Math Analysis Computations of Inverse Trigonometric Functions

Ok we are trying some more examples of computing values of the inverse trigonometric functions.0000

Here we have to find the arctangent of the following common values.0006

I will start with my unit circle.0013

I remember that when arctangent I am always looking for values between –pi/2 and pi/2 because arctangent is always between pi /2 and –pi/ 2.0029

I wrote less than or equal to there, but it should really be less than because arctangent never actually hits – pi/2 or pi/2.0050

That is because tangent itself cannot be evaluated on pi/2 or –pi/2 because you are basically trying to divide by 0 there.0059

I’m going to make a list here of x and arctan _z 1 and root 3.0071

It is really useful if you can remember some common values of the tangent function.0098

For example tan (pi/0) is 0.0105

Tan (pi/6) is sine of pi/6 divided by cosine of pi/6, which works out to root 3/30113

Tan (pi/4) is sine of 4 divided by cosine.0126

Well in both are 2/2 so it is one.0132

Tan (pi/3) is sin/cosin of pi/ 3, that is root 3.0136

That tells me what the values of tangent are at these common values, pi/6, pi/4, and pi/3.0143

There is another way to remember the values of the tangent functions, which is to draw this line at x=1.0155

The tangent function is actually where these lines hit that line. That is 1, root 3/3, and root 3.0164

So if you remember your common values of tangent, it is easy to figure out the values of arctangent.0178

Let me start from the bottom because that is the easiest.0186

Root 3 here, what angle has tangent root 3?0190

Well, we just figured that out, that is pi/3.0194

Also, we get that from here.0199

Pi/3 is the angle that has tangent root 3.0201

What angle has tangent 1?0206

Well, there it is right there, it is pi/4.0210

What angle has root 3/3? It is pi/6.0214

From getting these common values over here.0220

What angles has tangent 0? Well, that is 0.0223

The negative ones are little trickier but if you just remember that the values in the fourth quadrant down here.0227

They are just the negatives of the same values we had up in the first quadrant.0235

That is –root 3, -1, and –root 3/3.0242

For –root 3, what is an angle that has that as its tangent? it is –pi/3.0252

For -1, what is an angle that has that as its tangent? It is –pi/4.0259

Finally for –root 3/3, what is an angle that has that as its tangent? It is –pi/6.0264

There are really two things that are keys to being able to find our tangents.0274

The first is knowing your tangents of the common values.0278

Again, if you can not remember those, just remember that tangent is sine over cosine.0282

If you remember the sine and cosine of the common values, it all comes back to those triangles.0295

Then you can figure out the tangents.0299

It is good to remember the common values that you do get from tangent namely root 3, 1, and root 3/3.0302

It is good to remember those numbers. But if you can not remember them, you can work them out from sine and cosine.0311

Once you figure those out, then you can figure out the common values of tangent of the negative angles.0316

You have to remember that arctangent always takes values between –pi/2 and pi/2.0324

Looking for angles between –pi/2 and pi/2 that have the tangents that we have been given.0334

When we are given these tangents, we just remember values between –pi/2 and pi/2 that have those particular tangents.0341

On our last example here, we are given various angles.0000

We have to find the arcsine of their sine or the same for cosine and tangent.0005

It is very important that you would be able to graph this on the unit circle.0012

That is really the way you start the problem like this.0015

There is my unit circle and that is 0, pi/2, pi, 3 pi/2, and 2 pi.0021

Now, -7 pi/6. The negative means you are going the downward direction from 0.0040

If you go a little bit past pi and -7 pi/6 puts you right there, -7 pi/6.0046

So, arcsine of sine of -7 pi/60056

The sine of that angle is the y coordinate and it is positive there.0068

So, it is arcsine and that is a 30, 60, 90 triangle.0076

I know the common values there are ½. It is positive because the y coordinate is positive.0083

It is arcsine of 1 ½ but remember that arcsine is always between – pi/2 and pi/2.0088

I want an angle between –pi/2 and pi/2.0105

Whose sign is ½. Well, sign is the y coordinate.0108

There is an angle who is y coordinate is ½ and that angle is pi/6.0117

My x are, there is pi/6.0126

It is an angle between –pi/2 and pi/2 who is sign is ½.0132

Let us try the next one, arcosine of 4 pi/3.0138

Well, I find 4 pi/3, that is between pi and 2 pi.0148

In fact, that is down here.0153

Its cosine is x coordinate, its cosine is x coordinate and there again we have a 30, 60, 90 triangle.0158

Its cosine of 4 pi/3 is ½.0174

But since we are on the left side of the page, it is negative because the x coordinate is negative there.0183

Now I’m trying to find the arcosine of – ½.0189

Remember in arcosine, you will always look for an angle between 0 and pi.0196

I’m looking for an angle between 0 and pi whose cosine is – ½.0200

There it is right there.0209

That is 2 pi/3.0212

It is between 0 and pi, and its cosine is – ½.0215

So, our answer there is 2 pi/3.0219

Finally, we have the arctangent of the tangent of -5 pi/4.0229

Let us figure out where that angle is on the unit circle.0239

It is negative so we are going around in the downward direction from 0.0242

5 pi/4 is just past pi.0246

We go down on the downward direction and we end up over here at 5 pi/4.0250

So, -5 pi/4. I’m sorry that is -5 pi/4. Let me go back carefully.0260

Its tangent is sine/cosine.0270

That is a 45, 45, 90 triangle.0276

Its tangent is sine/cosine.0285

Sine is root 2/2 because it is positive.0288

Cosine is –root 2/2 that simplifies down to -1.0293

We want the arctangent of -1.0299

Now, arctangent always takes values between -pi/2 and pi/2 arctangent.0302

We are looking for an angle between –pi/2 and pi/2 whose tangent is -1.0317

So, I’m looking for an angle whose tangent is -1.0327

There it is at that 45, 45, 90 triangle.0339

There is my angle whose tangent is -1.0344

And that angle is –pi/4.0348

To recap, what really made it possible to this problem were finding sine, cosine, and tangent to start with.0361

The first thing you need to is your common values of sine, cosine, and tangent.0369

You need to be able to graph these things on the unit circle.0374

You need to identify the 30, 60, 90 triangles and the 45, 45, 90 triangles and know the common values.0378

Remember the common values for the 30, 60, 90 triangles are always ½ and root 3/2.0388

The common values for a 45, 45, 90 triangles are root 2/2 and root 2/2.0398

You need to know those common values and identify which triangle you are working with.0414

And figure out what quadrant you are in to figure out whether the sine and cosine are positive or negative.0423

The way you remember that is by all students take calculus.0430

In the first quadrant, they are all positive.0434

In the second quadrant, sine is positive only.0437

In the third quadrant, tangent is positive.0440

And in the fourth quadrant, cosine is positive.0443

You figure out which quadrant all these things are positive or negative and then you can find your sine and cosine.0448

And then what we have to do is find arcsines, arcosines, and arctangents.0456

The key to that was remembering the common values but also remembering these ranges.0462

Arcsine, it has to be between –pi/2 and pi/2.0470

Arcosine, between 0 and pi.0474

And arctangent, between –pi/2 and pi/2.0477

What we are doing in this step for all three problems was we had the sine or the cosine of the tangent.0482

We were trying to find an angle in the appropriate range that had the correct sine, cosine, or tangent.0490

That is where our arcosine, arsine, and arctangent are.0498

We are looking for an angle whose sign, cosine, and arctangent is the value that you are given.0501

We are looking for an angle inside this range.0510

In each case, we have a value and we found angles inside that range that have the right sine, cosine, or tangent.0514

That is what the answers are.0521

That is the end of our lecture on computations of inverse trigonometric functions.0524

These are the trigonometry lectures for www.educator.com.0529

Hi these are the trigonometry lectures for educator.com and today we're talking about computations of inverse trigonometric functions.0000

In the previous lecture, we learned the definitions and we practiced a little bit with arcsin, arccos, and arctan, you might want to review those a little bit before you go through this lecture.0009

I'm assuming now that you know a little bit about the definitions of arcsin, arccos, and arctan, and we'll practice using them and working them out for some common values today.0020

The key thing to remember here is where these functions are defined and what kinds of values you're going to get from them.0032

Arcsin, remember you start out with the number between -1 and 1, and you always get an answer between -π/2 and π/2.0040

It's very helpful if you remember the unit circle there.0052

Arcsin always gives you an angle in the fourth and the first quadrant between -π/2, 0, and π/2.0055

You're looking for angles in that range that have a particular sine.0068

Arccos, also between -1 and 1, produces an answer between 0, and π.0075

Again, it's helpful to draw the unit circle and keep that in mind.0081

There's 0, π/2, and π.0088

You're trying to find angles between 0 and π that have a given cosine.0095

Arctan, you can find the arctan of any number.0101

Again, you're trying to find an angle between -π/2 and 0, and π/2 that has a given tangent.0110

I say exclusive here because you would never actually give an answer of -π/2 or π/2 for arctan because arctan never actually hits those values.0129

If you think of that coming from the other direction, we can't talk about the tangent of π/2 or -π/2, because those involve divisions by 0.0143

When we're talking about arctan, we'll never get -π/2 or π/2 as an answer.0153

Let's practice finding some common values.0161

Here's some common values that we should be able to figure out arc sines of.0166

Let me start by drawing my unit circle.0172

There's -π/2 and 0, and π/2, remember, our answers always going to be in that range.0183

Let me just graph those common values and see what angles they correspond to.0191

I'll make a little chart here.0197

x and arcsin(x), so we got -1, negative root 3 over 2, negative root 2 over 2, -1/2, 0, 1/2, root 2 over 2, root 3 over 2, and 1.0198

Remember, sin(x), sine is the y-coordinate.0222

I'm looking for an angle that has y-coordinate of -1 to start with, so I want to find an angle that has y-coordinate down there at -1, and that's clearly -π/2.0227

That's the answer.0238

Negative root 3 over 2, that's an angle down there, so the angle that has sine of negative root 3 over 2, must be -π/3.0240

Negative root 2 over 2, that's the one right there, so that's a -π/4.0258

-1/2, the y-coordinate -1/2, is right there, that's -π/3.0267

Arcsin(0), what angle has sin(0)?0277

Well, it's 0.0280

What angle has sin(1/2)?0283

Well, what angle has vertical y-coordinate 1/2?0285

That's π/3.0288

Root 2 over 2, that's our 45-degree angle, also known as π/4.0294

Root 3 over 3, that's our 60-degree angle, also known as π/3.0303

Finally, we know that the sin(π/2) is 1, so the arcsin(1)=π/2.0313

In each case, it's a matter of looking at the value and thinking, "Okay, that's my y-coordinate, where am I on the unit circle?"0321

"What angle between -π/2 and π/2 has sine equal to that value?"0332

Of course, if you know your values of sine very well, then it's not too hard to figure out the arcsin function.0339

You really don't need to memorize this.0345

You just need to know your common values for sin(x) very well, and to know when they're positive or negative.0348

Then you can figure out the values for arcsin(x).0354

In our second problem here, we're asked to find which of the arcsin, acrccos, and arctan functions are odd, even or neither.0360

It's really, the key to thinking about this one is probably to think about the graphs and not so much to think about the original definitions of odd or even.0369

Let me write down the important properties to remember here.0377

An odd function has rotational symmetry around the origin.0382

The way I remember that is that 3 is an odd number and x³, y=x³ has rotational symmetry around the origin.0406

Even functions have mirror symmetry across the y-axis and the way I remember that is that 2 is an even number, and the graph of y=x² has mirror symmetry across the y-axis.0423

That's how I remember the pictures for odd or even functions.0458

Now, let me draw the graphs of arcsin and arccos, and arctan, and we'll just test them out.0464

Arcsin, remember you take a piece of the sine graph, there's sin(x) or sin(θ).0473

Arcsin, I'll draw this one in blue.0482

It's the reflection of that graph in the y=x line, that's arcsin(x) in blue.0485

Now, if you check that out, that has rotational symmetry around the origin.0497

So, arcsin(x) is odd.0507

Let's take a look at cosine and arccos.0517

Cosine, remember, you got to snip off a piece of cosine graph that will make arccos into a function.0522

There's cos(θ), and now in blue, I'll graph arccos.0530

There's arccos(x) in blue.0546

Now, that graph is neither mirror symmetric across the y-axis nor is it rotationally symmetric around the origin.0552

So, it's not odd or even, which is a little bit surprising, because even though if you remember cos(θ) is even.0565

It turns out that arccos(x) is not odd or even, even though cos(θ) was an even function.0590

Finally, arctan, let me start out by drawing the tangent graph, or at least the piece of it that we're going to snip off.0597

It kind of looks like the graph of y=x³, but it's not the same as the graph of y=x³.0609

One big difference is that tan(x) has asymptotes at π/2 and -π/2, and of course y=x³ has no asymptotes at all.0617

What I just graphed here is tan(θ) and then in blue I'll graph arctan(θ).0627

I'm flipping it across the line y=x.0638

It also has asymptotes neither horizontal asymptotes.0642

That blue graph is arctan(x), and if you look at that, that is rotationally symmetric around the origin.0648

So that's also an odd function.0666

This problem is really kind of testing whether you know the graphs of arcsin, arccos, and arctan look like.0675

If you don't remember those, then you go back to sine, cosine and tangent, and you snip off the important pieces of those graphs, and you flip them around y=x to get the graphs of arcsin, arccos, and arctan.0685

Those are the graphs that I have in blue here, arcsin, arccos, and arctan.0700

The other thing that this problem is really testing is whether you remember the graphical characterizations of odd and even functions.0705

If you know that odd functions have rotational symmetry around the origin, even functions have mirror symmetry across the y-axis, it's easy to check these graphs to just look at them and see whether they have the right kind of symmetry.0717

Of course, what you find out is that arcsin(x) has rotational symmetry, arccos(x) doesn't have either one, arctan(x) also has rotational symmetry.0732

For our third example here, we're trying to find arccos of the following list of common values.0746

Again, it's useful to start with a unit circle here.0753

Once you start with a unit circle, remember that with arccos, you're looking for values between 0 and π.0766

Arccos is always between 0 and π.0779

We're looking for angles between 0 and π that have cosines equal to this list of values.0786

I'll make a little chart here.0796

-1, negative root 3 over 2, negative root 2 over 2, -1/2, 0, 1/2, root 2 over 2, root 3 over 2 and 1.0805

Remember, cosine is the x-value.0823

I'm going to draw each one of these values as an x-value, as an x-coordinate, and then I'll see what angle has that particular cosine.0826

So, -1, the x-coordinate of -1 is over here, clearly that's π, that's the angle π.0834

Negative root 3 over 2, I'll draw that as the x-coordinate, and that angle is 5π/6.0845

Negative root 2 over 2, I'll draw that as the x-coordinate, I know that's a 45-degree angle, so that's 3π/4, that's the arccos of negative root 2 over 20863

-1/2, if we draw that as the x-coordinate, that's a 30-60-90 triangle, that's 2π/3.0879

What angle has cos(0)?0894

That means what angle has x-coordinate 0, that's π/2.0896

What angle has cos(1/2)?0903

Again, a 30-60-90 triangle, that must be π/3.0909

Root 2 over 2, that's a 45-degree angle, so that's π/4.0917

Root 3 over 2, that's a 30-60-90 angle again, that's π/6.0927

Finally, what angle has x-coordinate 1?0939

That's just 0.0943

The trick here is remembering your common values of cosine on the unit circle.0947

I know all the common values of cosine on the unit circle very well because I remember my 30-60-90 triangles, and I remember my 45-45-90 triangles.0952

I know which ones are positive and which ones are negative.0965

Finally, I remember that arccos is always between 0 and π.0968

So, I'm looking for angles between 0 and π that have these cosines.0973

These are the angles that have the right cosines and are in the right range.0981

We'll try some more examples of these later.0985