Unit circle definition of trigonometric functions
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Unit circle definition of trig functions
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Example: Unit circle definition of sin and cos
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Example: Using the unit circle definition of trig functions
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Example: Trig function values using unit circle definition
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Example: The signs of sine and cosecant
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Unit Circle Manipulative
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Unit circle
Example: Using the unit circle definition of trig functions Worked example using our knowledge of 30-60-90 triangles and similarity to evaluate tan, cos and sin.
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- It looks like this exercise is asking use to do many
- things and i encourage you to pause this video and try to do it on your own
- first and then see maybe how I tackle the same exact thing
- so the first thing they're saying is Draw
- a 60 degree angle in standard position together
- with a unit circle.Use the triangle
- below, so this triangle right over here, to find the x- and y- coordinates
- of the point of intersection of the terminal side and
- the circle.So let me draw a unit circle here
- so that is my y-axis
- and then this could be my x-axis
- and then I'm gonna draw a unit circle, which is a circle of radius one
- so I'm gonna do my best to draw
- its obviously hard to freehand
- draw a circle, but I have a little easier time when I draw it as
- a dotted line, I can later make it undotted
- okay!...so that's pretty close, my best attempt
- at freehand drawing a unit circle
- so there you go...filling the dots
- now that i need it to be a solid line
- so that's my best attempt...fair enough.so I've drawn
- the unit circle..."Unit circle"
- it has radius One, so this point right over here
- is going to be the point (1,0)
- the x-coordinate is 1 and the y-coordinate is 0
- This point right over here, on the Y-coordinate.So thats the point (o,1)
- and then this over here is the point (-1,0)
- and this point right over here is a zero x-coordinate and a negative one y-coordinate (0,-1)
- there you go.So those are those 3 points or those 4 points
- right over there.So that is my unit circle
- now they say draw a 60 degree angle in standard
- position together with a unit circle. so standard position,
- you start to draw a ray, its centered at the origin
- its along the positive x-axis, that's your initial side of your angle
- and then you...since we're going positive 60 degree, we'll go in a
- counter clockwise direction, so we're going positive
- 60 degrees, counter clockwise direction and so your terminal side
- would be a ray that looks something like that, so that's
- our 60 degree angle in standard position
- Use the triangle below, so that's this triangle to find the x- and y-coordinates
- of the point of intersection of the terminal side and the circle.
- so it sounds very complex but all they're saying is
- what are the x- and y-coordinates of that point right over there
- so what is the x-coordinate and what is the y-coordinate
- and I'll give you a little bit of a hint here
- we can drop...we can construct a right triangle by taking a line
- from that point and going straight down, since it went straight down
- we know that this is a 90 degree angle
- we also know that if this is a 60 degree angle right over here
- so this is a 60 degree angle ..and this is a 90 degree angle
- we know that all the angles of a triangle have to add up to 180 degrees
- so we know that this up here is going to be a 30 degree angle
- so using that information..and this triangle right over here
- and what we know about similar triangle, can we now figure out what x- and
- y- are going to be.. well lets think about it
- lets think about it a little bit
- so what is...what is the x-coordinate of this
- that's just the distance we have to go along the x-axis over here
- this is the x and then we would go y up that much. so this distance right over here
- essentially the bottom of this 30, 60 90 triangle
- this is going to be..this length is going to be equal to
- our x-coordinate.and then this height over here
- the height of this triangle, the length of this side
- that's going to be equal to our y-coordinate...
- so if we just figure out the lengths of these sides, then we know what the x- and y-coordinates
- are going to be, and we also know that these are similar triangles
- you have a 60 degree angle here ...you have a 60 degree angle here
- you have a 30 degree angle up here...you have a 30 degree angle up here.
- you have a 90 degree angle over here...you have a 90 degree angle over here
- so all the angles are the same, these are similar
- triangles, which tell you the ratio of corresponding sides are going to be the same
- well lets think about the ratios of the corresponding sides.
- well what sides do we know on this triangle right over here, here they gave us all the sides
- what sides do we know here?. well we knew that this was a unit circle
- so what is the distance between our origin and this intersection point
- so what is the length ..essentially of the hypotenuse of this right triangle
- well the hypotenuse is a radius
- of the unit circle,a unit circle is a circle which has radius 1
- so the hypotenuse here has the radius 1
- and so hopefully we can now use this information and knowing that the ratio
- of the corresponding sides are going to be to same, to come up with
- what x- and y- are.
- so, one thing we can say is that the ratio of the short side of this triangle
- to the long side should be the same as the ratio of the short side of this triangle
- to the long side, or the ratio of the side opposite the 30 degree angle
- to the hypotenuse should be the same as the ratio of the side opposite to the 30 degree angle to the
- hypotenuse.so we could say the ratio of 1 to 2
- the ratio of 1 to 2 would be the same as the ratio of
- x to 1 ..to the hypotenuse
- or this could immediately get solved because x divided by 1 is just equal to x
- or we get x is equal to one half (x = 1/2)
- lets use that same logic again.
- The ratio of this side opposite the 60 degree side to the hypotenuse
- so square root of 3 to the hypotenuse
- should be the same as the ratio of the side opposite the 60 degree angle
- which in this case is y to this hypotenuse ...to 1
- well once again this is essentially Y divided by 1 (y/1) is just Y
- so we get y is equal to square root of 3 over 2 (y=√3/2)
- so we now know the x- and y-coordinates, this we can rewrite
- as (1/2,√3/2)
- now the last part they ask use to
- confirm that the x-coordinate is equal to the cosine of 60 degree (cos60°)
- let me do that in different color..confirm that the x-coordinate is equal to cos60°
- well ..lets take out our SOH CAH TOA
- again, just to remember what sine,cosine and tangent definitions are
- SOH.... CAH
- SOH CAH ...let me use a different color
- SOH CAH TOA
- So whats the Cosine of 60° going to be equal to?
- So the cosine of this angle.
- so the cosine of 60°(cos60°) is going to be equal to the adjacent over the hypotenuse(Adj/Hyp)
- its going to be equal to x over the hypotenuse.
- its going to be equal to x over the hypotenuse which is just 1, this is a unit circle
- so x is going to be equal to the cosine of 60°(x=cos60°)
- our x-coordinate is the same thing a cosine of 60°(cos60°)
- and we already figured out that x is equal to one half (x=1/2)
- so that all checks out. We can do the same thing for the sine.
- We must make sure that the y-coordinate is equal to the sine of 60°(sin60°)
- so sine of 60° ..based on our definition of these trig functions
- is equal to the opposite over hypotenuse(Opp/Hyp)
- so its equal to the opposite ..is our y-coordinate
- the length here is the same as our y-coordinate
- so its equal to y over our hypotenuse..well that's definitely equal to y
- which we already got with square to 3 over 2 which is equal to square root of 3 over 2
- so that checks out
- and that y over x is equal to tangent of sixty degrees
- well tangent is opposite over adjacent (opp/adj)
- so let me right it over here ..running out of real estate
- tangent of 60 degrees
- TOA tells us its opposite over adjacent
- the opposite side is y ..the adjacent side is x
- so that also works out
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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