Trigonometry Half-Angle Formulas

Ok we are going to try some examples of the half angle formulas.0000

Now, we are going to find the sin and cos of pi/8 and we are going to check that our answer satisfies the pythagorean identity sin² + cos² = 1.0005

Let me remind you of the half angle formulas, we have sin(1/2x) is equal to + or – the square root of ½, 1 – cos X, we will be using that and cos(1/2x).0015

Same formula just for the plus, square root of 1/2 , 1 + cos X.0033

In this case, we want to find the sin and cos of pi/8, now that is not one of the common values that we need to memorize.0045

I do not remember the sin and cos of pi/8, what I do know is that pi/8 is ½ of pi/4.0052

And pi/4 is a common value, I know the sin and cos of pi/4, that is my starting point.0065

I’m going to take the X to be pi/4 and I’m going to plug into the half angle formulas.0073

Sin(pi/8) is equal to + or – the square root of ½, of 1 – cos(pi/4), because my X is pi/4.0080

Now, cos(pi/4) is a common value that I remember, I know that by heart it is square root of 2/2.0099

I’m going to put this over common denominator, so ½ is 2/2, (2 – root 2/2), (+ or – the square root of 2 – root 2/4), (+ or – the square root of 2 – root 2), the square root of 4 is just 2 there.0111

Now, I need to deal with this + or -, figure out whether it is positive or negative.0135

Let me draw a quick unit circle here, I know that pi/4 is over here, pi/8 is half of that so it is down here.0140

Definitely in the first quadrant there, both sin and cos will be going to be positive, remember sin and cos, and the x and y coordinates.0157

I know that I want to take the positive square root here in quadrant 1, sin(pi/8) must be positive.0167

So, sin(pi/8) then must be equal to the + square root of (2 – root 2/2).0185

Now let us figure out the cos(pi/8), I will do that on the new page here.0201

cos(pi/8) is equal to + or – the square root of 1/2 ((1 + cos(pi/4)), the X is pi/4 and then we are finding the cos(x/2) here.0207

This is + or – the square root of ½ (1 +… now cos(pi/4), we remember that very well, that is a common value, root 2/2).0228

Put this over the common denominator so we get (2 + root 2)/2, combine those fractions so we get (2 + root 2)/4.0241

Finally, we can split up the square root into numerator and denominator, (2 + root 2)/4.0255

Again, we already said that pi/8 is in quadrant 1, the first quadrant, so cos(pi/8) must be positive, that is the X value.0264

We have cos(pi/8) equal to the + square root of (2 + root 2).0284

I simplify slightly wrong here the square root of 4 of course is 2, I accidentally wrote it that down again as 4.0300

Cos(pi/8) is the + square root of (2 + root 2)/2.0311

Now we figured out the sin and cos of pi/8 using our half angle formulas.0321

The last thing the problem asks us to do is to check that those sin and cos verify the pythagorean identity sin² + cos² = 1.0336

Let us work those out, sin²(pi/8) + cos²(pi/8).0337

The sin was (root 2 – root 2)/2, we need to square that, the cos was the same thing as the facts (2 + root 2)/2.0351

If we square that out in the numerator region, we will get square root², so they cancelled each other we get ((2 – root 2)/denominator 2² is 4) + ((the numerator is (2 + root 2)/4).0367

If we combined those the minus root 2 and plus root 2 cancel, so we get 4/4, of course simplifies down to 1.0384

That does check then that sin²(pi/8) + cos²(pi/8) = 1, it checks the pythagorean identity.0395

They key to that problem was really just remembering the half angle formula for sin and cos, sin(1/2x) and cos(1/2x).0402

To get pi/8 we just noticed that it is half of pi/4 and we plugged it into the formulas and we get this answer which has a + or – square root.0414

We got to remember that pi/8 is in quadrant 1, its sin and cos, its x and y values are both going to be positive, so we take the positive square root.0424

So, let us try one more example here. We have to prove the trigonometric identity involving some half angle formulas, tan (x) × tan (1/2x) = sec (x) – 1.0000

This one is a little tricky but it is going to be a lot easier if you remember an example we did in the earlier part of the lecture.0013

What we figured out the trigonometric for tan (1/2x), we figured out a half angle formula from the earlier example.0020

We proved that tan(1/2x) = 1 – cos(x) / sin(x), that was a little bit of work to prove.0039

Actually, that involves removing some + or – and some square roots, that was tricky.0054

But having done that work, it would be pretty easy to prove this trigonometric identity.0059

Let me start with the left hand side because that one looks a little more complicated and I will try to manipulate it into the right hand side.0066

The left hand side is tan(x) × tan (1/2x), now this tan(x), I’m not sure exactly what to with that so for a lot of better option.0075

I will convert that in to sin(x) / cos(x), that is the definition of tan, you can always convert tan in to sin / cos.0088

Tan(1/2x) that is not so obvious but I remember this previous example where we derived this identity 1 – cos (x) / sin(x).0097

I’m going to convert that in to 1- cos(x) / sin(x).0108

Now the sin(x) is cancelled and we are left with 1 – cos(x) / cos(x), if I put that up in to 1 / cos(x) – cos(x) / cos(x)0116

Then 1 – cos by definition is sec(x) and cos / cos is of course 1.0135

And look we transformed it in to the right hand side.0142

Again, as with most trigonometric identities it is not always obvious to proceed with them but a general rule of the thumb is try to attack the more complicated looking side first.0149

In this case, that was the left hand side, tan(1/2x) I remember my half angle identity for tan(1/2x) that we proved earlier on in the earlier part of the lecture.0160

I expand that out, I do not know what to do with the tan so I just convert it in to sin and cos, that is something you do if you can not think of anything else to do, convert everything to sin and cos.0173

Then it starts to cancel nicely, separates out, and suddenly converts in to the right hand side once I remember the definition of sec(x).0185

The general rule is there, go for the more complicated side, convert into sin and cos, invoke any half angle or double angle or that you already know.0194

That is the end of the half angle formulas as part of the trigonometry lectures on www.educator.com, thanks for watching.0207

Hi, these are the trigonometry lectures for educator.com and today we're going to talk about the half-angle formulas.0000

The main formulas that we're going to be using today, we have a formula for sin(1/2 x) and cos(1/2 x).0007

They're a little bit cumbersome, sin(1/2 x) is equal to plus or minus the square root of 1/2 of 1-cos(x).0017

Cos(1/2 x), same formula, except there's a plus in it.0027

They're a little bit cumbersome but we'll practice using them and you'll see that they're not so bad.0033

What makes them difficult is the plus or minus and the square root signs.0037

That's probably the confusing part.0042

Actually, what's inside the square root sign isn't bad at all, the 1-cos(x), 1+cos(x) aren't too bad.0044

Let's try them out right away with some examples.0050

Our first example is to find the sine and cosine of 15 degrees and then we'll check that our answers satisfy the Pythagorean identity sin²+cos²=1.0054

The first thing to notice here is 15 degrees is not a common value, it's not one that where we've memorized the sine and cosine.0065

We'll have to use the half-angle formulas here.0076

We'll start with 15 is 1/2 of 30, so we're going to use the sine formula.0079

Remember, sin(1/2 x) is equal to plus or minus the square root of 1/2(1-cos(x)).0089

The x in question here is 30, we're trying to find the sin(15), sin(15) is equal to plus or minus the square root of 1/2(1-cos(30)).0102

That's plus or minus the square root of 1/2.0120

Now, 30 degrees is a common value, that's π/6, and I've got all the sines and cosines of the common values memorized.0125

Cos(π/6), cos(30) is root 3/2.0134

I'm just going to do a little bit of algebra with this expression here, plus or minus the square root of 1/2.0140

I'm going to put 1 in root 3 over 2 over a common denominator, that will be 2-3/2.0148

I'm just writing 1 there as 2/2.0154

If I combine these, I get 2 minus root 3 over 4.0158

I still have this plus or minus which is not very good because I want to give a single answer, I don't want to give two different answers.0168

Let's remember where 15 degrees is.0175

Here's 0 degrees, and here's 90 degrees, 15 degrees is way over here right in the first quadrant.0180

Since sine and cosine are the x and y values, actually sine is the y-value, and cosine is the x-value.0188

They're both positive in the first quadrant.0197

15 degrees is in quadrant 1.0201

The sine is its y-value, it's positive, sin(15) is greater than 0, it's positive.0212

Our answer then, is sin(15) must be the positive square root there, it's the square root of 2 minus root 3 over 4.0220

Now, if you've been paying really close attention to the educator.com trigonometry lectures, you'll know that we've actually solved this problem before.0236

15 degrees if you convert it to radians, is actually π/12, and we worked out the sin(π/12) before not using the half-angle identities but using the addition and subtraction formulas.0246

We worked out the sine and cosine of π/12 by realizing it as π/4-π/6.0264

When we worked it out, sin(π/12) using the subtraction formulas, we got the answer square root of 6 minus the square root of 2 over 4.0273

There's a little bit of a worry here, because it's seems like we did the same problem using two different sets of formula and we got two quite different looking answers.0290

We got the square root of 6 minus the square root of 2 over 4 last time we did it.0302

This time we have the square root of 2 minus the square root of 3 over 4.0306

Actually, that can be simplified a little bit into, if we just take the square root of the top part, then the square root of 4 is just 2.0314

We have this two different answers here, or at least they seem to be different.0324

Let me show you that these two answers can actually be reconciled.0329

How do these answers agree?0336

Let me start with the old answer, the one we did in a previous lecture on educator.com.0346

Our old answer was root 6 minus root 2 over 4.0353

What I'm going to do is square the numerator, root 6 minus root 2 squared.0361

To pay for that, I have to take a square root later.0370

Remembering an algebra formula (a-b)² is a²-2ab+b², I remember that algebra formula.0373

I've got a quantity that I'm squaring here, root 6 squared is 6 minus 2 root 6 root 2 plus root 2 squared is just 2, all over 4.0385

This simplifies down to 8 minus 2 root 12 over 4.0405

But root 12, I could pull a 4 out of that, and it turns into a 2 on the outside.0413

But I already had a 2 on the outside, you combine those and you get 4 root 3 over 4.0424

Now, I can factor 4 out of the numerator.0427

When it comes outside it becomes a 2, left on the inside will be, the 8 turns into a 2, and the minus 4 root 3 becomes a minus root 3 over 4.0432

That simplifies down to 2/4 cancel into 1/2, square root of 2 minus root 3 over 2.0446

Look at that, that's our new answer that we've just arrived using the half-angle formula.0454

We could do that problem using the addition and subtraction formulas as we did a couple of lectures ago or we could do it using our new half-angle working it out from what we now about 30 degrees.0465

Either way, we get down to the same answer.0481

We still have to find the cos(15).0486

We're going to use the half-angle formula for cosine, cos(1/2 x) is equal to plus or minus the square root of 1/2 times 1 plus cos(x).0490

The cos(15), 15 is 1/2 of 30, this is square root of 1/2 times 1 plus cos(30).0506

Cos(30) is a common value that I remember very well, 1 plus root 3 over 2.0520

If I put those, combine those over a common denominator, I get 1/2 2 plus root 3 over 2.0529

That simplifies down to 2 plus root 3 over 4, or if I take the square root of the bottom 2 plus root 3 over 2.0541

I still have that plus or minus but remember 15 degrees is safely there in the first quadrant.0556

It's sine and cosine, it's x and y values.0570

They're both positive, 15 degrees is in quadrant 1, cos(15) is positive.0571

The cos(15) must be the positive square root 2 plus root 3 over 2.0590

There's my answer.0600

Last thing we were supposed to check there was that the answer satisfies the Pythagorean identity sin²+cos²=1.0602

Sin²(15)+cos²(15), sin²(15) was square root of 2 minus root 3/2.0613

We're going to square that.0627

Plus cos²(15) is 2 plus root 3, square of that, over 2.0630

We'll square that one out.0636

Now, in the top, the square root and the square will cancel each other away.0642

We get 2 minus root 3.0647

In the bottom, we have 2, squared is 4, plus 2 plus root 3/4.0650

When we add those together, the root 3s cancel.0657

We just get 4 over 4 which is 1.0662

When we worked out sin²(15)+cos²(15), we did indeed get 1, showing that it does confirm the Pythagorean identity.0668

The key to that problem was really just recognizing that 15 is 1/2 of 30, and then invoking the sine and cosine half-angle formulas plugging in x=30 working them through doing a little bit of algebra, and getting our answers there.0677

The only other step that was a little bit tricky was recognizing whether we wanted to use the positive or the negative square root.0694

That's the matter of recognizing that 15 degrees is in the first quadrant, in both cases sine and cosine are both positive.0702

For our second example here, we're asked to use, to prove a trigonometric identity, (cos(1/2 x)+sin(1/2 x))/(cos(1/2 x)-sin(1/2 x))=sec(x)+tan(x).0712

That's a pretty complicated identity.0727

It's not really obvious where to start.0730

You might want to jump into the half-angle formulas because you see cos(1/2 x), sin(1/2)x.0733

I'm going to say, let's try to avoid the half-angle formulas here if we can.0741

Here's why.0745

Remember that cos(1/2 x) is equal to plus or minus the square root of something or other, so is sin(1/2 x) is equal to plus or minus the square root of something or other.0746

If we start putting those in, we're going to have plus or minuses, or lots of square roots, it's going to get complicated.0761

I'm going to try to avoid those.0766

Instead, I have another strategy which we've seen before in proving trigonometric identities.0769

If you have (a+b)×(a-b), remember from algebra, that's the difference of squares formula.0775

That's a²-b².0781

That can be really useful if you have an (a+b) in the denominator or an (a-b) in the denominator.0783

You multiply both sides by the conjugate, by the other one, and then you get the difference of squares.0790

Let's try that out on this one.0796

The left-hand side, I'm going to work with the left-hand side because I see that (a-b) in the denominator.0799

That's cos(1/2 x)+sin(1/2 x)/(cos(1/2 x)-sin(1/2 x)).0811

Now, I'm going to multiply top and bottom by the conjugate of the denominator.0823

That means where I saw a minus before, I'm going to multiply by the same expression with a plus in it, sin(1/2 x).0826

Of course, I have to multiply the top by the same thing, (cos(1/2 x)+sin(1/2 x).0837

Let's see where we'll go with that.0845

In the numerator, we actually have cos((1/2 x)+sin(1/2 x))², so that's cos²(1/2 x)+2sin(1/2 x)×cos(1/2 x)+ sin²(1/2 x).0848

We can invoke this difference of squares formula.0873

We get cos²(1/2 x)-sin²(1/2 x).0877

There's several good things that are going to happen right now but they will only happen if you remember the double angle identities.0886

Let me write those down for you.0894

I'm going to write them in θ instead of x.0896

Remember that sin(2θ)=2sin(θ)×cos(θ), cos(2θ)=cos²(θ)-sin²(θ).0901

Now, look at what we have here.0922

There's several good things that are going to happen.0923

First of all, cos² and sin², those combined, and those give me a 1.0926

Now, we have 2 sine of something, cosine of something, and the something is 1/2 x.0933

If you look back at our sin(2θ), 2 sine of something and cosine of something is equal to sine of 2 times that thing.0941

We have sin(2×1/2 x).0951

Now, I have cos² of something minus sin² of something, and I know that cos² of something minus sin² of something is equal to cosine of 2 times that something, so cos(2×1/2 x).0958

You can simplify this a little bit, this is (1+sin(x))/cos(x).0979

I'll split that up into 1/cos(x) + sin(x)/cos(x).0990

Those are expressions that I recognize, 1/cos(x) is sec(x), sin(x)/cos(x) is tan(x).0998

Look, now we've got the right-hand side of the original trigonometric identity.1008

That was the right-hand side right there.1014

That was a pretty tricky one.1018

There were several key steps involved there.1020

The first is looking at the left-hand side and noticing that we have something minus something in the bottom, so we're going to use this difference of squares formula.1024

We're going to multiply top and bottom by the conjugate.1036

Once we multiply top and bottom by the conjugate, we get something that looks pretty messy, but we start invoking these identities all over the place.1039

First of all, sin²+cos² gives you 1.1046

Secondly, 2 sine of something cosine of something, that's the double angle formula for sine.1051

Then, cos² of something minus sin² of something, that's the double angle formula for cosine.1058

That simplifies it down to (1+sin(x))/cos(x).1065

Those split apart and convert easily into secant and tangent, and all of a sudden we have the right-hand side.1070

You may have to experiment a bit with different techniques when you're proving these trigonometric identities.1077

The ones that I'm using for example, these are ones I've worked out ahead of time, so I know right away which technique I'm going to use.1083

Even when I'm working on this, I'll try multiplying a few different things together, maybe splitting up things differently and invoking different half-angle formulas, double angle formulas, and finally I find the sequence that works.1090

When you're asked to prove this trigonometric identities, go ahead and experiment a little.1103

If it seems like it's getting really complicated, maybe go back and try something else.1108

Eventually, you'll find something that converts to one side of the equation into the other.1113

For our next example, we have to prove a half-angle formula for tangents, and we're told to be careful about removing plus or minus signs.1119

The reason for that is we're going to be using the sine and cosine half-angle formulas, and those both have plus or minus in them.1129

Let me remind you what those are.1139

The cos(1/2 x) is plus or minus the square root of 1/2 times 1+cos(x).1144

The sin(1/2 x) is plus or minusthe square root of 1/2 times 1-cos(x).1157

Those are the formulas we're going to be using.1173

We're given the tan(1/2 x).1177

Let me start with that, I'll call that the left-hand side.1179

Left-hand side is tan(1/2 x).1186

Now, I don't have a formula yet for tan(1/2 x), I'm going to split that up into sin(1/2 x)/cos(1/2 x) because I do have formulas for those.1192

Those are my half-angle formulas.1204

In the numerator, I get plus or minus the square root of 1/2.1207

Sin(1/2 x) is 1-cos(x).1213

In the denominator, cos(1/2 x), the same thing except that I get 1/2 (1+cos(x)).1219

Here's the thing, you might think that you can cancel plus or minus signs, but you really can't.1228

The reason is that plus or minus signs means that both the top and bottom could be positive or could be negative.1234

When you divide them together, you don't know if the answer's going to be positive or negative.1241

I'm going to put one big plus or minus sign on the outside but I can't just cancel those away.1246

I'm also going to combine everything under the square root here.1252

I get (1/2 (1-cos(x)))/(1/2 (1+cos(x))).1257

The obvious thing to do there is to cancel the (1/2)'s, so you get (1-cos(x))/(1+cos(x)).1270

It's not so obvious where to go from here but remember that we've been practicing this rule (a-b)×(a+b)=a²-b².1282

That comes up all over the place for trigonometric identities and with other algebraic formulas as well.1295

The trick is when you have either one of those in the denominator, you multiply by the conjugate.1302

Here, we have 1-cos(x) in the denominator, I'm going to multiply by 1+cos(x).1308

Of course, I have to multiply the numerator by the same thing.1317

Sorry, I have 1+cos(x) in the denominator, so the conjugate would be 1-cos(x).1322

Multiply top and bottom by 1-cos(x).1327

The point of that is to invoke this difference of squares formula in the denominator.1331

This is all taking place under a big square root.1337

1-cos(x), I'll just write that as (1-cos(x))².1342

I don't need to multiply that out.1350

In the bottom, I got 1-cos(x) times 1+cos(x), that's the difference of squares formula, that's 1-cos²(x).1351

Now, I'm going to separate out the top and the bottom part here, because in the top, I've got a square root of a perfect square.1367

On the top, I'm just going to write it as 1-cos(x), because I had the square root of (1-cos(x))².1382

In the bottom, I still have a square root, 1-cos²(x), that's something that should set off some warning bells in your brain.1389

Certainly doesn't mind, because I remember that sin²+cos²=1, the Pythagorean identity.1398

If you move that around, if you see 1-cos², that's equal to sin².1407

So, 1-cos²(x)=sin²(x), that cancels with the square root.1415

I've already got a plus or minus outside, I don't need that another one, (1-cos(x))/sin(x).1423

I almost got what I want, I've almost got the right-hand side (1-cos(x))/sin(x).1435

The problem is this plus or minus.1444

The directions of this exercise said we have to be very careful about why we can remove any plus or minus signs.1446

Let me write the big question here, "Why can we remove this?"1455

That actually takes a bit of explanation.1464

I'm going to go on a new slide to explain that.1467

From the last slide, we figured out that the left-hand side is equal to plus or minus (1-cos(x))/sin(x), but we aren't sure if we can remove the plus or minus from the right-hand side.1470

Let's think about that.1497

First of all, I know that cos(x) is always less than 1.1499

That's because cos(x), remember, is the x values on the unit circle so it's always between -1 and 1.1509

1-cos(x) is always greater than 0, the numerator here, the 1-cos(x), the numerator 1-cos(x) is always positive.1517

That part isn't really affected by the plus or minus.1543

What about the sin(x)?1547

I know that that is not always positive.1548

What about sin(x)?1555

Let me draw a unit circle, because this really depends on where x lies on the unit circle.1560

There are several different cases depending on where x lies on the unit circle.1575

Let me write down the four quadrants, 1, 2, 3, and 4.1580

Let's try and figure out where x could lie on the unit circle.1589

There's sort of 4 cases.1593

If x is in quadrant 1, then remember, sin(x) is its y-value, so sin(x) is positive.1600

And x/2, if x is in quadrant 1, if that's x right there, then x/2 will also be in quadrant 1.1620

Tan(x) will also be positive.1636

Remember All Students Take Calculus, they're all positive in the first quadrant, second quadrant, only sine is, third quadrant only tangent is, and fourth quadrant only cosine is.1640

If x is in quadrant 1, then they're both positive.1652

Both sides here, the tan(1/2 x) and the sin(x), Oops, I said tan(x) and I should have said the tan(1/2 x) or x/2, they're both positive.1654

Let's check the second quadrant, if x is in quadrant 2, then sin(x) will still be positive, and x/2 ...1666

Well if x is over here in quadrant 2, then x/2 will be in quadrant 1 because it's half of x.1686

It's tangent, will still be positive.1699

Again, both the sin(x) and the tan(x) will both be positive.1704

Third case is if x is in quadrant 3, then sin(x) is less than 0 because x is down here in quadrant 3.1709

Where will x/2 be?1730

If x is in quadrant 3, that means x is bigger than π, x/2 is bigger than π/2.1734

x/2 will be over here in quadrant 2.1748

Tan(x/2), in quadrant 2, only the sine is positive, the tangent is negative.1760

Sine is negative because x is in quadrant 3, and tan(x/2) is also negative.1769

Finally, if x is in quadrant 4, then sin(x) is less than 0 because it's still below the x-axis, its y-coordinate is negative.1775

If x is somewhere over here in the fourth quadrant, x is between π and 2π, x/2 will be between π/2 and π.1797

So x/2 is still in quadrant 2, tan(x/2) is still negative.1812

Now, there's 4 cases there.1825

In the first 2 cases, sine was positive and tangent was positive, tan(x/2) is positive.1826

In the second 2 cases, in the last 2 cases, sine was negative and tan(x/2) was also negative.1834

Sin(x) and tan(x/2), they're either both negative or both positive, that means they have the same plus or minus sign, have the same sign in terms of positive and negative not as any have the same sign.1844

We can drop the plus or minus, and finally say tan(x/2) or tan(1/2 x)=(1-cos(x))/sin(x).1872

In all 4 cases, the tan(x/2) has the same plus or minus as sin(x).1897

Then remember the 1-cos(x) is always positive, the left-hand and the right-hand side will always have the same plus or minus, we don't need to attach another plus or minus.1905

That was a pretty tricky one.1916

The secret to that was starting with the tan(x/2), expanding it out using the formulas for cos(1/2 x) and sin(1/2 x) or x/2.1918

We worked it down, we did some algebra, simplifying a square root.1934

Then we still have that plus or minus at the end.1938

What we had to do was this sort of case by case study of each of the four quadrants to say when is sine positive or negative, when is tan(x/2) is positive or negative.1940

The last term, 1-cos(x) was always positive.1951

Finally, we figured out that sin(x) and tan(x/2) could be positive or negative, but they'll always be the same so we don't need to put in a plus or minus there.1956

We'll try some more examples later on.1966

The examples we're going to be using later, we're going to be using this formula for tan(x/2), so it's worth remembering this formula for tan(x/2).1969

Later on, we'll be using that to solve some more trigonometric identities.1982