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• - [Instructor] What we're going to do in this video

• is prove that the limit

• as theta approaches zero

• of sine of theta

• over theta is equal to one.

• or trigonometric construction that I have here.

• So this white circle, this is a unit circle,

• that we'll label it as such.

• So it has radius one,

• unit

• circle.

• So what does the length

• of this salmon-colored line represent?

• Well, the height of this line would be the y-coordinate

• of where this radius intersects the unit circle.

• And so by definition, by the unit circle definition

• of trig functions, the length of this line

• is going to be sine of theta.

• If we wanted to make sure that also worked for thetas

• that end up in the fourth quadrant, which will be useful,

• we can just insure that it's the absolute value

• of the sine of theta.

• Can I express that in terms of a trigonometric function?

• Well, let's think about it.

• What would tangent of theta be?

• Let me write it over here.

• Tangent of theta

• is equal to opposite over adjacent.

• So if we look at this broader triangle right over here,

• this is our angle theta in radians.

• This is the opposite side.

• The adjacent side down here, this just has length one.

• Remember, this is a unit circle.

• So this just has length one,

• so the tangent of theta is the opposite side.

• The opposite side is equal to the tangent of theta.

• And just like before, this is going to be a positive value

• for sitting here in the first quadrant

• but I want things to work

• in both the first and the fourth quadrant

• for the sake of our proof,

• so I'm just gonna put an absolute value here.

• So now that we've done,

• I'm gonna think about some triangles

• and their respective areas.

• So first, I'm gonna draw a triangle

• that sits in this wedge, in this pie piece,

• this pie slice within the circle,

• so I can construct this triangle.

• And so let's think about the area

• of what I am shading in right over here.

• How can I express that area?

• Well, it's a triangle.

• We know that the area of a triangle

• is 1/2 base times height.

• We know the height

• is the absolute value of the sine of theta

• and we know that the base is equal to one,

• so the area here is going to be equal to 1/2

• times our base, which is one,

• times our height,

• which is the absolute value of the sine of theta.

• I'll rewrite it over here.

• I can just rewrite that

• as the absolute value of the sine of theta over two.

• Now let's think about the area of this wedge

• that I am highlighting in this yellow color.

• So what fraction of the entire circle is this going to be?

• If I were to go all the way around the circle,

• it would be two pi radians,

• so this is theta over to two pis of the entire circle

• and we know the area of the circle.

• This is a unit circle, it has a radius one,

• so it'd be times the area of the circle,

• which would be pi times the radius square,

• the radius is one, so it's just gonna be times pi.

• And so the area of this wedge right over here,

• theta over two.

• And if we wanted to make this work

• for thetas in the fourth quadrant,

• we could just write an absolute value sign right over there

• 'cause we're talking about positive area.

• in this blue color, and this is pretty straightforward.

• The area here is gonna be 1/2 times base times height.

• So the area, and once again, this is this entire are,

• that's going to be 1/2 times our base, which is one,

• times our height,

• which is the absolute value of tangent of theta.

• And so I can just write that down

• as the absolute value of the tangent of theta over two.

• Now, how would you compare the areas

• of this pink or this salmon-colored triangle

• which sits inside of this wedge

• and how do you compare that area of the wedge

• to the bigger triangle?

• Well, it's clear that the area of the salmon triangle

• is less than or equal to the area of the wedge

• and the area of the wedge is less than or equal to

• the area of the big, blue triangle.

• The wedge includes the salmon triangle

• plus this area right over here,

• and then the blue triangle includes the wedge

• plus it has this area right over here.

• So I think we can feel good visually

• that this statement right over here is true

• and I'm just gonna do

• a little bit of algebraic manipulation.

• Let me multiply everything by two

• so I can rewrite that the absolute value

• of sine of theta is less than or equal to

• the absolute value of theta

• which is less than or equal to the absolute value

• of tangent of theta, and let's see.

• Actually, instead of writing the absolute value

• of tangent of theta, I'm gonna rewrite that

• as the absolute value of sine of theta

• over the absolute value of cosine of theta.

• That's gonna be the same thing

• as the absolute value of tangent of theta.

• And the reason why I did that

• is we can now divide everything

• by the absolute value of sine of theta.

• Since we're dividing by a positive quantity,

• it's not going to change the direction of the inequalities.

• So let's do that

• I'm gonna divide this

• by an absolute value of sine of theta.

• I'm gonna divide this

• by an absolute value of the sine of theta

• and then I'm gonna divide this

• by an absolute value of the sine of theta.

• And what do I get?

• Well, over here, I get a one

• and on the right-hand side, I get a one

• over the absolute value of cosine theta.

• These two cancel out.

• So the next step I'm gonna do

• is take the reciprocal of everything.

• And so when I take the reciprocal of everything,

• that actually will switch the inequalities.

• The reciprocal of one is still going to be one

• but now, since I'm taking the reciprocal of this here,

• it's gonna be greater than or equal to

• the absolute value of the sine of theta

• over the absolute value of theta,

• and that's going to be greater than or equal to

• the reciprocal of one

• over the absolute value of cosine of theta

• is the absolute value of cosine of theta.

• approaching zero from that direction

• or from that direction there,

• so that would be the first and fourth quadrants.

• So if we're in the first quadrant and theta is positive,

• sine of theta is gonna be positive as well.

• And if we're in the fourth quadrant and theta's negative,

• well, sine of theta is gonna have the same sign.

• It's going to be negative as well.

• And so these absolute value signs aren't necessary.

• sine of theta and theta are both positive.

• In the fourth quadrant, they're both negative,

• but when you divide them,

• you're going to get a positive value, so I can erase those.

• If we're in the first or fourth quadrant,

• our X value is not negative,

• and so cosine of theta, which is the x-coordinate

• on our unit circle, is not going to be negative,

• and so we don't need the absolute value signs over there.

• Now, we should pause a second