I always use these formulas: Cos(a)=cos(b) => a=2npi+-b. Sin(a)=sin(b) => a=npi+b(-1)^n Tan(a)=tan(b) => a=npi + b Really useful and a life changer for me.
I haven't watched bprp for almost 2 years back when the sub count wasnt this high. Really happy this channel is doing well. And also really surprised about the beard, but you do look wizard though
The usage of +2πn and n E Z allows the cancellation of the sin function by maintaining a representation of all valid solutions which would otherwise be lost when trying to take the inverse of a function that is not one to one.
What i have learnt in school, using the unit circle, was what you showed on the second equation. I always use that strategy solving trig equations sin(a)=sin(b) a=b+2pik or a=pi-b+2pik, k being an integer Similar thing for cosine and tangent
You would have shifted sin(theta) on the left you will get sin(πtheta)-sin(theta) solve according to sin(a)-sin(b) formula and generalize the solution your way is also amazing but this is also a way of solving
Yes. I have a question, why is legitimate to apply arcsine (cancel the sines) after he added the period (2πn)? I tought you could not apply inverse functions if the function is not injective.
@@MrRogordo First of all, you actually need a function to be bijective (together injective and subjective) to be sure it is invertible. But if a function isn't bijective in its whole domain, it doesn't mean you can't invert it, you just can't do it in the domain. If you reduce the domain to the interval [-π/2, π/2] and the codomain to [-1, 1] then you can invert y = sin(x). We say that this function is invertible within that interval. So if you apply the arcsin function you just find solutions in a restricted interval (indeed, [-π/2, π/2]) and what you have to do to include the other infinite solutions is to eventually just add the period. (I hope it makes sense, I'm not a major or anything!)
@@MrRogordo Because this way you consider the periodicty of the sine fuction, just like you add +- when you take square root as just like sine, x^(2) is also not injective.
@@MrRogordo you are right that we can take inverse of function only when it is injective But first by taking the value of the fiction that is the range for example sinx=1/2 you took it's inverse and got that is x= π/6rads now it's OK but now you want to generalize as we know sinx is defined for every real value as input so to generalize he added 2πn and in question it is not mentioned the value of theta should lie in certain interval nice question be curious
sina=sinb implies that: a = b + 2kpi or a = pi - b + 2kpi. (sine function is symmetric with respect to y-axis, which has polar equation of theta=pi/2, so a=b or a=2theta -b.... with a period of 2pi) cosa = cosb implies that: a = b + 2kpi. or a = -b + 2kpi. (Cosine function is symmetric with respect to x-axis which has the equation theta =0...that is why we get + or - theta) tana= tanb implies that a = b + kpi. (tangent function is symmetric wrt to origin O) no need to use double angle formula.... no need for sketches also.. but the sketches help see it, rather than memorizing techniques if solving. Example: sin2x = sinx, 2x=x+2kpi. or 2x=pi-x+2kpi x=2kpi. or x=pi/3 +2kpi/3
The problem with that solution is that it is tough to implement it in exams. BTW this solution is used during times of EXTREME desperation. Speaking from experience.
Trust me Graphical method is the best way to get it done. Cleared class 12 exam in 2007 after 2 previous attempts. 😄😄😄 Worked for me like a charm. Don't ask me my maths score. 😁😁
Hey there! Yes we have an identity that can be used to solve the second problem. Whenever we reach to sinø=sin@ , we can simply say ø = nπ +(-1)ⁿ@ . Use this for your second problem and you'll get the answer real quick. It will automatically cover both the cases depending upon whether n is even or odd.
BPRP! I have an easy solution for this. When two sines are equal, it means either that their arguments are equal or that they add up to π. 1) sin(2x) = sin(x) So 2x = x + 2πn, which means x = 2πn or 2x = π - x + 2πn, which means x = π/3 + 2πn/3 = π(2n + 1)/3. 2) sin(πx) = sin(x) So πx = x + 2πn, which leads to x = 2πn/(π - 1) or πx = π - x + 2πn, which leads to x = π/(π + 1) + 2πn/(π + 1) = π(2n + 1)/(π + 1).
@@omarfahmy5880 i know that calls (Сумма и разность тригонометрических функций) on russian language. You can copy text in brackets and see what is it
0:26 In trigonometric equations chapter class 11 We are taught identity for sin (x) = sin (y) Then x = n(pi) + (-1)to power n Using x and y it is theta and alpha
Here are the 3 solutions for solving sin A = sin B I have seen so far: 1. (from video) B=2nπ+A or B=(2n+1)π-A 2. (from a comment) B=nπ+((-1)^n)A 3. (from several comments) 0=sin A - sin B =2cos((A+B)/2)sin((A-B)/2) ["sum to product" formula]; now solve cos((A+B)/2)=0 and sin((A-B)/2)=0 [straight forward]
Solve: sin(αx) = sin(βx) If α = β, clearly the equation hold for all x. Otherwise: Case 1: αx = βx + 2πn (α-β)x = 2πn x = 2πn/(α-β) Case 2: αx = π - βx + 2πn (α+β)x = (2n+1)π x = (2n+1)π/(α+β) Gives rise to extra solutions, as long as β =/ -α
I made like this the first one : sin(2x) =sin(x) 2x = x + 2kPi or 2x = (Pi-x)+2kPi x= 2kPi or x = PI/3 + 2kPi/3 (same result, 4 points + periods : 0 ; PI/3 ; PI ; 5PI/3 [= -PI/3])
I was expecting euler formula to breakdown ( e^i theta one ) and use that to get sin (pi*theta) to simpler form. (I would have done like this if the question was for more marks... ) This is simpler and nice.....
Greetings from Peru 🇵🇪 and nice video as always! I solved it by using the subtraction to product identity for sines, but I think the method used in the video is sooo much better, because not many people know or remember the formula (i had to google it 😂). It would be cool if you talked about statistics, please, I barely know anything about that branch of mathematics :(.
At 1:20 my mistake was to divide both sides by sin(Θ) so I get 2 cos (Θ) = 1 and solve it, and NOT remembering I assumed sin(Θ) was not zero, so I missed the sin(Θ) = 0 solutions. I learned something.
@@Kdd160 I am actually talking in general.You can't cancel out the sine simply. What I mean is when u cancel both sides of an equation , u ignore one solution.
And also sin theta in this case is zero , which cant divide anything. Not even itself . So this method ain't recommended. Your answer may be right but the process certainly aint.
@@Kdd160 And if you know the general equation for sin alpha = sin theta , you can do it too , without any trouble but this channel focuses on the basic method.
heres the answer for the challenge:2npi is the numerator and the denominator can be1-pi or 1+pi btw love yoor videos they are school math but at the same time they arent
Why can you "cancel" the "sin" after adding 2·n·π on the "inside" and account for the "sign" of the "inside" if the function is not injective? To explain what is going on, first I want to use a simpler, far more familiar example, and build up the explanation from there by analogy. Consider the function f : R -> R+, t |-> t^2. This function is not injective, so it cannot be inverted, and this is because f(t) = f(-t). Therefore, if I want to solve the equation f(x) = y for x, I cannot simply apply an inverse function f^(-1) and get x = [f^(-1)](y) to solve the equation. So what can we do? Notice that, in this particular instance, if we restrict the domain of the function to R+ or to R-, then in either case, the restricted function is injective, and does have an inverse. If we restrict it to R+, then its inverse is given by t |-> sqrt(t), while if we restrict to R-, then its inverse is given by t |-> -sqrt(t). Furthermore, R- and R+ partition the actual domain R, except for the double counted 0 shared by both sets. This matters, because when we say the solutions to the equation are x = sqrt(y) and x = -sqrt(y), the fact that both "inverse" functions evaluated at y are the only solutions is guaranteed by the fact that their domains partition the domain of the domain of the original function they "inverted." So with this in mind, you know have an actual method of solution: if x is an element of R+, then x^2 = y ==> x = sqrt(y), while if x is an element of R-, then x^2 = y ==> x = -sqrt(y). So the solutions are sqrt(y) and -sqrt(y): to shorten things, we tend to abbreviate this with a "plus-minus" symbol in front, which is how most people would typically write the final draft. However, there is a different way of encoding both solutions into a single expression: by indexing the cases with an expression that depends on the index. In this case, you can do this by noticing that x = (-1)^n·sqrt(y), allowing n to be any integer, which encodes both cases, by realizing that the first solution is given by an even parity on the index, and the second solution is given by an odd parity on the index. You could take this a step further, and just say that [(-1)^n·x]^2 = x^2, so you can replace x^2 = y with [(-1)^n·x]^2 = y. If I "cancel" the squares now, I get (-1)^n·x = sqrt(y), which implies x = (-1)^n·sqrt(y), getting back to where we should be. The "cancellation" works, because the expression (-1)^n encodes all the possible cases at once with the index, so that for any given n, there is a restriction to x that makes the "left-hand side" injective. What BPRP did in this video is literally the same thing, except he used the function x |-> sin(x) instead of x |-> x^2, and he partitioned the domain into subdomains [m·π - π/2, m·π + π/2] instead of partitioning into R+ and R-. Why? Because the equation uses the sine function, and the function is injective when its domain is restricted into those intervals, provided that m is an arbitrary integer. More importantly, for each m, you get a unique interval of restricted injectivity, so that you can ensure this partitioning gives you every solution, and no extraneous solution. So you can write the equation sin(π·x) = sin(x - m·π). This may not initially look at all like what BPRP has done here, but the distinction merely boils down to reindexing or renaming quantities. How? If m is even, then m = -2·n, so sin(π·x) = sin(x + 2·n·π). Here, if I "cancel" the "sin," I get π·x = x + 2·n·π. This works because, for each distinct value of n, I get a distinct restriction on x that guarantees restricted injectivity, allowing the "cancellation" for that particular value of m. But this only gives us exactly half of the solutions, and the other half occurs when m is odd. You can get odd values of m by realizing that sin(t) = sin(π - t), so that sin(x + 2·n·π) = sin(-x + [-2·n + 1]·π) = sin(π·x), and again, "cancelling the sin" here works for the same reasons already explained. This gives the solutions x = [-2·n + 1]·π/(π + 1), which is not quite was had in the video, but you can just rename -n into n to get that, and you can do this because n is an arbitrary integer, and n |-> -n is bijective. This is exactly the idea behind what BPRP did, although maybe it could have been simpler to execute its explanation differently, and it may have been less confusing for some in that case. For example, I think it may have been productive to actually approach this by using the arcsin[sin(x)] approach instead, and breaking it into cases, and explaining the cases in indexed form, so that the viewers can have understanding why the symbolic manipulations work in the first place. Why do we do this, if it is so contrived in the end? We do it because you cannot explicitly write out infinitely many cases without indexing and without these symbolic techniques, since, well, you know, it would take an infinite amount of time to do so, an amount of time we certainly do not have available. It is just useful to be able to do this. This concept can be generalized to any noninjective function f : X -> Y. If you want to solve f(x) = y, with x an element of X and y and element of Y, then you cannot invert f, but you can do restricted inversion, whereby you partition X (with maybe some double counting at a single point per pair of sets) into sets X(i), where i is an index, such that when the domain f is restricted to each one of those sets, the restricted f is injective. You can thus define a restricted inverse g(i) : Y -> X(i) for each i, so that x = [g(i)](y) encodes every solution to the equation, and each value of i gives you a true solution (not extraneous). The index family to which i belongs to can also always be chosen so that the value of [g(i)](y) is unique to i. How you go about doing the symbolic manipulation, however, depends very much on the properties of f. Of course, if f is injective, the same procedure works trivially: you partition X into a single set: the set X, and you define a single inverse g : Y -> X that works on all of the domain of X, so that x = g(y) is the sole solution to the equation. So this solving process is actually a special case of the general, noninjective case. In a reply to my own comment, I will explain how to choose a g(i) for the case when f(x) = sin(π·x), detailing the alternative approach BPRP could have taken that I mentioned earlier.
Suppose x is an element of [-π/2, +π/2], n is an integer. So x + 2·n·π is an element of [2·n·π - π/2, 2·n·π + π/2], arcsin[sin(x + 2·n·π)] = x ==> (-1)^(2·n)·arcsin[sin(x + 2·n·π)] + 2·n·π = x + 2·n·π, -arcsin(sin[x + (2·n + 1)·π] = x ==> (-1)^(2·n + 1)·arcsin(sin[x + (2·n + 1)·π]) + (2·n + 1)·π = x + (2·n + 1)·π. Therefore, if x is an element of [m·π - π/2, m·π + π/2] & m is an integer, then m·π + (-1)^m·arcsin[sin(x)] = x. Since x |-> sin(x) : [m·π - π/2, m·π + π/2] -> [-1, +1] is bijective for every m, it has an inverse x |-> m·π + (-1)^m·arcsin(x) : [-1, +1] -> [m·π - π/2, m·π + π/2]. Therefore, if x is an element of [m - 1/2, m + 1/2] & m is an integer, then m·π + (-1)^m·arcsin[sin(π·x)] = π·x. Since x |-> sin(π·x) : [m - 1/2, m + 1/2] -> [-1, +1] is bijective for every m, it has an inverse x |-> m + (-1)^m·arcsin(x)/π : [-1, +1] -> [m - 1/2, m + 1/2]. Given sin(π·x) = sin(x), you can conclude that m·π + (-1)^m·arcsin[sin(π·x)] = π·x & n·π + (-1)^n·arcsin[sin(x)] = x for integers m & n imply arcsin[sin(π·x)] = (-1)^m·π·(x - m) & arcsin[sin(x)] = (-1)^n·(x - n·π), hence (-1)^m·(π·x - π·m) = (-1)^n·(x - n·π), equivalent to (-1)^m·π·x - (-1)^m·m·π = (-1)^n·x - (-1)^n·n·π. Thus [(-1)^m·π - (-1)^n]·x = [(-1)^m·m - (-1)^n·n]·π, and finally, x = [(-1)^m·m - (-1)^n·n]·π/[(-1)^m·π - (-1)^n]. Let m = 2·μ and n = 0, 1, and this recovers the solutions in the video.
Can you do lim_(x->infinity)(lnx / W(x))? Maybe a Math for fun type of video. L'Hôpital's rule suggests that the limit is 1 which it is not. What's the matter and what's the actual limit?
sin 2x = sin x but if u use "sum and product formulae" it will be more hard sin 2x - sin x = 0 2cos(3x/2)sin(x/2)=0 3x/2= pi/2+2pi*n, 3pi/2+2pi*n x=pi/3+4/3*pi*n, pi+4/3*pi*n if n = 1 ,then x=2pi-pi/3, 2pi+pi/3 if n = 2 ,then x=3pi, 4pi-pi/3 do it repeatedly so x=n*pi,pi/3+2pi*n,-pi/3+2pi*n hmmm.. it seem really hard than using sin2x=2sinxcosx
It's not another case. Sine is an odd function, so -sin(theta+kπ)=sin(-(theta+kπ))=sin(kπ-theta)=sine(π-theta+(k-1)π) This is equivalent to his second case when (k-1)=2n and is *false* at any other value.
*Sir Steve Chow* please read this comment for a golden equation. Your videos are very amazing. I have derived a very beautiful equation in which there are all amazing things like Phi,π,e,i and even Fibonacci series(All five in one equation). I request you to please make a video on it. The beautiful golden equation is difficult to type here but still I will try to type. It is as follow: ϕ(ϕ^(e^(πi)-n)+-(n+1)th term of Fibonacci series) = +-(n+2)th term of Fibonacci series where n is a non-negative integer and if n is even consider only + sign and if n is odd consider only - sign. You can check it on calculator it will work. If you want proof of it then reply my comment and I will give my number and I can easily explain it's proof on a call as it will be difficult to explain it in comment. Thank you for reading this comment. I hope you will make a video on it.
@@pravindahiya719 actually I am just a student and I don't have a UA-cam channel. I also don't get time to make a video due to homework and other things. So I am trying to tell bprp about it. I hope he will make a video.
@@Ari-pq4db there are so many formulas with no applications like formula for general term of Fibonacci series or formula for arc trigonometric functions. There's no application for general term of Fibonacci series, What would someone do by even finding 100th term of Fibonacci series and it is even difficult to solve. And formula for arc trigo functions are also useless as in their formulas there is another trigo function whose value we must know and in other formula for arc trigo function there is iota which is very difficult to eliminate. So it's not necessary that a formula should have an application.
@@mathevengers1131 No Sir , Any mathematical formulas like Fibonacci series , inverse trigo functions or any other you name it carries application , i.e. in computer programming we have to use algorithms based on Fibonacci series , any programmed machine (complex or simple) have to use sin , cos or inverse functions to perform the required actions . There might be few other applications also ! And if you agree with me so far , find out what your equation can be used for/in . I think it's a huge thing to create your own equation ( "PROVIDED IT WORKS" , you mentioned that in your comment so I believe you 🤠) .
sin(t) = t iff t = 0. However, if t ~ 0, then sin(t) ~ t. Here, "~" means "is approximately equal to." Therefore, if, for example, t = 0.000001, then you in practical applications, you can simply have sin(t) ~= t, because the difference between the two is so small as to be immeasurable.
@@luisaguilera3242 If you want to solve the equation x^2 = y^2 for x, then you want to take the square root, and you get sqrt(x^2) = sqrt(y^2), which is equivalent to |x| = |y|. This equation has two solutions in R, because x |-> |x|: R -> R+ is an even function, so it is not injective. How would you deal with this equation? You would realize that R+ and R- partition the set R into two "opposite" sets, with 0 being double counted, and when the domain is restricted into either of the sets, the restricted function does become injective in both intervals. So you break the equation-solving process into two cases: x >= 0, and x =< 0. If x >= 0, then |x| = x, so x = |y|, and the equation has already been solved. If x =< 0, then |x| = -x, so -x = |y|, thus x = -|y|, and the equation has been solved. Since there are no other cases to analyze, you are done, and the equation has two solutions, x = |y| or x = -|y|. Here, you can also rewrite this as x = y or x = -y, since it you get both solutions regardless of whether y =< 0 or y >= 0. However, there is one way to concisely express both solutions in one expression: this is encoded by realizing that |x| = |-x|, so you can write |(-1)^n·x| in place of |x|, allowing yourself to substitute integer values of n. Then you get |(-1)^n·x| = |y|, and if you "cancel" (not the correct terminology, but this is what you and others refer to it as, so we will go along with it) the absolute values, you get (-1)^n·x = y, which implies x = (-1)^n·y. This gives you both solutions in a single expression, but why does this work? Because the breakdown of cases into separate equations is encoded in the indexed expression (-1)^n. By substituting different parities of n, you get every case allowed. When solving sin(π·x) = sin(x), the same concept applies. x |-> sin(x) : R -> [-1, +1] is not injective. However, if you restrict the domain to an interval [m·π - π/2, m·π + π/2], where m is an arbitrary integer, then the restricted function is injective for each m in a mutually exclusive fashion, except for endpoints, which are double counted. Furthermore, the function has period 2·π, and for odd m in the above intervals, the function has the opposite growth as with even m, so you can consider sin(π·x) = sin(x - m·π) for even m and sin(π·x) = sin(x - m·π) for odd m separately. If m is even, then m = -2·n, so sin(π·x) = sin(x + 2·n·π), and if m is odd, then m = -2·n - 1, so sin(π·x) = sin(x + 2·n·π + π), but sin(t + π) = sin(-t) for any t, so sin(π·x) = sin[-(x + 2·n·π)]. Thus the two cases are sin(π·x) = sin(x + 2·n·π) or sin(π·x) = sin[-(x + 2·n·π)], with n being an arbitrary integer, and this is precisely what the video did. Each value of n will give a unique interval, in such a way that both cases combined partition the set R into intervals in which, when you restrict the function's domain to those intervals, the function becomes injective, and n is indexing those intervals. This why the function becomes "cancellable" once you do break the equation into those two cases: by doing so, the function becomes "injective for each value of n" in both cases.
I did a method that involved converting sin(pi x) into (e^(i pi x) - e^-(i pi x))/2 which seems like it should still work. you end up being able to factorise stuff and doing it as a quadratic. I only wanted to get one of the solutions and i got pi/(pi+1), but it might still be possible to get the other solutions? the method is: e^(i pi x) - e^(-i pi x) = e ^(ix) - e ^-(ix) (e^ix)^pi - (e^ix)^-pi - e^ix + (e^ix)^-1 = 0 substitute y = e^ix y^pi - y + (1/y - 1/y^pi) = 0 y^pi - y + (y^pi-y)/y^(pi+1) = 0 y^(pi+1) (y^pi - y) + (y^pi - y) = 0 (y^(pi +1) + 1)(y^pi - y)=0 y^(pi+1) +1 = 0 or y^pi -y = 0 y*(pi+1) = -1 or y^(pi-1) = 1 ln(-1) on wolframalpha give i pi which is the result that lead to x= pi/(pi+1), and the other one with ln(1) leads to x = 0. I think if i did the whole thing taking into account the multiple solutions with the 2pi n stuff it would have lead to the general solutions maybe? I'm iffy about if i should have been allowed to do the ln(-1) = i *pi thing.
The sine function has a period of 2·π, hence if x = a solves sin(x) = a, then necessarily, a + 2·n·π solves it too; otherwise, it would contradict the periodicity.
I just use 2π periodicity property of the sin(X) function: sin(πθ)☰sin(θ+2nπ) gives BPRP's first solution then consider sin(X) is an odd function, i.e. sin(X)☰-sin(-X)☰sin(π-X) sin(πθ)☰sin(π-θ+2nπ) gives second BPRP solution. No need for them pesky CAST circles.
Is it possible to solve the second equation using the euler's formulas ? It we put this equation as : ( e^(iOpi) - e^(-iOpi) )/2i = sin (O), where O is theta ; then we simplify as ( (e^(i*pi))^O - (e^(-i*pi))^O)/2i = sin O ( (-1)^O - (-1)^O )/2i = sin O So we get Sin O = 0 So O = n*pi, n being an integer. Is this way correct or am i wrong somewhere ?
Is it possible to sum up a single term infinitely many times and have it equal to pi? I don’t mean sum up different terms. I mean sum up the same number over and over. It might be more reasonable if we let the signs alternate but what if we don’t even allow that to happen. Is it possible?
Wanna learn more interesting math? Visit brilliant.org/blackpenredpen/ and start exploring. Try their trig & precalc courses after this video!
Hey bprp...
Can you proof arcsin(2) + arccos(2)=π/2
Must reply...
Can you integrate y=sin(x)/ln(x) from 0 to 1?
For the 2nd question does just θ = π/(π+1) not work?
I always use these formulas:
Cos(a)=cos(b)
=> a=2npi+-b.
Sin(a)=sin(b)
=> a=npi+b(-1)^n
Tan(a)=tan(b)
=> a=npi + b
Really useful and a life changer for me.
ㅠ
@@gyejung848if youre gonna use a different keyboard you can just use the greek keyboard and get actual pi!! π
@@gyejung848 ∏
2nπ
I haven't watched bprp for almost 2 years back when the sub count wasnt this high. Really happy this channel is doing well. And also really surprised about the beard, but you do look wizard though
Haha thanks.
Pre-calc feels so long ago yet it's only been 2 years but I'm doing abstract algebra and analysis now
Time flies when you're having fun
The usage of +2πn and n E Z allows the cancellation of the sin function by maintaining a representation of all valid solutions which would otherwise be lost when trying to take the inverse of a function that is not one to one.
What i have learnt in school, using the unit circle, was what you showed on the second equation.
I always use that strategy solving trig equations
sin(a)=sin(b)
a=b+2pik or a=pi-b+2pik,
k being an integer
Similar thing for cosine and tangent
Same. I think more you learn and more you forget the simplest way ;-)
You would have shifted sin(theta) on the left you will get sin(πtheta)-sin(theta) solve according to sin(a)-sin(b) formula and generalize the solution your way is also amazing but this is also a way of solving
Yes.
I have a question, why is legitimate to apply arcsine (cancel the sines) after he added the period (2πn)?
I tought you could not apply inverse functions if the function is not injective.
@@MrRogordo First of all, you actually need a function to be bijective (together injective and subjective) to be sure it is invertible. But if a function isn't bijective in its whole domain, it doesn't mean you can't invert it, you just can't do it in the domain. If you reduce the domain to the interval [-π/2, π/2] and the codomain to [-1, 1] then you can invert y = sin(x). We say that this function is invertible within that interval. So if you apply the arcsin function you just find solutions in a restricted interval (indeed, [-π/2, π/2]) and what you have to do to include the other infinite solutions is to eventually just add the period.
(I hope it makes sense, I'm not a major or anything!)
@@MrRogordo Because this way you consider the periodicty of the sine fuction, just like you add +- when you take square root as just like sine, x^(2) is also not injective.
@@MrRogordo you are right that we can take inverse of function only when it is injective
But first by taking the value of the fiction that is the range for example sinx=1/2 you took it's inverse and got that is x= π/6rads now it's OK but now you want to generalize as we know sinx is defined for every real value as input so to generalize he added 2πn and in question it is not mentioned the value of theta should lie in certain interval nice question be curious
@@christianpalumbo8278 Thanks, I got it
sina=sinb implies that:
a = b + 2kpi or a = pi - b + 2kpi. (sine function is symmetric with respect to y-axis, which has polar equation of theta=pi/2, so a=b or a=2theta -b.... with a period of 2pi)
cosa = cosb implies that:
a = b + 2kpi. or a = -b + 2kpi. (Cosine function is symmetric with respect to x-axis which has the equation theta =0...that is why we get + or - theta)
tana= tanb implies that a = b + kpi. (tangent function is symmetric wrt to origin O)
no need to use double angle formula.... no need for sketches also.. but the sketches help see it, rather than memorizing techniques if solving.
Example:
sin2x = sinx,
2x=x+2kpi. or 2x=pi-x+2kpi
x=2kpi. or x=pi/3 +2kpi/3
cos(pi*O)=cos(O) (O is theta)
pi*O= O+2k*pi or pi*O= -O+2k*pi (k is integer)
(pi-1)*O=2k*pi or (pi+1)*O=2k*pi
O=(2k*pi)/(pi-1) or O=(2k*pi)/(pi+1)
I solved it by thinking of the intersection of a horizontal line with the sine curve. That allowed me to catch both cases.
The problem with that solution is that it is tough to implement it in exams. BTW this solution is used during times of EXTREME desperation. Speaking from experience.
Please can you explain more ?
Yes even I solved by using the sine curve. Graphical method is always best
Trust me Graphical method is the best way to get it done. Cleared class 12 exam in 2007 after 2 previous attempts. 😄😄😄 Worked for me like a charm. Don't ask me my maths score. 😁😁
To quote a song
Pre calculus didn't help me prepare for calculus, for calculus, help meee
Sounds like when Freddie Mercury took calculus
Yes it did. How were you planning to do calculus without knowing algebra and trig?
@@pbj4184 it's a song, it probably implies that pre calculus wasn't enough to fully prepare
@@harshvardhanpandey8057 when Brian discovered* calculus
@@harshvardhanpandey8057 I understood that reference.
Hey there! Yes we have an identity that can be used to solve the second problem. Whenever we reach to sinø=sin@ , we can simply say
ø = nπ +(-1)ⁿ@ . Use this for your second problem and you'll get the answer real quick. It will automatically cover both the cases depending upon whether n is even or odd.
1. Theta=60
2.Theta=180/(π+1)
You may check in calculator for
1.sin(theta)=sin(2theta)
2.sin(πtheta) =sin(theta)
I’m by no means a precalc student, but I didn’t use calculus, so here’s the set of solutions I found: {k ∈ ℤ | 2kπ/(π-1) ∨ π - 2kπ/(π-1)}.
BPRP! I have an easy solution for this.
When two sines are equal, it means either that their arguments are equal or that they add up to π.
1) sin(2x) = sin(x)
So 2x = x + 2πn, which means x = 2πn
or 2x = π - x + 2πn, which means x = π/3 + 2πn/3 = π(2n + 1)/3.
2) sin(πx) = sin(x)
So πx = x + 2πn, which leads to x = 2πn/(π - 1)
or πx = π - x + 2πn, which leads to x = π/(π + 1) + 2πn/(π + 1) = π(2n + 1)/(π + 1).
Woke: Theta = 0
Yeah
That's what I thought
Another woke: theta = kpi, k €Z
That's what I also thought
That's woke, but it can be woker :
Sin(∅) ≈ ∅
For second one,
sin(πa)-sina= 0
2 cos( (πa+a)/2) sin((πa-a)/2)= 0
so,
either , sin term or cos term is zero . Hence I get 2 solution sets .
what is this?
@@omarfahmy5880 i know that calls (Сумма и разность тригонометрических функций) on russian language.
You can copy text in brackets and see what is it
Hey,I have one
π(theta)=nπ + ((-1)^n)theta
Where n is an integer
class 11th NCERT😂
@@theshibu214 😭🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣😂🤣😂🤣🤣😂🤣🤣😂😂🤣🤣🤣😂🤣🤣😂🤣🤣😭🤣🤣🤣😂🤣🤣🤣🤣😂😂🤣🤣🤣🤣😭😭🤣😂😂🤣🤣🤣🤣🤣🤣🤣😂😂😂😂🤣🤣🤣😭😂🤣🤣😭😂🤣🤣😂🤣😅😂😂🤣🤣😂🤣🤣😂😂🤣🤣🤣🤣😂😂😂🤣😂😂😂😂😂😂😂😂😂😂😂😂😂😭😭🤣🤣🤣😂😂😂🤣😂🤣🤣😂😂😂😂😂😅😂😂🤣😂😂🤣🤣😂😂😂😂😂
we can also use SinA-SinB = 2Cos((A+B)/2)Sin((A-B)/2) then compute..Lil bit of overkill i guess ...
Not overkill at all, this is an excellent alternative solution.
0:26
In trigonometric equations chapter class 11
We are taught identity for sin (x) = sin (y)
Then x = n(pi) + (-1)to power n
Using x and y it is theta and alpha
That evil laugh at 6:46 😂😂😂
Mathematicians have a (a+bi) sense of humour.
Its not imaginary its just complex😎
@@SR-qe7rj yeah 😎
Some have a/b ( where a and b are integers and b≠0) thinking as well
When b does not equal 0:
a=0
Case two a very easy approach is to use sinA = sin(pi-A) by identity, with A being (pi)(theta)....works quickly
Here are the 3 solutions for solving sin A = sin B I have seen so far:
1. (from video) B=2nπ+A or B=(2n+1)π-A
2. (from a comment) B=nπ+((-1)^n)A
3. (from several comments) 0=sin A - sin B =2cos((A+B)/2)sin((A-B)/2) ["sum to product" formula]; now solve cos((A+B)/2)=0 and sin((A-B)/2)=0 [straight forward]
could we use sum to product for question 2 as well?
@noob master I think he wished to use the formula for
sin(A) - sin(B)
@@boka_314 ya that's better
Yup
I used that method 😂
But honestly, his method is more visual and concept-based rather than the application of a formula
his beard is the source of his power
it is easy when 1you understand what is a circle and 2 you understand what is a sinus.
nice job done.
sin 2θ = sin θ
θ = πκ or π/3 ± 2πκ
κ ∈ ℤ
sin πθ = sin θ
θ = 2πκ/(π-1) or (2πκ+π)/(π-1)
κ ∈ ℤ
At 1:26, you could cancel out the sin thetas at that point and get cosine theta equals 1/2.
cos( pi * x ) = cos( x )
x = 2 * n * pi / ( pi - 1 )
and x = 2 * n * pi / ( pi + 1)
Solve:
sin(αx) = sin(βx)
If α = β, clearly the equation hold for all x. Otherwise:
Case 1:
αx = βx + 2πn
(α-β)x = 2πn
x = 2πn/(α-β)
Case 2:
αx = π - βx + 2πn
(α+β)x = (2n+1)π
x = (2n+1)π/(α+β)
Gives rise to extra solutions, as long as β =/ -α
I made like this the first one :
sin(2x) =sin(x)
2x = x + 2kPi or 2x = (Pi-x)+2kPi
x= 2kPi or x = PI/3 + 2kPi/3 (same result, 4 points + periods : 0 ; PI/3 ; PI ; 5PI/3 [= -PI/3])
8:54 The angle you drew is -pi not pi, but at the end it doesn't make any difference.
Thanks for the video !
I was expecting euler formula to breakdown ( e^i theta one ) and use that to get sin (pi*theta) to simpler form.
(I would have done like this if the question was for more marks... )
This is simpler and nice.....
Bring more such problems, highly needed.
Please provide the link to the video of sine not being injective
sin x= sin y if x-y=2*pi*n or x+y=pi + 2*pi*n for some integer n
Greetings from Peru 🇵🇪 and nice video as always! I solved it by using the subtraction to product identity for sines, but I think the method used in the video is sooo much better, because not many people know or remember the formula (i had to google it 😂).
It would be cool if you talked about statistics, please, I barely know anything about that branch of mathematics :(.
At 1:20 my mistake was to divide both sides by sin(Θ) so I get 2 cos (Θ) = 1 and solve it, and NOT remembering I assumed sin(Θ) was not zero, so I missed the sin(Θ) = 0 solutions. I learned something.
Same here
we can use next formula: sin(a)-sin(b)=2sin((a-b)/2)cos((a+b)/2). Result will be same, but this solution is more quick.
Well, even if we cancel out the sines in the first equation, theres a way we can get all the infinite values, just add 2pi*n on either side.
Cancelling sines on both sides with theta equals 0 (for example) is a division by zero, which I would not recommend...
@@yatogami7393 no, the actual solution is theta = 2pi*n, where n is any integer. If n=0, theta = 0, we get that solution, isnt it? 🤔🤔
@@Kdd160 I am actually talking in general.You can't cancel out the sine simply. What I mean is when u cancel both sides of an equation , u ignore one solution.
And also sin theta in this case is zero , which cant divide anything. Not even itself . So this method ain't recommended.
Your answer may be right but the process certainly aint.
@@Kdd160 And if you know the general equation for sin alpha = sin theta , you can do it too , without any trouble but this channel focuses on the basic method.
Now we are adults so we'll say π/3 .
Correct 👍🏿
First result cos case could also be written
Theta = 2m.pi +/- pi/3
I will share this with my students! I will also include a related one: sin (67x) = cos (33x). I'll post a video if anyone is interested.
heres the answer for the challenge:2npi is the numerator and the denominator can be1-pi or 1+pi
btw love yoor videos they are school math but at the same time they arent
Why can you "cancel" the "sin" after adding 2·n·π on the "inside" and account for the "sign" of the "inside" if the function is not injective? To explain what is going on, first I want to use a simpler, far more familiar example, and build up the explanation from there by analogy.
Consider the function f : R -> R+, t |-> t^2. This function is not injective, so it cannot be inverted, and this is because f(t) = f(-t). Therefore, if I want to solve the equation f(x) = y for x, I cannot simply apply an inverse function f^(-1) and get x = [f^(-1)](y) to solve the equation. So what can we do? Notice that, in this particular instance, if we restrict the domain of the function to R+ or to R-, then in either case, the restricted function is injective, and does have an inverse. If we restrict it to R+, then its inverse is given by t |-> sqrt(t), while if we restrict to R-, then its inverse is given by t |-> -sqrt(t). Furthermore, R- and R+ partition the actual domain R, except for the double counted 0 shared by both sets. This matters, because when we say the solutions to the equation are x = sqrt(y) and x = -sqrt(y), the fact that both "inverse" functions evaluated at y are the only solutions is guaranteed by the fact that their domains partition the domain of the domain of the original function they "inverted." So with this in mind, you know have an actual method of solution: if x is an element of R+, then x^2 = y ==> x = sqrt(y), while if x is an element of R-, then x^2 = y ==> x = -sqrt(y). So the solutions are sqrt(y) and -sqrt(y): to shorten things, we tend to abbreviate this with a "plus-minus" symbol in front, which is how most people would typically write the final draft. However, there is a different way of encoding both solutions into a single expression: by indexing the cases with an expression that depends on the index. In this case, you can do this by noticing that x = (-1)^n·sqrt(y), allowing n to be any integer, which encodes both cases, by realizing that the first solution is given by an even parity on the index, and the second solution is given by an odd parity on the index. You could take this a step further, and just say that [(-1)^n·x]^2 = x^2, so you can replace x^2 = y with [(-1)^n·x]^2 = y. If I "cancel" the squares now, I get (-1)^n·x = sqrt(y), which implies x = (-1)^n·sqrt(y), getting back to where we should be. The "cancellation" works, because the expression (-1)^n encodes all the possible cases at once with the index, so that for any given n, there is a restriction to x that makes the "left-hand side" injective.
What BPRP did in this video is literally the same thing, except he used the function x |-> sin(x) instead of x |-> x^2, and he partitioned the domain into subdomains [m·π - π/2, m·π + π/2] instead of partitioning into R+ and R-. Why? Because the equation uses the sine function, and the function is injective when its domain is restricted into those intervals, provided that m is an arbitrary integer. More importantly, for each m, you get a unique interval of restricted injectivity, so that you can ensure this partitioning gives you every solution, and no extraneous solution. So you can write the equation sin(π·x) = sin(x - m·π). This may not initially look at all like what BPRP has done here, but the distinction merely boils down to reindexing or renaming quantities. How? If m is even, then m = -2·n, so sin(π·x) = sin(x + 2·n·π). Here, if I "cancel" the "sin," I get π·x = x + 2·n·π. This works because, for each distinct value of n, I get a distinct restriction on x that guarantees restricted injectivity, allowing the "cancellation" for that particular value of m. But this only gives us exactly half of the solutions, and the other half occurs when m is odd. You can get odd values of m by realizing that sin(t) = sin(π - t), so that sin(x + 2·n·π) = sin(-x + [-2·n + 1]·π) = sin(π·x), and again, "cancelling the sin" here works for the same reasons already explained. This gives the solutions x = [-2·n + 1]·π/(π + 1), which is not quite was had in the video, but you can just rename -n into n to get that, and you can do this because n is an arbitrary integer, and n |-> -n is bijective. This is exactly the idea behind what BPRP did, although maybe it could have been simpler to execute its explanation differently, and it may have been less confusing for some in that case. For example, I think it may have been productive to actually approach this by using the arcsin[sin(x)] approach instead, and breaking it into cases, and explaining the cases in indexed form, so that the viewers can have understanding why the symbolic manipulations work in the first place.
Why do we do this, if it is so contrived in the end? We do it because you cannot explicitly write out infinitely many cases without indexing and without these symbolic techniques, since, well, you know, it would take an infinite amount of time to do so, an amount of time we certainly do not have available. It is just useful to be able to do this.
This concept can be generalized to any noninjective function f : X -> Y. If you want to solve f(x) = y, with x an element of X and y and element of Y, then you cannot invert f, but you can do restricted inversion, whereby you partition X (with maybe some double counting at a single point per pair of sets) into sets X(i), where i is an index, such that when the domain f is restricted to each one of those sets, the restricted f is injective. You can thus define a restricted inverse g(i) : Y -> X(i) for each i, so that x = [g(i)](y) encodes every solution to the equation, and each value of i gives you a true solution (not extraneous). The index family to which i belongs to can also always be chosen so that the value of [g(i)](y) is unique to i. How you go about doing the symbolic manipulation, however, depends very much on the properties of f. Of course, if f is injective, the same procedure works trivially: you partition X into a single set: the set X, and you define a single inverse g : Y -> X that works on all of the domain of X, so that x = g(y) is the sole solution to the equation. So this solving process is actually a special case of the general, noninjective case. In a reply to my own comment, I will explain how to choose a g(i) for the case when f(x) = sin(π·x), detailing the alternative approach BPRP could have taken that I mentioned earlier.
Suppose x is an element of [-π/2, +π/2], n is an integer. So x + 2·n·π is an element of [2·n·π - π/2, 2·n·π + π/2], arcsin[sin(x + 2·n·π)] = x ==> (-1)^(2·n)·arcsin[sin(x + 2·n·π)] + 2·n·π = x + 2·n·π, -arcsin(sin[x + (2·n + 1)·π] = x ==> (-1)^(2·n + 1)·arcsin(sin[x + (2·n + 1)·π]) + (2·n + 1)·π = x + (2·n + 1)·π.
Therefore, if x is an element of [m·π - π/2, m·π + π/2] & m is an integer, then m·π + (-1)^m·arcsin[sin(x)] = x. Since x |-> sin(x) : [m·π - π/2, m·π + π/2] -> [-1, +1] is bijective for every m, it has an inverse x |-> m·π + (-1)^m·arcsin(x) : [-1, +1] -> [m·π - π/2, m·π + π/2].
Therefore, if x is an element of [m - 1/2, m + 1/2] & m is an integer, then m·π + (-1)^m·arcsin[sin(π·x)] = π·x. Since x |-> sin(π·x) : [m - 1/2, m + 1/2] -> [-1, +1] is bijective for every m, it has an inverse x |-> m + (-1)^m·arcsin(x)/π : [-1, +1] -> [m - 1/2, m + 1/2].
Given sin(π·x) = sin(x), you can conclude that m·π + (-1)^m·arcsin[sin(π·x)] = π·x & n·π + (-1)^n·arcsin[sin(x)] = x for integers m & n imply arcsin[sin(π·x)] = (-1)^m·π·(x - m) & arcsin[sin(x)] = (-1)^n·(x - n·π), hence (-1)^m·(π·x - π·m) = (-1)^n·(x - n·π), equivalent to (-1)^m·π·x - (-1)^m·m·π = (-1)^n·x - (-1)^n·n·π. Thus [(-1)^m·π - (-1)^n]·x = [(-1)^m·m - (-1)^n·n]·π, and finally, x = [(-1)^m·m - (-1)^n·n]·π/[(-1)^m·π - (-1)^n]. Let m = 2·μ and n = 0, 1, and this recovers the solutions in the video.
I subbed in cosθ for its complex exponential identity, used your quadratic method from the last sinz video and got ∓iπ/3.
Can you do lim_(x->infinity)(lnx / W(x))? Maybe a Math for fun type of video. L'Hôpital's rule suggests that the limit is 1 which it is not. What's the matter and what's the actual limit?
Use sina-sinb=2cos(a+b)/2sin(a-b)/2
sin 2x = sin x
but if u use "sum and product formulae" it will be more hard
sin 2x - sin x = 0
2cos(3x/2)sin(x/2)=0
3x/2= pi/2+2pi*n, 3pi/2+2pi*n
x=pi/3+4/3*pi*n, pi+4/3*pi*n
if n = 1 ,then
x=2pi-pi/3, 2pi+pi/3
if n = 2 ,then
x=3pi, 4pi-pi/3
do it repeatedly so x=n*pi,pi/3+2pi*n,-pi/3+2pi*n
hmmm.. it seem really hard than using sin2x=2sinxcosx
Sin(πtheta)= -sin(theta +nπ) also a case
It's not another case. Sine is an odd function, so -sin(theta+kπ)=sin(-(theta+kπ))=sin(kπ-theta)=sine(π-theta+(k-1)π)
This is equivalent to his second case when (k-1)=2n and is *false* at any other value.
Hey , can you plz solve this , integration of (sec^2 x - 7) / sin^7 x dx
How to cancel your sin?
Repent.
Or if someone has nothing better to do than digging an edgy joke the sin made decades ago
*Sir Steve Chow* please read this comment for a golden equation.
Your videos are very amazing. I have derived a very beautiful equation in which there are all amazing things like Phi,π,e,i and even Fibonacci series(All five in one equation). I request you to please make a video on it. The beautiful golden equation is difficult to type here but still I will try to type. It is as follow:
ϕ(ϕ^(e^(πi)-n)+-(n+1)th term of Fibonacci series) = +-(n+2)th term of Fibonacci series where n is a non-negative integer and if n is even consider only + sign and if n is odd consider only - sign.
You can check it on calculator it will work. If you want proof of it then reply my comment and I will give my number and I can easily explain it's proof on a call as it will be difficult to explain it in comment.
Thank you for reading this comment. I hope you will make a video on it.
Dear Kathan !
YOU should also make a video on it !
@@pravindahiya719 actually I am just a student and I don't have a UA-cam channel. I also don't get time to make a video due to homework and other things. So I am trying to tell bprp about it. I hope he will make a video.
It's pretty amazing that your have derived an equation by yourself but , can you explain me it's applications ( I mean what can we find from it ) ?
@@Ari-pq4db there are so many formulas with no applications like formula for general term of Fibonacci series or formula for arc trigonometric functions. There's no application for general term of Fibonacci series, What would someone do by even finding 100th term of Fibonacci series and it is even difficult to solve. And formula for arc trigo functions are also useless as in their formulas there is another trigo function whose value we must know and in other formula for arc trigo function there is iota which is very difficult to eliminate. So it's not necessary that a formula should have an application.
@@mathevengers1131 No Sir , Any mathematical formulas like Fibonacci series , inverse trigo functions or any other you name it carries application , i.e. in computer programming we have to use algorithms based on Fibonacci series , any programmed machine (complex or simple) have to use sin , cos or inverse functions to perform the required actions . There might be few other applications also ! And if you agree with me so far , find out what your equation can be used for/in . I think it's a huge thing to create your own equation ( "PROVIDED IT WORKS" , you mentioned that in your comment so I believe you 🤠) .
Or you could write π - theta + 2nπ and then simplify by factoring a π out of the first and third term.
Which is the video about sin not being injective? I would like to see it.
ua-cam.com/video/VlorTPQ6ZBA/v-deo.html this one :)
Thank you so much!
I feel like crying just by seeing the question omg
are you ok? that’s not normal
u good homie?
In an exam, would there be extra time to allow students to list every separate value ;)
Sir can you explain some times we take
Sin(theta) = theta ?
How is this possible please explain !
sin(t) = t iff t = 0. However, if t ~ 0, then sin(t) ~ t. Here, "~" means "is approximately equal to." Therefore, if, for example, t = 0.000001, then you in practical applications, you can simply have sin(t) ~= t, because the difference between the two is so small as to be immeasurable.
@@angelmendez-rivera351 thnx 👍
@6:01 how is that you can legitimately cancel sin? where is before we could not?
Because now all the possible solutions have been accounted for by the addition of 2·n·π, so noninjectivity will no longer be an issue.
@@angelmendez-rivera351 thankyou will try and understand that
@@angelmendez-rivera351 i dont understand it at all
@@luisaguilera3242 If you want to solve the equation x^2 = y^2 for x, then you want to take the square root, and you get sqrt(x^2) = sqrt(y^2), which is equivalent to |x| = |y|. This equation has two solutions in R, because x |-> |x|: R -> R+ is an even function, so it is not injective. How would you deal with this equation? You would realize that R+ and R- partition the set R into two "opposite" sets, with 0 being double counted, and when the domain is restricted into either of the sets, the restricted function does become injective in both intervals. So you break the equation-solving process into two cases: x >= 0, and x =< 0. If x >= 0, then |x| = x, so x = |y|, and the equation has already been solved. If x =< 0, then |x| = -x, so -x = |y|, thus x = -|y|, and the equation has been solved. Since there are no other cases to analyze, you are done, and the equation has two solutions, x = |y| or x = -|y|. Here, you can also rewrite this as x = y or x = -y, since it you get both solutions regardless of whether y =< 0 or y >= 0.
However, there is one way to concisely express both solutions in one expression: this is encoded by realizing that |x| = |-x|, so you can write |(-1)^n·x| in place of |x|, allowing yourself to substitute integer values of n. Then you get |(-1)^n·x| = |y|, and if you "cancel" (not the correct terminology, but this is what you and others refer to it as, so we will go along with it) the absolute values, you get (-1)^n·x = y, which implies x = (-1)^n·y. This gives you both solutions in a single expression, but why does this work? Because the breakdown of cases into separate equations is encoded in the indexed expression (-1)^n. By substituting different parities of n, you get every case allowed. When solving sin(π·x) = sin(x), the same concept applies. x |-> sin(x) : R -> [-1, +1] is not injective. However, if you restrict the domain to an interval [m·π - π/2, m·π + π/2], where m is an arbitrary integer, then the restricted function is injective for each m in a mutually exclusive fashion, except for endpoints, which are double counted. Furthermore, the function has period 2·π, and for odd m in the above intervals, the function has the opposite growth as with even m, so you can consider sin(π·x) = sin(x - m·π) for even m and sin(π·x) = sin(x - m·π) for odd m separately. If m is even, then m = -2·n, so sin(π·x) = sin(x + 2·n·π), and if m is odd, then m = -2·n - 1, so sin(π·x) = sin(x + 2·n·π + π), but sin(t + π) = sin(-t) for any t, so sin(π·x) = sin[-(x + 2·n·π)]. Thus the two cases are sin(π·x) = sin(x + 2·n·π) or sin(π·x) = sin[-(x + 2·n·π)], with n being an arbitrary integer, and this is precisely what the video did. Each value of n will give a unique interval, in such a way that both cases combined partition the set R into intervals in which, when you restrict the function's domain to those intervals, the function becomes injective, and n is indexing those intervals. This why the function becomes "cancellable" once you do break the equation into those two cases: by doing so, the function becomes "injective for each value of n" in both cases.
Answer is (-1)^n + 2 n 0
me yesterday: eh, I'll watch this tomorrow.
*watches it the next day after mocking the AMC12A*
me today: should've watched this video...
gracias bprd... eres un crack bro
Who else immediately thought of the trivial solution theta = 0?
I do that all the time. *sees a complex differential equation* oh y=0
Im having trouble solving an equation similar to sin(4+x)=2sin(2+x) Is there any hope? (The numbers were different but I dont remember them)
@ゴゴ Joji Joestar ゴゴ That worked so well! Thankyousomuch BESTO FRENDO
Pretty easy with sin(180-x)=sin(x) but it was fun!
For question 2, shouldn’t there be more cases?
The 2 in sin(2 theta): Oh no mesa disappearing
Hello! És sikerült! Világegyetem protokol l !
Can I use sum to product formula in the second question?
Where are u from bro?
Outstanding way of teaching
I haven't really understand why when the input is different we can't cancel the sin ?? We cancel the sin and we add 2kpi
if sin(theta) = sin(alpha)
then,
theta = n * pi() + (-1)^(n) pi() * alpha
@Are you sure ABOUT THAT ? alpha is (pi theta) and n is 1
I'm in Precalc, I may be wrong, but for number 2, could we just say that theta is 0?
That's Good 🔥
I did a method that involved converting sin(pi x) into (e^(i pi x) - e^-(i pi x))/2 which seems like it should still work. you end up being able to factorise stuff and doing it as a quadratic.
I only wanted to get one of the solutions and i got pi/(pi+1), but it might still be possible to get the other solutions?
the method is:
e^(i pi x) - e^(-i pi x) = e ^(ix) - e ^-(ix)
(e^ix)^pi - (e^ix)^-pi - e^ix + (e^ix)^-1 = 0
substitute y = e^ix
y^pi - y + (1/y - 1/y^pi) = 0
y^pi - y + (y^pi-y)/y^(pi+1) = 0
y^(pi+1) (y^pi - y) + (y^pi - y) = 0
(y^(pi +1) + 1)(y^pi - y)=0
y^(pi+1) +1 = 0 or y^pi -y = 0
y*(pi+1) = -1 or y^(pi-1) = 1
ln(-1) on wolframalpha give i pi which is the result that lead to x= pi/(pi+1), and the other one with ln(1) leads to x = 0.
I think if i did the whole thing taking into account the multiple solutions with the 2pi n stuff it would have lead to the general solutions maybe? I'm iffy about if i should have been allowed to do the ln(-1) = i *pi thing.
Is it that hard?
sin(πθ)=sinθ
+)case 1: πθ=θ+2πn
+)case 2: πθ=π-θ+2πn
Then solve them
Ezy hv been doin calculus for 2 years, answer is 2
Simply let theta be zero so sin(n theta) will always be sin ( theta ) which will give you 0
Is this possible seriously or its just for fun as i am new on your channel please i need an answer for myself
The point is u can cancel out the sine once you consider those two cases. 😃
Why do you use a blue pen? Are you allowed to do that?
No, not really. Every time I use it, I lose a subscriber. : (
lol
Use the identity sina-sinb=2sin...
Why can we cancel 'sine' in example 2?
How to solve integral of (1/underroot(1+x7) dx from 1 to infinity? Please give the solution.
I didn’t understand the part with adding 2n(pi) does anyone have any videos that explain it more in depth? ty
The sine function has a period of 2·π, hence if x = a solves sin(x) = a, then necessarily, a + 2·n·π solves it too; otherwise, it would contradict the periodicity.
How did you do that? amazing
Yo can ye do old IITJEE sums? You will get millions of views with the right thumbnail and Indian in the title
This channel is for math lovers not for some nerds that do not understand the beauty of it.Our mindset has to change.
Also,he did solve some iit problems.
@@tonyk7509 You mean to say JEE problems are for "nerds who dont understand beauty of maths"
@@yatogami7393 Thanks for correcting me sir.YES.
@@yatogami7393 Even students have become the promoters of business based education. Pathetic. 🤬🤬🤬.
Could you please tell me the application of this expression
The equations themselves are not important. The critical thinking involved in solving them is what matters here.
@@angelmendez-rivera351 we are growing a tree for fruits only
Solutions to: (x pi) - x + ((x pi)^3 - x^3)/3! + ((x pi)^5 - x^5)/5! - ((x pi)^7 - x^7)/7! ...
I just use 2π periodicity property of the sin(X) function:
sin(πθ)☰sin(θ+2nπ) gives BPRP's first solution
then consider sin(X) is an odd function, i.e. sin(X)☰-sin(-X)☰sin(π-X)
sin(πθ)☰sin(π-θ+2nπ) gives second BPRP solution.
No need for them pesky CAST circles.
I like how he switches pen
Hey can you make a video about Ramanujan pi series?
is there a way to get the 2 or 4 possible solutions for and equation on the casio fx 115
I did the second equal the first most i took Φ=πθ/2 And then I decided to the 1 is right to do this?
Is it possible to solve the second equation using the euler's formulas ?
It we put this equation as :
( e^(iOpi) - e^(-iOpi) )/2i = sin (O), where O is theta ; then we simplify as
( (e^(i*pi))^O - (e^(-i*pi))^O)/2i = sin O
( (-1)^O - (-1)^O )/2i = sin O
So we get
Sin O = 0
So O = n*pi, n being an integer.
Is this way correct or am i wrong somewhere ?
what if you replace pi*theta with pi*(theta+2n)? would that work?
But if you switch the side, isn’t the sin theta negative now
Is it possible to sum up a single term infinitely many times and have it equal to pi? I don’t mean sum up different terms. I mean sum up the same number over and over. It might be more reasonable if we let the signs alternate but what if we don’t even allow that to happen. Is it possible?
Not sure how to represent this but one idea would be to take an infinite sum of An/n where A is a constant and n is summed up from 1 to infinity.
Answer i got is theta = √20
I know its not possible but i got this answer from fourier series.
At 9:04 why not just say pi - theta + 2n*pi?