1/x = 4n-1(pie/2) where n is an interger since tanx is in denominator so we have to remove inegral pie from domain and solving these conditions you will get the answer
What about the system of equations: y = x+1 and y^x = x^y +1? Just interested in finding an algebraic solution that requires no graphing, cuz no online tool could do it. Also, the solutions (real value solutions) to this system of equations are (x,y)=(0,1),(1,2) and (2,3). :D If you could do a video on this, thanks!
try integrating (1dx/tanx )
i could have definitely used this in my pre-calc class what!! this was explained so well, thank you
LOVE IT!!! THANKS SIR!
4:30 interesting way to write π
I did the exact same thing with inverse trignometric functions. Just didn't added the nPI. Certainly, this version of equation is more correct since there should always be infinite solns for a trignometric equation. This is how I did it:
tan(1/x) = cot(x)
1/x = tan(-1)(tan(PI/2-x))
1/x = PI/2 - x.
this would only give solns in range of 0 to 180deg since this is the natural domain of Tan.
One question for smatest teacher
Is it possible to find a generalized form for all sin, cos and tan?
Is there a geometric interpretation of the solution?
Just throwing it out there for the math gods since I spent the last 5 days trying it
trigonometry functions can all be generalized in terms of the angle by expanding them in Taylor series, if that is what you are looking for
Can even give you a hint:
e^ix = cos(x) + isin(x), and e^k = 1 + k + k^2/2! + k^3/3! + ...
In trying the tan problem for myself, I'm getting pi/4+(sqrt(4-(pi^2)/4)/2)*i or pi/4-(sqrt(4-(pi^2)/4)/2)*i. At least when I restrict arccos(0) to pi/2
some advanced in-your-head math here. it's not easy to have above and below factors in your head all at once, but it's good to practice. nice video.
Complex answers at n=0. Interesting! I will have to plug them in and see if they work.
4:30 all equal to pi lmao, I love the confidence with which he says it although there’s a 0 there
Even the engineers' π = 3 is closer :)
we can assume solution of x as e^i(theta), so x+1/x is just 2 cos (theta) and solve from there :)
The answer is complex if n=0 or n=-1, right? because then (1+2n)²=1, so we have pi²-16 inside the sqrt, which is negative.
4:32 “We all make mistakes in the heat of passion, Jimbo”
Another way to get to the main equation is the property that if two angles are such that one's tangent is the other's reciprocal then their sum must be π/2 + kπ.
haven't watched your video for a year, looks like you've done some things new to them. it's time for me to get back and learn some math cuz I've been slacking off for a year. there's almost nothing left inside my head now.😔
I just love it when he says “Now this is soo cool, because…”, because it’s so cool.
If you don't add nπ, you get two complex solutions. because in solution, π^2-16 is negative and in square root.
Great! I watched yt just now and saw the premiere!
1:41 we are all enlightened adults now so we will say tau/4 :p
4:33 particularly if pi = 0
tauism for the win #tauismftw
tangent function circles (or cycles?) with a period of pi. So it's actually 1/x+x-pi/2=k*pi k belongs to integer.
Method 2:
tan(x)tan(1/x)=1
So we can say tan(x+1/x)=infinity
B/w 1-tan(x)tan(1/x) on denominator will be zero
In other words x+1/x=(2n+1)pi/2
And then we get the same quadratic
I don't know why my teachers never taught that cot{co tan} thing in the class, but I figured out that myself. cosine of @=sine of complementary angle of @.
Another way to get to the quadratic equation
tan(1/x) = 1/tan(x)
sin(1/x)/cos(1/x) = cos(x)/sin(x)
sin(x)*sin(1/x) = cos(x)*cos(1/x)
cos(x)*cos(1/x) - sin(x)*sin(1/x) = 0
Using the identity
cos(α+β) = cos(α)cos(β)-sin(α)sin(β)
cos(x+1/x) = 0
x+1/x = +-π/2 + 2πn
Which can also be written as
x+1/x = π/2 + πn
About n=0 in \sqrt{(1+2n)^2\pi^ 2-16}? In this case the roots is complex now? Thanks
Hello?
When you took arctan{tan(1/x)}
=arctan{tan(pi/2-x)}
You gotta make sure 1/x and pi/2-x both belongs to (-pi/2,pi/2)
Upon solving this inequality we get x belongs to null set..
So tan(1/x) is not equal to 1/tan(x) for any value of x belongs to R(Real no.)
@@jash21222 because it assumes that it includes the domain,it does it by default, but a cautious man should take every possibility.
@@jash21222 okay I'll solve it again,thanks for clarifying
@@jash21222 and let you know
This looks like the identity
arctan(x)+arctan(1/x)=pi/2
tauism for the win #tauismftw
There is no solutions to the Q1. given in the end..right?
I did it like this...
tan(x).tan(1/x)=1
1- tan(x).tan(1/x) =0
We know, tan(A+B) = (tanA+ tanB)/ 1- tanAtanB
Hence, tan(x+1/x) = undefined, since the denominator is 0
x+1/x = (2n-1)π/2
Then we get the ans...
Solve for X: sqrt(sin(x))/(e^x) = sqrt(sin(x)/(e^x))
2: From e^(x-e^a) = ln(x), come up with a function/formula to calculate all possible values of a for a given value x.
3: Solve for x, x^(1/e) = ln(x)
4: x = d/dx(sin(x)/pi)
5 **(CAREFULLY MADE)**: integral from 0 to inf of ((e^sin(x))*(cos(x)^a)/(1+sin(x)^2)) dx
6: explain integrals in a minute
they're pretty hard
Helo! Is it any possible to take square root of ( (1+2n)pi )^2 - 4^2 using this formula: (a+b)(a-b) = aa - bb ?
My approach:-
tan(1/x)=1/tan(x)
=> tan(1/x).tan(x)=1 ----> Equation (1)
Now, we know:-
tan(x+1/x)= [tan(x)+tan(1/x)]/[1-tan(x).tan(1/x)]
Now, using Equation (1):-
tan(x+1/x)= [tan(x)+tan(1/x)]/0 ---->(Possible only if RHS is tan(pi/2)
Then,
tan(x+1/x)=tan(pi/2)
=> x+1/x = n*pi+(pi/2) where, n is an integer
Now, solve further for final solution 🙃
sin(1/x) = 1/sin(x)
|sin(1/x)| = 1
they could only be equal when sin(1/x) = sin(x) = 1 or sin(1/x) = sin(x) = -1
however when sin(x) = 1 (so x = pi/2 + 2kpi), sin(1/x) is not equal to 1. same thing when sin(x) is negative 1
no solutions in R
Other easy way:
tan(a)tan(b)=0
Implies
cos(a)cos(b)-sin(a)sin(b)=0
Implies
cos(a+b)=0
Hence
a+b= pi/2 mod pi
The vocabulary lesson at the start blew me away
I have a question that wolframalpha couldn't solve but I know has solutions. Maybe they aren't nice closed forms idk but here goes:
For x in ]0, pi/2[
tan(x)=arctan(2x)
solve for x, good luck
Why cannot we do this,
1.First take tan(x+(1/x))
2. This will give (tan(x)+tan(1/x))/(1-(tan(x))(tan(1/x)))
3. When we substitute tan(1/x)=1/tan(x) in the denominator, we will be getting infinity
4. From the above we can say that x+1/x = 90
5. Solve the quadratic equation to get the answer.
Please correct me if you find any mistakes
he got to that final equation in the vid, but you have to account for the fact that there are infinitely many angles for which this equation holds (since angles can be greater than 360), so two answers from the quadratic aren't enough. what he did was sum 180 (pi in radians) multiplied by an integer, and this doesn't alter the value because tg(180+x)= tg(x).
So by taking into account the infinite solutions, the final result can only be x expressed in terms of n.
Thank you for teaching me what CO prefix means. That really helps in many problems!!
Happy to help!
We can put x= cotinverse y in both lhs and rhs and ques easily solved
4:34 "and this equal to pi", its not 0, he draw the circumference of the circle
where can I get the "identities for you" board ?
If x+1/x=pi/2
We know from AM GM Inequality
Minimum value of x+1/x is 2
Here,pi/2 less than 2...
Is this x really a solution here sir?
I dont know if I did something wrong but cant you do this;
First multiply both sides by tan(x): ___tan(1/x) x tan(x) = 1
convert tan into sin and cos:_________(sin(1/x) x sin(x)) / (cos(1/x) x cos(x)) = 1
times by (cos(1/x) x cos(x)):_________sin(1/x) x sin(x) = cos(1/x) x cos(x)
subtract by (sin(1/x) x sin(x)):______(cos(1/x) x cos(x)) - (sin(1/x) x sin(x)) = 0
This is in the form cos(A+B), so:____cos((1/x) + x) = 0
Then,
(1/x) + x = arccos(0)
(1/x) + x = pi/2
(x^2 +1)/x = pi/2
x^2 + 1= (pi/2)x
(x^2) - (pi/2)x +1 = 0 And you have a quadratic,
Solving this give you 2 complex numbers.
@@jash21222 oh i see.
Where can I buy that trigonometric identities board?!?
Since we have tan(x)*tan(1/x)=1 I though of the formulas for tan(a+b) and tan(a-b), for the sum you get tan(x+1/x) = [tan(x)+tan(1/x)]/0 => x+1/x=π/2 like the solution in the video, but for the difference you’d get tan(x-1/x) = [tan(x)-tan(1/x)]/2 which apparently has no solutions but idk why
In real domain
-1=1
so sin(1/x) = 1 is left to check
but 1/sin(2/pi) is not equal to one
so equation sin(1/x)=1/sin(x) has no solutions in real domain
my brain actually imploded when you said that co is compimentary
0:34 typical asian parents when they ask u to solve a question:
Why split 1 by infinity variable x, and sin cos or tan it.
What is it exactly, because I haven't learned anything with math since 2018, and just trying it in a program like desmos seems graphically pleasing, and if I could really understand coding like I understand desmos it'd be good too, but I'm still stuck on how an abstract concept like trig functions explain curves so well, but without a smoothing of x, it becomes boxy.
I know desmos is a program that allows me to see the numbers, and what they're equal to but I sometimes wish I can do a linear progression that shows from the point of selection, another 0 - tau rotation with index that explains how the next bit of the lines movement from the selected point is influenced in its momental path, with a choosable arc length, and curve strength. Basically write loops like a bezier curve, without the bezier curve.
How do i find that board? I need it if some question asks for proof. I will just write it on the board.
Sir I'm from Bangladesh and in my calculas entrance exam I found a very interesting question which is integral tan(x^e)+tan(e^x)/e^2π-2 limits are from -π/2 to π/4 could you please make a video on this sir pls make a video 😭❤️
God I love these videos never stop making them
I think you need to evaluate the domain. When you add nπ ... there seems to me we would have to add some constraints like n=odd integers to satisfy 1/x ; x>0 I'm amateur, but am i wrong?
tauism for the win #tauismftw
The odd number constraint is why you see 2n+1 in the end instead of just n. Everything he did was correct
Oh yes, actually a case that can be described as without loss of generality
n π can be either added or minused
hello Mr.
just wanna ask when to use certain method for solving a quadratic equation like when to use quadratic formula method or factoring method or completing the square method.
These methods are all equal. You can use whatever method you like most.
I tend to try factorising when the coefficients are integers and look like they could easily factorise, if they don't look nice or I can't factorise I use the quadratic equation. Completing the square is a nice method but it tends to be tedious and slow to do in practice (also, doing this method with variables for coefficients yields the quadratic equation anyway).
1:23 These meanings are related, though. The complementary angle goes with the original angle to make up 90°, sorry, π/2. So too a complementary breakfast or whatever goes with your hotel room to make up the complete package. Something like that.
What an insane trick! I never even knew that!
No pikachu on the micro?!! Just realized you drop that feature 😛
(4:33) _...and this is all equal to pi_
tauism for the win #tauismftw
What is arcsin[ (sinx)^2]
Integration of tan(x^2) please do it i didn't find any were on Google also please do it your calculas is good 👍🏻😊
Dumb question: If the derivative of arcsin x is *1/sqrt(1-x^2)*, and the derivative of arccos x is *-1/sqrt(1-x^2)*, then what will you obtain if you integrate *1/sqrt(1-x^2)*? Will you get arcsin x or will you get -arccos x?
Either, since they differ by a constant.
Only one question at what point of life we are gonna use this
Sir please integration of x^xdx
Can you solve this via graphical method?
I should be n 0 and n -1 or we have pi^2 - 16 < 0
can i also have an image of those identities?
Then Does tan(1/x)dt=1/tan(x)dt ?
Hi thank's for videos
How you solve I = ∫xtan(x)dx ???
So it isn't an identity, but it has infinitely many solutions
Love maths
I know right! 😆
Solutions for this equation are countably infinite but for an identity they are uncountably infinite 😊
@@davidhayesirl countable uncountable have nothing to do with it. An equation can have uncountable solutions and still not be always true.
Well the thing is when it comes to identities they satisfy every freaking value for x, but this equation doesn't satisfy every value for x.
In my country we don't teach what co in cosine and cotangent is. And it's actually very cool name
Actual Identity: tan(pi/2 - x) = 1 / tan(x)
(Yep this works for all values of x, where x is in radians.)
it took about 16 years of public education and finally someone finally explains what the "co" meant good job america
4:26 All equal to Pi :D
sin(x)sin(1/x) = cos(x)cos(1/x)
cos(x)cos(1/x) - sin(x)sin(1/x) = 0
cos(x+1/x) = cos((1/2+n)π)
x + 1/x = (1/2+n)π
x² - (1/2+n)πx + 1 = 0
Just solve this equation
why "co" about tri fcn
> if diff one time, minus add
Can someone explain why you need to add npi instead of just multiplying both sides by x?
For f(x) = f(a) where f(x) is a periodic function of period k, sure x = a will give you one solution to your equation. But in fact owing to the periodicity, there are actually infinite solutions to the equation f(x) = f(a) which can be written as x = a + nk (n belongs to integers).
In our case f(x) is nothing but tan x and the period k is π.
and sometimes when one method doesnt work the other works. Why is this so ?take factoring(doesn't work) and quadratic formula (does work) for example
MY BRO do you have a pdf of that board? I want it so bad
Loved it thanks so much !!