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Sal finds the expressions that together represent all possible solutions to the equation sin(x)=1/3. Created by Sal Khan.
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- At5:35, is there some reason that -0.34+pi*n is left out of the answer? It seems like it belongs as well.(92 votes)
- Well at5:25Sal explains why -0.34+2pi*n won't work: sin(-0.34)= -1/3 and not 1/3 and if we add a multiple of 2pi we are gonna wind up there again. But if we add n pi to it there are two cases which might happen: we choose an uneven n and we are gonna end up with (pi - 0.34) + (n-1)pi, take sine of that: sin(2.80) = 1/3, that (n-1) is always gonna be even so it won't change anything, however when we chose an even n, we will get -0.34 + 2pi*n, now sine(-0.34) = -1/3, and we don't want that. So for half of the Ns we would get a wrong answer.(105 votes)
- Why am i getting 19.4712 when im taking sin^-1 (1/3) and not .3398 like sal https://www.youtube.com/watch?v=NC7iWEQ9Kug#t=172 ? (got a casio fx-9750 if that matter)
Thanks in advance(22 votes)
- 19.4712 is the angle in degrees.
0.3398 is the angle in radians.
So, you need to set your calculator to radians.(90 votes)
- Is it just me, or are these two videos not that useful for solving the problems given in the practice. I've missed every one of them so far, and feel like the videos failed to cover something. The explanations given when I click "tips" or whatever are mystifying. They mention identities that are not covered in these videos.(34 votes)
- I was having the same issue and this video helped immensely. https://www.youtube.com/watch?v=eRigtAyW_DI
(If for some reason that link doesn't work it's youtube: Solving a Sinusoidal Equation (general))(12 votes)
- I don't understand this, previously Sal said that arcsin outputs can only be in the first and fourth quadrant, how come we take 2.8 radians as an answer if it's in the second quadrant?(30 votes)
- Since x is the unknown value, we are looking for x. With inverse trig functions, x represents the domain of a "vanilla" sin function and the range of the inverse function: they are the same thing.
Now although we need to restrict the range of arcsin to get a valid inverse function, of course other angle options are true. We need to be careful to see what the question is asking us to find. The solution set of an equation will contain all possible answers for that equation.
https://youtu.be/NC7iWEQ9Kug?t=19 on the unit circle, the principal value is in QI, being in the range 'negate' 'pi/2' 'less than or equal to 'theta' 'less than or equal to' 'pi/2'.(3 votes)
- Do you tell the 1/3 is in radians just because there is no degree notation?(6 votes)
- Yes. Basically, once you get introduced to radians, that is the end of using degrees. All higher math is done in radians only.
There is no unit in radians because it is just a ratio. The angle in radians is the ratio of the arc length to the radius.(21 votes)
- I'm confused as to when you're supposed to limit the range of the inverse trig functions. In this video, shouldn't 2.80 + 2pi(n) be invalid because sin^-1 is limited to [-pi/2, pi/2]?(5 votes)
- The range of the arcsine function has to be bounded because otherwise the input would map to several outputs.
But this doesn't mean that the domain of the sine function must be bounded because several inputs can map to the same output.
So, if we are asked what arcsin(1) is, then the answer is 𝜋∕2,
but if we are asked what value of 𝑥 gives us sin(𝑥) = 1, then the answer would be 𝜋∕2 + 𝑛∙2𝜋, where 𝑛 is an integer.(13 votes)
- How would would solve something that is this complicating?
- OK, hmmm
2sin²x + cos x - 1 = 0
The key here would be to use the trig identities.
The most basic one is
cos²x + sin²x = 1
(This is sometimes called the Pythagorean Identity)
How does it help? It gets rid of the messy 2sin²x that makes your equation such a complicated thing to solve.
First solve the identity for sin²x
sin²x= 1 - cos²x
so, 2 (
sin²x) = 2(1 - cos²x)
Now we plug that into your challenge problem:
1 - cos²x) + cos x - 1 = 0
2 - 2cos²x + cosx - 1 = 0
- 2cos²x + cosx - 1 + 2 = 0
- 2cos²x + cosx + 1 = 0 → switch signs (multiply by -1)
2cos²x - cosx - 1 = 0
this is the same pattern as 2a² - a - 1, which factors into (2a + 1)(a - 1)
2cos²x - cosx - 1 = 0 factors into (2cosx + 1)(cosx - 1) = 0
Now you can solve using the zero property
(2cosx + 1) = 0; so 2cosx = -1 and cosx = -1/2
(cosx - 1) = 0; so cosx = 1
Now, finally, you can use arccos to solve for possible x values (measure of the angle x).
If cosx = -1/2, arccos (-1/2) = 120 degrees or 2π/3
If cosx = 1, arccos (1) = 0 degrees or 0 radians
The rest of the answer depends on exactly what they were asking--the principal value or all possible values.
Hope that helps(18 votes)
- hey we just set a limit to values of sin from pi/2 to 3pi/2 then why are we considering the angle in 2 quadrant at3:19(6 votes)
- The value of sin is set to 1/3, which is within the domain of sin values. However, the angle measures that make up a sin value can be in any quadrant, because sin is only a y value.(4 votes)
- sir, -0.34+(pi)n should also be an answer(2 votes)
- I'm incredibly confused when it cones to solving these equations. I've been doing practice questions for over an hour trying to pick it apart.
First it asks me to select one or more expressions that together represent all solutions to the equation. And then it gives me sin(x)=0.4 to work with and a bunch of answers (e.g. 21.80∘ +n⋅360∘, 66.42∘+n⋅180∘) How do I know when to append the n*180 or n*360? When I click a hint, it shoots me a rang of -180∘ < x < 180.
Where did it get that range from? That's not even considering when I'm using radians. How can I tell if a decimal is in the range of, say, 3pi/2 and 9pi/2?
I'm incredibly confused where they are getting the ranges from, and I've never encountered these "identities" of sin, cos, tan, etc.
Any help would be greatly appreciated.
- This might help.
So n is used to represent multiple solutions.
I presume when you are referring to 2pi radians you are referring to period.
Suppose we have the question: 3cos(2pi*t/3) + 10 = 11
Don't be overwhelmed if you don't understand where these theorems come from. They just come from the unit circle. Nothing is stopping you from applying these theorems even if you don't know the proof.
To solve these kinds of question we need to know the following:
The general solution for:
cos(theta) = a
2*n*pi +/- alpha (1)
sin theta = a
theta = n*pi +(-1)^n * alpha (2)
tan(theta) = a
theta = n*pi + alpha (3)
Of course you are not going to be able to apply theorem (1)/(2)/(3) in these types of questions because of the coefficent of t is 2pi/3 not 1.
So let u = 2pi/3 * t. With this substitution you can now apply the theorem as the coefficent is now 1 which is coefficent of theta.
So you have
u = n*pi +(-1)^n * alpha # note alpha = sin^-1(a)
2pi/3 * t = n*pi +(-1)^n * alpha # make t the subject and your done.
Note: To save time in a exam you don't need to actually substitute u however if you feel uncomfortable you may do so.(1 vote)
Voiceover:Which of these are contained in the solution set to sine of X is equal to 1/3? Answers should be rounded to the nearest hundredths. Select all that apply. I encourage you to pause the video right now and work on it on your own. I'm assuming you've given a go at it. Let's think about what this is asking. They're asking what are the X values? What is the solution set? What are the possible X values where sine of X is equal to 1/3? To help us visualize this, let's draw a unit circle. That's my Y axis. This right over here is my X axis. This is X set one. This is Y positive one. Negative one along the X axis, and negative one on the Y axis. The unit circle, I'm going to center it at zero. It's going to have a radius of one, a radius of one and we just have to remind ourselves what the unit circle definition of the sine function is. If we have some angle, one side of the angle is going to be a ray along the positive X axis, if we do this in a color you can see, along the positive X axis. Then the other side, so let's see, this is our angle right over here. Let's say that's some angle theta. The sine of this angle is going to be the Y value of where this ray intersects the unit circle. This right over here, that is going to be sine of theta. With that review out of the way, let's think about what X values, and we're assuming we're dealing in radians. What X values when if I take the sine of it are going to give me 1/3? When does Y equal 1/3 along the unit circle? That's 2/3, 1/3 right over here. We see it equals 1/3 exactly two places, here and here. There's two angles where, or at least two, if we just take one or two on each pass of the unit circle. Then we can keep adding multiples of two pi to get as many as we want. We see just on the unit circle we could have this angle. We could have this angle right over here. Or we could go all the way around to that angle right over there. Then we could add any multiple of two pi to those angles to get other angles that would also work where if I took the sine of them I would get 1/3. Now let's think about what these are. Here we can take our calculator out, and we could take the inverse sine of 1/3. Let's do that. The inverse sine of 1 over 3. We have to remember what the range of the inverse sine function is. It's going to give us a value between negative pi over 2 and pi over 2, so a value that sticks us in either the first or the fourth quadrant if we're thinking about the unit circle right over here. We see that gave us zero point, if we round to the nearest hundredth, 34. Essentially they've given us this value. They've given us 0.34. That's this angle right over here. How did I know that? Well, it's a positive value. It's greater than zero, but it's less than pi over two. Pi is 3.14, so pi over two is going to be 1.57 and we can go on and on and on so this right over here is 0.34 radians. But what would this thing over here be? It's going to be whatever, if we go to the negative X axis and we subtract 0.34 so we subtracted 0.34. This is 0.34 We're going to get to this angle. It's going to be if we take pi minus the previous answer, it gets us we round to the nearest hundredth, is 2.8 radians. This is 0.34 radians, and then this one, let me do it in this purple color, this one right over here if we were to go all the way around, it's pi minus 0.34 which is 2.80 radians rounding to the nearest hundredth. Now that's not all of the values. We can add multiples of two pi to each of these. So 2.80 plus any multiple of two pi's, so two pi N where N is an integer. N is an integer. Or we could take 0.34 and add any multiple of two pi. So two pi N where N is an integer. Our solution set here just to rewrite it kind of outside of this messiness is going to be 2.80 radians plus two pi N where N is an integer, and 0.34 plus two pi N where N is an integer. Let's see which of these are at least a subset of this. We look at 0.34 plus two pi N where N is an integer. That's exactly what we wrote over here. That's 0.34. If N is a positive integer we'll go around this way and we keep getting back to the same point. If it's a negative integer we go around that way. We keep getting to the same point, but that's definitely in the solution set. 0.34 plus pi N for N an integer. So if we have 0.34 and if we were to not add two pi, but just pi where would we get to? Well, we would get to right over there. The sine of this isn't going to be positive 1/3. It's going to be negative 1/3. So we could rule that out. Negative 0.34, well that's this angle right over here. The sine of that's going to be negative 1/3. If you add a multiple of two pi to that, you're still going to get negative 1/3 so that doesn't work. Same thing for this one right over here. 2.8 plus two pi N, that's what we wrote right over here. 2 point going all the way to 2.8 and any multiple of it is going to get you back to that same point so that one works. 2.8 plus pi N so if you're here and if you added pi you're going to get over here, and the sine of that isn't going to be positive 1/3. It's going to be negative 1/3 so we can rule this one out as well. These are the only two apply, and if you actually take them together you have the entire solution set to this equation right over here.