WEBVTT

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OK.

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Let's take a look at how to solve

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some composition of trig functions with their inverses.

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So, all of these functions today are going to be

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a trig function inverse of a trig function of an angle.

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Often when you evaluate composition of functions in the inverses,

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it's very easy.

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You just take whatever that input is as your answer.

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That's not going to work here because we've restricted

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the range of

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the inverse trig functions so much.

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So,

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these first three examples we have the same angle 5 pi over 8.

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Let's see where that is on a unit circle.

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It's a good idea whenever you're doing these to

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draw a circle

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and

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draw in the angle.

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That's about approximately 5 pi over 8.

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We can't say that sine inverse of sine of 5 pi over 8 is just 5 pi over 8 because

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our answer to any sine inverse question needs to be in the 4th or 1st quadrant,

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specifically between negative pi over 2 and positive pi over 2.

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So, we need to figure out which angle

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has the same sine value as 5 pi over 8.

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So, if we just come straight across,

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reflect over the y-axis,

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that will give us a point that has the same y

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value. Place on the unit circle with the same y value.

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We just want to figure out

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what is this angle here.

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Well,

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to get to 5 pi over 8,

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we could have started on the negative

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x-axis and rotated backwards

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3/8 of pi because 5 is 3 from 8.

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So, let's just go forward 3 pi over 8

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from the positive x-axis so we get 3 pi

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over 8.

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Tangent inverse of tangent 5 pi over 8

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is not going to be the same thing.

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We need to find a place in the unit circle where y over x is the same

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as it is at

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this point up here where the angle is 5 pi over 8.

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We need therefore to be in the 4th quadrant because

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tangent is negative in the 2nd quadrant and also in the 4th quadrant.

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So, what we're going to do is we're going to actually rotate

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pi radians a 180 degrees around.

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So,

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this angle here must be negative

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3 pi over 8.

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I figured that out by again saying

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I'm 3 pi over 8 away from the negative x-axis,

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so I need a backup 3 pi over 8 from the positive x-axis.

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What about cosine inverse of cosine of 5 pi over 8?

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What do we need to change there?

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Well,

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turns out nothing.

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Unlike sine and tangent inverse,

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cosine inverse

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is

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always going to give us an answer in the first or second quadrant.

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5 pi over 8 is already in the second quadrant,

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so we don't need to change anything at all.

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OK.

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What about

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this second angle 12 pi over 7?

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Let's draw that in.

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That's going to be in the 4th quadrant,

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maybe about right there.

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And

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inverse of sine of 12 pi over 7,

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you might think it's just 12 pi over 7 because we're already in the fourth quadrant.

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But we need to name the angle in such a way that the

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angle is between negative pi over 2 and positive pi over 2.

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So, we don't have to change the position on the unit circle.

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We're in the right place.

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We need to give it the right name, and the right name in this case

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is negative 2 pi over 7.

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I know that because

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12 pi over 7 is just 2/7

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away from being

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14 pi over 7,

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which would be 2 pi a complete revolution.

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How about tangent inverse of tangent of 12 pi over 7?

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Well,

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same deal.

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We're in the correct place.

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We don't need to rotate at all or reflect over the y-axis.

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So, we just need to give it the right name,

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which is -2 pi over 7.

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But what about cosine inverse of cosine 12 pi over 7?

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Here we do have to do something.

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We have to change the point

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because we're not in the first or second quadrant.

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How do we get to the right place?

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Well,

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we want to find a place in the unit circle that is the same x value.

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So, let's go straight on up,

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and it must be right around there.

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So, what is this angle here?

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That angle there,

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we went backwards 2 pi over 7 to get down to 12 pi over 7.

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So, let's go forwards

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the same amount,

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so this will be positive 2 pi over 7.

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OK.

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So that's how you solve those.

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Why don't you try a couple on your own here?

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Let's move this over.

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I have a few for you to try out.

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Try those.

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Pause the video,

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try those,

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and then I'll come back and tell you if you have the right answer.

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All right,

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let's take a look.

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6 pi over 5.

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Good idea to quickly sketch a circle,

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unit circle and see where is that

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6 pi over 5.

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It's just over 5 pi over 5,

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which is half a circle.

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So, we're right around there.

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Now,

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we're in the 3rd quadrant,

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so we do need to change the position of the point

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for

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sine inverse.

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We want to reflect over the y-axis,

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so we'll have the same y value.

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So, we need to be right there.

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So, instead of going 1/5 of pi beyond the x-axis,

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we're going to come back a 5th of pi.

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This should be negative

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pi over 5.

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For tangent,

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we also need to change,

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but now we want to be in the first quadrant because the

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tangent will be positive like it is in the 3rd quadrant.

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So, in this case, it will be positive pi over 5.

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And for cosine inverse of cosine of 6 pi over 5,

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again we need to change the position because

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we're not in the first or second quadrant.

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So, we'll reflect over the x-axis that'll keep.

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It'll keep the x value the same.

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And so, now instead of going 1 more fifth of pi from the negative x-axis,

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we need a backup one.

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So, instead of 5 pi we go back one,

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that would be

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4

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pi over 5.

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All right,

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that's how you solve problems like this.

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I hope this has helped,

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and thanks for watching.