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We know that if we take all of the points
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in the *X, Y plane where x^2 + y^2 = 1*,
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we get ourselves the unit circle.
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Let me draw the unit circle.
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That's my y-axis; this is my x-axis.
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And the unit circle has the circle with the radius one.
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So that's x=1, that's x=-1, that's y=1, that's y=-1
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the unit circle looks something... let me draw it...
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something like this, I think you get the point
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Let's see if I can fill it in a little bit better.
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So you realize that it's not a dotted circle.
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There.
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That's my best attempt at drawing the unit circle.
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And we also know that the traditional trig functions,
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or maybe we should call them the circular trig functions
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are actually defined so that if you parameterize
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so if you were to take *x=cos t*
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and *y=sin t*
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and you pick any t, right over here
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and by definition it's going to sit on the unit circle
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by definition, *x^2 + y^2 = 1*
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so if you pick any t it's going to sit some place
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on this unit circle.
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Or another way to think of it is
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if you vary t it's going to start tracing out this circle
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And we know that t corresponds to the angle
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with a positive x-axis, in this case,
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that right over there is t.
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Now wouldn't it be neat if there were a similar
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analogy for, not the unit circle, but
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something we could call the unit hyperbola?
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So that's our little review of trigonometry right there;
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our traditional trigonometry, now let's think about
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the unit hyperbola.
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Well, *x^2 + y^2 = 1* is a unit circle, I'll say that
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*x^2 - y^2 = 1, I'm going to call this my unit hyperbola*.
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Or a unit rectangular hyperbola. Hyperbola.
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This is just a little bit of review from *Conic Sections*,
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but it would look something like this:
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It would look... something... that's my y-axis,
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this is my x-axis, and then we can say,
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well if y is 0, x can be ±1,
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so you can think of that as the unit part
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where it intersects the x-axis; that's +1, that's -1
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and it has asymptotes, y=x and y=-x
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We go through the intuition on that in the
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Conic Section videos, y=x is that dotted line,
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y=-x is that dotted line, right over there,
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and then this thing is going to look like this.
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It's going to have a right half that does something like this,
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and does something like this, all a review of
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*Conic Sections*, it gets closer to its asymptotes.
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To y=x or y=-x
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and the same thing on the left-hand side.
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It's going to do something like that.
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Wouldn't it be neat if we could parameterize
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x and y with analogous functions so that
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we get a similar type of property?
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And you might guess what those functions are,
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but let's actually try to verify it.
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What would happen if x is equal to our hyperbolic
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*cosine of t*, which is the same thing as
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*e^t + e^(-t)*, all of that over 2
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and y were to be equal to
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*hyperbolic sine* of t, which is equal to
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*e^t - e^(-t)* over 2. Wouldn't it be neat if there
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were an analogy here; over here you pick any t
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based on our circular trig functions, you ended up
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with a point on the unit circle. Wouldn't it be amazing
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if for any point t you ended up with a point on our,
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what we're calling our *unit hyperbola*?
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Well, in order for that to be true, with this parameterization
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*x^2 - y^2* would need to be equal to one.
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Let's see if that is the case!
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So *x^2 - y^2* is equal to, well let's square this business
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it's equal to e^(2t) plus
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two times the product of these two things
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*2e^t • e^(-t)*, this is e^0 here which is 1.
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Plus e^(-2t), e^(-t)^2, all of that over 4
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And then from that we will subtract y^2.
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Minus, so the numerator's going to be
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*e^(2t) - 2e^t • e^(-t) + e^(-2t)*, all of that over 4
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So, immediately, a couple of simplifications here.
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*e^t • e^(-t)*, that's just e^(t-t)
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which is equal to e^0, which is equal to 1
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This is going to be one, that's going to be one,
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so we're going to have a 2 in either of those cases
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and if we were to simplify it, all of this stuff over here
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I'll do a numerator, so this is going to be equal to
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over our [denominator] of 4
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*e^(2t) + 2 + e^(-2t) - e^(2t)*
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just distributing the negative sign
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Plus two, and then minus e^(-2t)
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Well this is convenient!
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(Oh, I was writing it in black, a hard color to see)
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This cancels with this,
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This and this also add up to zero and you're left with
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two plus two over four
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which is indeed equal to one!
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So this is a pretty good reason to call these two functions
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hyperbolic trig functions.
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These are the circular trig functions,
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you give me a t on these parameterizations
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we end up on the unit circle!
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You vary t, you trace out the unit circle.
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Here, for any real t, we're going to assume we're
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dealing with real numbers,
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for any real t we're going to end up on
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the unit hyperbola right over here
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and in particular we're going to end up on the right
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so it's not exactly... over here pretty much any of these
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points could be parameterized right here
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over here we're going to end up
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on a point on the right side of the unit hyperbola.
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The reason why it's the right side
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is... you go straight to the definition of
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*cosh t*, this thing can only be positive
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This thing can only be positive.
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e^t can only be positive, e^-t can only be positive
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so this is only positive.
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But you give any t you will end up on this hyperbola!
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Specifically the right side, if you want points on
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the left hand side, you'd have to take the
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*-cosh t and the sinh t*
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to end up right over there.
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But it's a pretty neat analogy.
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We're looking at Euler's identity and we
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kind of said, "oh, let's just start playing with these things!"
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There seems to be a similarity here if we were to
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remove the i 's and, all of a sudden, we've discovered
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another thing! That there is this relationship
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here there is this relationship between these trig functions
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and the unit circle, here between our newly defined
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hyperbolic trig functions and the unit hyperbola.
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And you'd also find if you were to vary t it's going
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to trace out... just as if you were to vary t here it
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traces out the unit circle... if you trace t here it will trace
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out the right-hand side, the right-hand side
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of the unit hyperbola.
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For this parameterization right here.