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Cosine equation solution set in an interval
Given the algebraic solution set for a cosine equation, find which solutions fall within an interval. Created by Sal Khan.
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- The interval (-pie/2,0),is arbitrary?. Could he have chosen any other interval?(11 votes)
- other people have asked the same problem check their answers and hopefully that will help you. I am asking myself the same question so I dont trust myself to answer(3 votes)
- 1:46, Sal encourages us to evaluate a portion of the equation on a calculator. Instead of coming up with approximately 0.22, I keep ending up at approximately 12.44. Can anyone explain why?(7 votes)
- 0.22 radians = 12.44°
Change the calculator mode to "RAD" instead of "DEG".(12 votes)
- Where did the interval [-π/2, 0] come from?(6 votes)
- That was just part of the given of the problem which puts our cos in the 3rd and 4th quadrant. So cos-1(x) will have positive values in the 1st and 4th quadrants and negative values in the 2nd and 3rd quadrants. Sal uses this limit to say we are only thinking about what is happening in the 3rd and 4th quadrant.(4 votes)
- Hi, why did SAL take [-Pi/2, 0] as the limit. Shouldn't the limit for the COS function be between 0 and Pi (0 and 3.14) ?
Considering the given value of -1/6 the coordinate turns out to be falling on second quadrant. But the same coordinate can be achieved on the vertically opposite side that is third quadrant. Since both these quadrants (second and third) give the stated value in the question (i.e. -1/6), I was assuming we should have restricted to one of these quadrant. Can someone help clarify this?(5 votes)
- The COS function (adj/hyp) on the unit circle is positive in the 1st and 4th quadrants, and negative in the 2nd and 3rd quadrants. If you went between the 1st and 2nd, the cos(first)=-cos(second), so you get two different answers.(2 votes)
- @0:52, why did he take -pi/2, was it just an arbitrary interval?(5 votes)
- 1/8*acos(-1/6) is 12.44??(2 votes)
- According to my calculator, 1/8 * acos(-1/6) is close to 12.45 in degrees and close to 0.22 in radians.(2 votes)
- The interval (-pie/2,0) refers to the left half of the unit circle?(2 votes)
- how did we know cos in the 3rd and 4th quadrant and limit is 4th quadrant what about 3rd?(1 vote)
- It's because of the interval. As x ranges from -pi/2 to 0, cosine is confined to be in the fourth quadrant.(1 vote)
- why is there a -pie/4n on one side but a +pie/4n on the other side? From what I understand the -pie/4 is because the interval is from [pie/2,0] where the pie/4 isn't included right? but I'm confused about the +pie/4(1 vote)
- 𝑛 is an arbitrary integer, which means that −𝑛 is also an arbitrary integer.
Thereby, +𝑛⋅𝜋∕4 and −𝑛⋅𝜋∕4 can be used interchangeably, because they are both arbitrary members of the same set.
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In the video Sal writes
𝑥 = 1∕8⋅cos⁻¹(−1∕6) − 𝑛⋅𝜋∕4
He then restricts 𝑥, which in turn restricts 𝑛 to 𝑛 = 1 or 𝑛 = 2,
which give 𝑥 ≈ −0.57 or 𝑥 ≈ −1.35
If he had instead written
𝑥 = 1∕8⋅cos⁻¹(−1∕6) + 𝑛⋅𝜋∕4
he would have gotten 𝑛 = −1 or 𝑛 = −2,
but he would still get the same 𝑥-values.(1 vote)
- I'm having a hard time following from the last video to this one. In the last video he specified that the value would be in the 2nd or 3rd quadrants (also using a diagram) because of the x value of -1/6. In this video, there is an added restriction of [-pi/2, 0] putting us in the 4th quadrant. Isn't there no solution in the 4th quadrant? If it's because of the division by 8, why are there multiple solutions +2(pi)n in the 4th quadrant?(1 vote)
- Remember how cos, sin, tan were defined in the unit circle and ASTC.
The negativity only holds for when the argument is x not 8x as in this example and the previous video.
This because the 8x is performing a horizontal stretch on the graph. More particularly it is compressing it by factor of 8. This means graph is going from negative values to positive values and vice versa a lot more quicker.(1 vote)
- [Instructor] In a previous video, we established the entire solution set for the following equation. And we saw that all the x's that can satisfy this equation are a combination of these x's and these x's here. The reason why I'm referring to each of them as numerous xs is that for any integer value of n, you'll get another solution. For any integer value of n, you'll get another solution. What I wanna do in this video is to make things a little bit more concrete. And the way that we're going to do it is by exploring all of the x values that satisfy this equation that sit in the closed interval from negative pi over two to zero. So I encourage you like always, pause this video and have a go at it by yourself before we work through it together. All right, now let's work through this together. So the first helpful thing is we have these algebraic expressions. We have things written in terms of pi. Let's approximate them all in terms of decimals. So even pi over two, we can approximate that. Let's see, if pi is approximately 3.14, half of that is approximately 1.57, so we could say this is approximately the closed interval from -1.57 to zero. - 1.57 isn't exactly negative pi over two, but it'll hopefully be suitable for what we're trying to do here. And now let's see if we can write the different parts of these expressions, or at least approximate them as decimals. So this could be rewritten as x is approximately, if you were to take 1/8 times the inverse cosine of -1/6, I encourage you to verify this on your own on a calculator, you would get that that's approximately 0.22. And then pi over four is approximately 0.785. So this expression would be approximately 0.22 minus 0.785 times n, where n could be any integer. And then this one over here on the right, let me do that in this yellow, x could be approximately equal to, well if this evaluates to approximately 0.22, then this is just the negative of it, so it's going to be -0.22. And then it's plus what approximately pi over four is, so 0.785n. And now what we could do is just try different n's and see if we're starting above or below this interval, and then see which of the x values actually fall in this interval. So let's just start here. If we just start at n equals zero, actually why don't I set up a little table here, we have n here and if we have the x value here, when n is zero, well, then you don't see this term, and you just get approximately 0.22. Now let's compare that to the interval. The upper bound of that interval is zero. So this does not sit in the interval. So this is too high and we would want to define the x's that sit in the interval. We wanna find lower values. So it's good that here, where you're subtracting 0.785, so I would use positive integer values of n to decrease this 0.22 here. So when n equals one, we would subtract 0.785 from that, and I'll round all of these to the hundredths place, and that would get us to -0.57, and that does sit in the interval. So this looks good. So this would be a solution in that interval right over here. And let's try n equals two. So we would subtract 0.785 again, and that would get us to -1.35, not 25, 35, and that also sits in the interval. It's larger than -1.57, so that looks good. Let's subtract 0.785 again, when n equals three, that would get us -2.14. Well, that's all of a sudden out of the interval because that's below the lower bound here. So this is too low. So using this expression, we've been able to find two x values that sit in the interval that we care about. Now let's use these x values right over here and I'll set up another table. So, let's see we have our n and then we have our x values. So let's start with n equals zero 'cause that's easy to compute, and then this term would go away, and we'd have -0.22, and that's actually in this interval here, it's below zero, it's larger than -1.57, so that one checks out. But now to really explore, we have to go in both directions. We have to increase it or decrease it. So if we wanted to increase it, we could have a situation where n equals one. So if n equals one, we're gonna add 0.785 to this. Now you immediately know that that's going to be a positive value, if you computed it, it'd be 0.57, which is larger than zero, so this is too high. So now we could try going lower than -0.22 by having negative values of n. So if n is equal to -1, that means we're subtracting 0.785 from this right over here which would get us to -1.01. Well, that one works out, so that's in our interval. And now let's subtract 0.785 again. So I'll have n equals -2. And so if I subtract 0.785 again, I could round that to -1.79, which is lower than -1.57, so it's out of our interval, so it's too low. So all of the x values that are in our interval that satisfy this equation are these two right over here. And this one and this one, and we are done.