how did you get Euler's e^(itheta)=cos(theta)+ isin(theta)

It is a very long process. It comes form Polynomial approximations of Functions...Check it out in the Calculus Playlist.

It kind of makes sense that the "i theta" got replaced by x, since theta could have used imaginary numbers only to create real values, however, how does the i in the denominator of the sin function go away.

The definition of hyperbolic sine doesn't include an imaginary part simply because it isn't how it was defined. Like regular sine and cosine are respectively the distances from the x and y axes for theta on the unit circle: y² + x² = 1, hyperbolic sine and cosine are the distances from the axes on the unit hyperbola (hence their names): y² - x² = 1. Hyperbolic functions were not developed by replacing theta by x * i^3 (which would cancel out the other i and get x). They only coincidentally (or not so coincidentally, as the function of unit hyperbola is rather similar to that of the unit circle) happened to result in similar exponential expressions. I hope that helps.

At 1:45, Why does cosine theta turn positive?

If you remember your trig value functions from the unit circle such as sin(pi/3) = sqrt(3)/2 etc. Then you will know that the cosine function is positive in the 1st and 4th quadrants and negative in the 2nd and 3rd quadrants (cosine is the x value in the unit circle; the 1st and 4th quadrants have positive x values). Therefore, let's say you have cos(pi/3) and cos(-pi/3). Pi/3 is in the 1st quadrant and -pi/3 is in the 4th quadrant. Because they both have positive x values, the cosine will be positive for both. Now let's say you have cos(5pi/6) and cos(-5pi/6). 5pi/6 is in the 2nd quadrant and -5pi/6 is in the 3rd quadrant. Because they both have negative x values, then the cosine will be negative for both.

At 1:46, how can cos(-theta) be the same as cos(theta)?

Cosine is an even function. That means that opposite inputs have the same output. You can see this from its graph because it is symmetric about the y-axis.

Is there a relationship between i, theta, and x?

e = lim n --> ∞ of (1 + 1/n)^n or lim n --> 0 (1 + n)^(1/n) i = √(-1) pi = ratio between a circle's circumference and its diameter x = theta, the angle in radians Euler's Formula usually relates e, i, pi, 1, and 0, but I think that it's better to understand that it relates more than just pi. Pi, as evaluated in trig functions, is oftentimes a neat answer, making the relationship between trigonometry and complex numbers beautiful. That's a mouthful to understand at once, but you'll get it when you get more in depth into math. The derivation for Euler's Formula cannot be done with the Precalculus's Toolbox of Mathematics provided to students (at least, not easily!) Euler's Formula: e^(ix) = cos(x) + i*sin(x) This is why when x=pi e^(i*pi) = cos(pi) + i*sin(pi) Well, sin(pi) is zero, so that get rids of the i, too. cos(pi) is -1 (look at the unit circle). So, e^(i*pi)=-1 Or, e^(i*pi) + 1 = 0 Ta da!

what is e and i? are they normal variables?

Sal explains what e is: https://www.khanacademy.org/science/core-finance/interest-tutorial/cont-comp-int-and-e/v/introduction-to-compound-interest-and-e Sal explains what i is: https://www.khanacademy.org/math/trigonometry/imaginary_complex_precalc/i_precalc/v/introduction-to-i-and-imaginary-numbers I hope this helps!

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Course: Staging content lifeboat > Unit 17

Lesson 4: Intro to hyperbolic trig functions

Hyperbolic function inspiration

Name: Hyperbolic function inspiration
Uploaded: 2016-03-03T01:11:34Z
Description: Exploring a motivation for even defining hyperbolic sine and cosine

Google Classroom

Exploring a motivation for even defining hyperbolic sine and cosine. Created by Sal Khan.

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anthony
Posted 11 years ago. Direct link to anthony's post “how did you get Euler's e...”
how did you get Euler's e^(itheta)=cos(theta)+ isin(theta)
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(27 votes)
Answer
- Anish Puthuraya
  Posted 11 years ago. Direct link to Anish Puthuraya's post “It is a very long process...”
  It is a very long process. It comes form Polynomial approximations of Functions...Check it out in the Calculus Playlist.
  Comment on Anish Puthuraya's post “It is a very long process...”
  (44 votes)
Jon Klatt
Posted 11 years ago. Direct link to Jon Klatt's post “It kind of makes sense th...”
It kind of makes sense that the "i theta" got replaced by x, since theta could have used imaginary numbers only to create real values, however, how does the i in the denominator of the sin function go away.
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(14 votes)
Answer
- Phi
  Posted 11 years ago. Direct link to Phi's post “The definition of hyperbo...”
  The definition of hyperbolic sine doesn't include an imaginary part simply because it isn't how it was defined.
  
  Like regular sine and cosine are respectively the distances from the x and y axes for theta on the unit circle: y² + x² = 1, hyperbolic sine and cosine are the distances from the axes on the unit hyperbola (hence their names): y² - x² = 1.
  
  Hyperbolic functions were not developed by replacing theta by x * i^3 (which would cancel out the other i and get x). They only coincidentally (or not so coincidentally, as the function of unit hyperbola is rather similar to that of the unit circle) happened to result in similar exponential expressions.
  
  I hope that helps.
  Button navigates to signup page
  (15 votes)
box 0f rox
Posted 11 years ago. Direct link to box 0f rox's post “At 1:45, Why does cosine ...”
At
1:45
, Why does cosine theta turn positive?
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(5 votes)
Answer
- Matthew Kuo
  Posted 11 years ago. Direct link to Matthew Kuo's post “If you remember your trig...”
  If you remember your trig value functions from the unit circle such as sin(pi/3) = sqrt(3)/2 etc. Then you will know that the cosine function is positive in the 1st and 4th quadrants and negative in the 2nd and 3rd quadrants (cosine is the x value in the unit circle; the 1st and 4th quadrants have positive x values). Therefore, let's say you have cos(pi/3) and cos(-pi/3). Pi/3 is in the 1st quadrant and -pi/3 is in the 4th quadrant. Because they both have positive x values, the cosine will be positive for both. Now let's say you have cos(5pi/6) and cos(-5pi/6). 5pi/6 is in the 2nd quadrant and -5pi/6 is in the 3rd quadrant. Because they both have negative x values, then the cosine will be negative for both.
  Button navigates to signup page
  (3 votes)
Rohini
Posted 10 years ago. Direct link to Rohini's post “Why would you have to exp...”
Why would you have to express sin and cos in an exponential form?
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(5 votes)
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Abu Zubair
Posted 10 years ago. Direct link to Abu Zubair's post “Is the value of i=√(-1)?”
Is the value of i=√(-1)?
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(2 votes)
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- Marvin Cohen
  Posted 10 years ago. Direct link to Marvin Cohen's post “You are correct. i^2 = -...”
  You are correct. i^2 = -1; therefore i = √(-1).
  Comment on Marvin Cohen's post “You are correct. i^2 = -...”
  (1 vote)
Carlos Palacio
Posted 11 years ago. Direct link to Carlos Palacio's post “Is there a relationship b...”
Is there a relationship between i, theta, and x?
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(1 vote)
Answer
- Eddie Schodowski
  Posted 11 years ago. Direct link to Eddie Schodowski's post “e = lim n --> ∞ of (1 + 1...”
  e = lim n --> ∞ of (1 + 1/n)^n or lim n --> 0 (1 + n)^(1/n)
  i = √(-1)
  pi = ratio between a circle's circumference and its diameter
  x = theta, the angle in radians
  Euler's Formula usually relates e, i, pi, 1, and 0, but I think that it's better to understand that it relates more than just pi. Pi, as evaluated in trig functions, is oftentimes a neat answer, making the relationship between trigonometry and complex numbers beautiful.
  
  That's a mouthful to understand at once, but you'll get it when you get more in depth into math. The derivation for Euler's Formula cannot be done with the Precalculus's Toolbox of Mathematics provided to students (at least, not easily!)
  
  Euler's Formula:
  e^(ix) = cos(x) + i*sin(x)
  
  This is why when x=pi
  e^(i*pi) = cos(pi) + i*sin(pi)
  Well, sin(pi) is zero, so that get rids of the i, too.
  cos(pi) is -1 (look at the unit circle).
  
  So, e^(i*pi)=-1
  Or,
  e^(i*pi) + 1 = 0
  
  Ta da!
  Comment on Eddie Schodowski's post “e = lim n --> ∞ of (1 + 1...”
  (5 votes)
mebble
Posted 10 years ago. Direct link to mebble's post “At 1:46, how can cos(-the...”
At
1:46
, how can cos(-theta) be the same as cos(theta)?
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(2 votes)
Answer
- Jesse
  Posted 10 years ago. Direct link to Jesse's post “Cosine is an even functio...”
  Cosine is an even function. That means that opposite inputs have the same output. You can see this from its graph because it is symmetric about the y-axis.
  Comment on Jesse's post “Cosine is an even functio...”
  (3 votes)
Boshasha S. AlFowzan
Posted 11 years ago. Direct link to Boshasha S. AlFowzan's post “what is e and i? are they...”
what is e and i?
are they normal variables?
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(1 vote)
Answer
- Noble Mushtak
  Posted 11 years ago. Direct link to Noble Mushtak's post “Sal explains what e is: h...”
  Sal explains what e is:
  https://www.khanacademy.org/science/core-finance/interest-tutorial/cont-comp-int-and-e/v/introduction-to-compound-interest-and-e
  Sal explains what i is:
  https://www.khanacademy.org/math/trigonometry/imaginary_complex_precalc/i_precalc/v/introduction-to-i-and-imaginary-numbers
  
  I hope this helps!
  Comment on Noble Mushtak's post “Sal explains what e is: h...”
  (4 votes)
maxpancho
Posted 10 years ago. Direct link to maxpancho's post “So how could he get rid o...”
So how could he get rid of those i's so easily?
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(2 votes)
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Wan-Zhan Lee
Posted 10 years ago. Direct link to Wan-Zhan Lee's post “At 7:33 why is it not e^x...”
At
7:33
why is it not e^x - e^x over 2x?
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(2 votes)
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- Sid
  Posted 10 years ago. Direct link to Sid's post “Why would it be ? sinh a...”
  Why would it be ? sinh and cosh are defined to be what he tells you they are.
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  (1 vote)

Video transcript

In the calculus playlist, we came up with a rationale for Euler's formula. And I guess that video has gotten somewhat famous with me talking about minds being blown and people not having emotion. But just as a reminder, Euler's formula is that e to the i theta-- Just the formula by itself is pretty amazing, but it becomes especially amazing when you substitute pi or tau for theta. And if you don't know what tau is, there are videos on that. But Euler's formula tells us that e to the i theta-- there's a very good rationale for saying that this is going to be equal to cosine of theta plus-- and I'll just change the color for fun-- plus i sine of theta. And so one thing I want to do is explore this a little bit. Can I express-- because of Euler's formula-- can I now express cosine of theta in terms of some combination of the other stuff? And can I now express sine of theta in terms of some other combination of the other stuff? Well, let's think about how we can isolate a cosine theta. If this is e to the i theta, let's see what e to the negative i theta would be. Actually, let me do it over here on the right-hand side. So e to the negative i theta. This is the same thing as e to the i times negative theta, which-- by Euler's formula-- it would be equal to cosine of negative theta plus i sine of negative theta. Well, cosine of negative theta-- this is the same thing as cosine of theta. And sine of negative theta is the same thing as negative sine of theta. So this whole thing simplifies to negative i sine of theta. So we can write e to the negative i theta. Let me do that pink color. We can write e to the negative i theta as being equal to cosine of theta minus i sine of theta. And now we have something pretty interesting going on. If we add these two expressions, these two equalities right over here, the left-hand side-- so let's just add them up. The left-hand side, we're going to be left with e to the i theta plus e to the negative i theta. And on the right-hand side, these i sine thetas are going to cancel out with each other. And we're going to be left with 2 times cosine of theta. And so now it's pretty straightforward. I just isolate the cosine theta, divide both sides by 2. Divide both sides by 2, and we get something pretty interesting. We get that cosine of theta is equal to this business-- e to the i theta plus e to the negative i theta, all of that over 2. And we could do a similar type of thing to try to isolate-- let me rewrite it over. So let me rewrite this little discovery we made. Cosine of theta can be rewritten as e to the i theta plus e to the negative i theta all of that over 2. So that's kind of interesting. We can now express cosine of theta in terms of a bunch of exponentials right over here. Let's see if we could do the same thing for sine of theta. Let's imagine e to the i theta-- and let's try to do it the other way. There's always these weird kind of parallels between cosine and sine. So let's just see what happens if we were to subtract this from that. So let's try doing that. If we were to take e to the i theta-- let me just rewrite it over again. e to the i theta is equal to cosine of theta plus i sine of theta. And from that, let's subtract e to the negative i theta. Well, now we're going to subtract this. So we're going to get negative cosine theta. And then, we're subtracting this. So plus i sine theta. And so we are left with-- when we do the subtraction-- the left-hand side, we have e to the i theta minus e to the negative i theta is equal to-- this cancels out-- and we're left with 2i times sine theta. If we want to solve for sine theta, we divide both sides by 2i. And we are going to be left with sine theta is equal to this business right over here. So let's do this. So this is our other discovery-- sine theta can be written as e to the i theta minus e to the negative i theta, all of that over 2i. So there's something interesting about these structures. I can write cosine of theta is the sum of these two exponentials that are dealing with imaginary numbers, sine of theta as the difference of these two exponential numbers dealing with imaginary numbers. Actually, over here, it's actually the average of the two functions. You're dividing by 2. Here we're dividing by 2i. And if you don't like that i in the denominator, you can multiply the numerator and the denominator by i. This will become a negative. You'll have the i in the numerator. But either way, you will have the same expression. So this is all I can say about this. This is just kind of interesting. It's just kind of a fun exploration of mathematics. But maybe there's something neat about the structure of these functions. What happens if we were to take functions like this, but if we were to essentially get rid of the i's right over here? They seem like there might be some kind of an analog between cosine of theta and sine of theta. And so let's define those functions. And in the next few videos, we'll explore them and see if they really are analogous in some bizarre, strange, and beautiful way to cosine of theta and sine theta. So we're just going to make new function definitions, and they're just being inspired by these new ways that we discovered to write cosine of theta and sine theta. And I'm going to call the first function cosh. And it really does come from hyperbolic cosine, and we'll explore more why we call it hyperbolic cosine. But let's just call it cosh. I'm just experimenting right over here. And we'll just say cosh of x is equal to-- it's going to be inspired by this, but we're going to get rid of all of the i's. So it's going to be e to the x plus e to the negative x, all of that over 2. Once again, just inspired by this right over here. Getting rid of all of the i's. And then, let's define another one-- hyperbolic sine. So I'll call it sinh. Sinh for hyperbolic sine. Sinh-- I guess you could also do-- of x. And I'm just to get rid of all of the i's. I'm just being inspired by this. So e to the x minus e to the negative x, all of that over 2. And in the next few videos, we're going to start exploring this. And we're going to start realizing that there is this strange, bizarre parallel between these things and our traditional trigonometric functions. Now, these will show up in some of your mathematics, but it is important. These aren't as core-- especially in traditional mathematics-- as these two characters are here, but it is a fun exploration.