Wanna learn more interesting math? Visit brilliant.org/blackpenredpen/​ and start exploring. Try their trig & precalc courses after this video!

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

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.

To quote a song Pre calculus didn't help me prepare for calculus, for calculus, help meee

Woke: Theta = 0

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

I solved it by thinking of the intersection of a horizontal line with the sine curve. That allowed me to catch both cases.

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)

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

Bring more such problems, highly needed.

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

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 :(.

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 .

8:54 The angle you drew is -pi not pi, but at the end it doesn't make any difference. Thanks for the video !

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.

I subbed in cosθ for its complex exponential identity, used your quadratic method from the last sinz video and got ∓iπ/3.

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)}.

1. Theta=60 2.Theta=180/(π+1) You may check in calculator for 1.sin(theta)=sin(2theta) 2.sin(πtheta) =sin(theta)

his beard is the source of his power

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

Hey,I have one π(theta)=nπ + ((-1)^n)theta Where n is an integer

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).

That evil laugh at 6:46 😂😂😂

Case two a very easy approach is to use sinA = sin(pi-A) by identity, with A being (pi)(theta)....works quickly

Thank you so much!

gracias bprd... eres un crack bro

we can also use SinA-SinB = 2Cos((A+B)/2)Sin((A-B)/2) then compute..Lil bit of overkill i guess ...

could we use sum to product for question 2 as well?

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?

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.....

it is easy when 1you understand what is a circle and 2 you understand what is a sinus. nice job done.

That's Good 🔥

Can I use sum to product formula in the second question?

Please provide the link to the video of sine not being injective

Where are u from bro? Outstanding way of teaching

sin x= sin y if x-y=2*pi*n or x+y=pi + 2*pi*n for some integer n

How did you do that? amazing

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.

Hey , can you plz solve this , integration of (sec^2 x - 7) / sin^7 x dx

Mathematicians have a (a+bi) sense of humour.

In an exam, would there be extra time to allow students to list every separate value ;)

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

is there a way to get the 2 or 4 possible solutions for and equation on the casio fx 115

Hey can you make a video about Ramanujan pi series?

Or you could write π - theta + 2nπ and then simplify by factoring a π out of the first and third term.

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?

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])

Use sina-sinb=2cos(a+b)/2sin(a-b)/2

First result cos case could also be written Theta = 2m.pi +/- pi/3

Which is the video about sin not being injective? I would like to see it.

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]

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.

How to solve integral of (1/underroot(1+x7) dx from 1 to infinity? Please give the solution.

cos( pi * x ) = cos( x ) x = 2 * n * pi / ( pi - 1 ) and x = 2 * n * pi / ( pi + 1)

what if you replace pi*theta with pi*(theta+2n)? would that work?

Hello! És sikerült! Világegyetem protokol l !

How to cancel your sin? Repent.

For question 2, shouldn’t there be more cases?

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

The 2 in sin(2 theta): Oh no mesa disappearing

Is this possible seriously or its just for fun as i am new on your channel please i need an answer for myself

I didn’t understand the part with adding 2n(pi) does anyone have any videos that explain it more in depth? ty

I like how he switches pen

Sir can you explain some times we take Sin(theta) = theta ? How is this possible please explain !

Pretty easy with sin(180-x)=sin(x) but it was fun!

Why can we cancel 'sine' in example 2?

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)

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.

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 ?

I feel like crying just by seeing the question omg

me yesterday: eh, I'll watch this tomorrow. *watches it the next day after mocking the AMC12A* me today: should've watched this video...

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.

Sin(πtheta)= -sin(theta +nπ) also a case

I'm in Precalc, I may be wrong, but for number 2, could we just say that theta is 0?

Now we are adults so we'll say π/3 . Correct 👍🏿

Could you please tell me the application of this expression

I want the example f(x) and f’(x) same other e^x and 0?

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?

Dat goatee.

@6:01 how is that you can legitimately cancel sin? where is before we could not?

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.

Better thumbnail XD

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

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

Hey sir i am from india My question is How to intigral of 1/e^tanx dx ?

Ezy hv been doin calculus for 2 years, answer is 2

*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.

Who else immediately thought of the trivial solution theta = 0?

🔥

When can we legitimately cancel sin? I didn't get it

Answer is (-1)^n + 2 n 0

I have this equation x to 4 power -13xsq +36=0

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.

At 9:04 why not just say pi - theta + 2n*pi?

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

Is it that hard? sin(πθ)=sinθ +)case 1: πθ=θ+2πn +)case 2: πθ=π-θ+2πn Then solve them

how can we solve this. ? ( (-3x²-2x+8)/(3x-4))