For more information, please see full course syllabus of Trigonometry
For more information, please see full course syllabus of Trigonometry
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HalfAngle Formulas
Main formulas:

Example 1:
Use the halfangle formulas to find the sine and cosine of 15^{°} . Check that the answers satisfy the Pythagorean identity sin^{2} x + cos^{2} x = 1.Example 2:
Prove the following trigonometric identity:

Example 3:
Prove the following halfangle formula for tangents. Be careful about removing any ± signs!

Example 4:
Use the halfangle formulas to find the sine and cosine of (π /8). Check that the answers satisfy the Pythagorean identity sin^{2} x + cos^{2} x = 1.Example 5:
Prove the following trigonometric identity:

HalfAngle Formulas
 Half  angle Formula: sin[1/2]x = ±√{[1/2](1 − cosx)}
 [1/2]x = [(5π)/8] x = [(5π)/4]
 sin[(5π)/8] = ±√{[1/2](1 − cos[(5π)/4])}
 sin[(5π)/8] = ±√{[1/2](1 − ( − [(√2 )/2])} ) = ±√{[1/2]([(2 + √2 )/2])} = ±√{[(2 + √2 )/4]} = ±[(√{2 + √2 } )/2]
 sin[(5π)/8] will be positive because [(5π)/8] is in quadrant II
 Half  angle Formula: cos[1/2]x = ±√{[1/2](1 + cosx)}
 [1/2]x = [5p/8] x = [5p/4]
 cos[(5π)/8] = ±√{[1/2](1 + cos[5p/4])}
 cos[(5π)/8] = ±√{[1/2](1 + ( − [(√2 )/2])} ) = ±√{[1/2]([(2 − √2 )/2])} = ±√{[(2 − √2 )/4]} = ±[(√{2 − √2 } )/2]
 cos[(5π)/8] will be negative because [(5π)/8] is in quadrant II
 Half  angle Formula: sin[1/2]x = ±√{[1/2](1 − cosx)}
 [1/2]x = 75^{°} ⇒ x = 150^{°}
 sin75^{°} = ±√{[1/2](1 − cos150^{°})}
 sin75^{°} = ±√{[1/2](1 − ( − [(√3 )/2])} ) = ±√{[1/2]([(2 + √3 )/2])} = ±√{[(2 + √3 )/4]} = ±[(√{2 + √3 } )/2]
 sin75^{°} will be positive because 75^{°} is in quadrant I
 Half  angle Formula: cos[1/2]x = ±√{[1/2](1 + cosx)}
 [1/2]x = 75^{°} ⇒ x = 150^{°}
 cos75^{°} = ±√{[1/2](1 + cos150^{°})}
 cos75^{°} = ±√{[1/2](1 + ( − [(√3 )/2])} ) = ±√{[1/2]([(2 − √3 )/2])} = ±√{[(2 − √3 )/4]} = ±[(√{2 − √3 } )/2]
 cos75^{°} will be positive because 75^{°} is in quadrant I
 Half  angle Formula: sin[1/2]x = ±√{[1/2](1 − cosx)}
 [1/2]x = 180^{°} ⇒ x = 360^{°}
 sin180^{°} = ±√{[1/2](1 − cos360^{°})}
 sin180^{°} = ±√{[1/2](1 − 1} ) = ±√{[1/2](0)} = 0
 Half  angle Formula: cos[1/2]x = ±√{[1/2](1 + cosx)}
 [1/2]x = 180^{°} ⇒ x = 360^{°}
 cos180^{°} = ±√{[1/2](1 + cos360^{°})}
 cos180^{°} = ±√{[1/2](1 + 1} ) = ±√{[1/2](2)} = ±√1 = ±1
 √{[(1 − cos6x)/2]} · [([1/2])/([1/2])]
 √{[(1 + cos4x)/2]} · [([1/2])/([1/2])]
 − √{[(1 − cos8x)/(1 + cos8x)]} · [([1/2])/([1/2])]
 − [sin4x/cos4x]
 [1/(sin2θ)] = [(cscθ)/(2cosθ)], Reciprocal Identity
 [1/(2sinθcosθ)] = [(cscθ)/(2cosθ)], Half  angle Identity
 [1/(sinθ)] ·[1/(2cosθ)] = [(cscθ)/(2cosθ)], Separate fractions
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
HalfAngle Formulas
Lecture Slides are screencaptured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
 Intro
 Main Formulas
 Example 1: Find Sine and Cosine of Angle using HalfAngle
 Example 2: Prove Trigonometric Identity using HalfAngle
 Example 3: Prove the HalfAngle Formula for Tangents
 Extra Example 1: Find Sine and Cosine of Angle using HalfAngle
 Extra Example 2: Prove Trigonometric Identity using HalfAngle
 Intro 0:00
 Main Formulas 0:09
 Confusing Part
 Example 1: Find Sine and Cosine of Angle using HalfAngle 0:54
 Example 2: Prove Trigonometric Identity using HalfAngle 11:51
 Example 3: Prove the HalfAngle Formula for Tangents 18:39
 Extra Example 1: Find Sine and Cosine of Angle using HalfAngle
 Extra Example 2: Prove Trigonometric Identity using HalfAngle
Trigonometry Online Course
I. Trigonometric Functions  

Angles  39:05  
Sine and Cosine Functions  43:16  
Sine and Cosine Values of Special Angles  33:05  
Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D  52:03  
Tangent and Cotangent Functions  36:04  
Secant and Cosecant Functions  27:18  
Inverse Trigonometric Functions  32:58  
Computations of Inverse Trigonometric Functions  31:08  
II. Trigonometric Identities  
Pythagorean Identity  19:11  
Identity Tan(squared)x+1=Sec(squared)x  23:16  
Addition and Subtraction Formulas  52:52  
Double Angle Formulas  29:05  
HalfAngle Formulas  43:55  
III. Applications of Trigonometry  
Trigonometry in Right Angles  25:43  
Law of Sines  56:40  
Law of Cosines  49:05  
Finding the Area of a Triangle  27:37  
Word Problems and Applications of Trigonometry  34:25  
Vectors  46:42  
IV. Complex Numbers and Polar Coordinates  
Polar Coordinates  1:07:35  
Complex Numbers  35:59  
Polar Form of Complex Numbers  40:43  
DeMoivre's Theorem  57:37 
Transcription: HalfAngle 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 sin^{2} + cos^{2} = 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 sin^{2} + cos^{2} = 1.0336
Let us work those out, sin^{2}(pi/8) + cos^{2}(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 root^{2}, so they cancelled each other we get ((2 – root 2)/denominator 2^{2} 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 sin^{2}(pi/8) + cos^{2}(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 halfangle 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 1cos(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 1cos(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 sin^{2}+cos^{2}=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 halfangle 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(1cos(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(1cos(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 23/2.0148
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 yvalue, and cosine is the xvalue.0188
They're both positive in the first quadrant.0197
15 degrees is in quadrant 1.0201
The sine is its yvalue, 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 halfangle identities but using the addition and subtraction formulas.0246
We worked out the sine and cosine of π/12 by realizing it as π/4π/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 (ab)^{2} is a^{2}2ab+b^{2}, 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 halfangle 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 halfangle 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 halfangle 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 sin^{2}+cos^{2}=1.0602
Sin^{2}(15)+cos^{2}(15), sin^{2}(15) was square root of 2 minus root 3/2.0613
We're going to square that.0627
Plus cos^{2}(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 sin^{2}(15)+cos^{2}(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 halfangle 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 halfangle formulas because you see cos(1/2 x), sin(1/2)x.0733
I'm going to say, let's try to avoid the halfangle 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)×(ab), remember from algebra, that's the difference of squares formula.0775
That's a^{2}b^{2}.0781
That can be really useful if you have an (a+b) in the denominator or an (ab) 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 lefthand side, I'm going to work with the lefthand side because I see that (ab) 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 cos^{2}(1/2 x)+2sin(1/2 x)×cos(1/2 x)+ sin^{2}(1/2 x).0848
We can invoke this difference of squares formula.0873
We get cos^{2}(1/2 x)sin^{2}(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)=cos^{2}(θ)sin^{2}(θ).0901
Now, look at what we have here.0922
There's several good things that are going to happen.0923
First of all, cos^{2} and sin^{2}, 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 cos^{2} of something minus sin^{2} of something, and I know that cos^{2} of something minus sin^{2} 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 righthand side of the original trigonometric identity.1008
That was the righthand 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 lefthand 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, sin^{2}+cos^{2} gives you 1.1046
Secondly, 2 sine of something cosine of something, that's the double angle formula for sine.1051
Then, cos^{2} of something minus sin^{2} 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 righthand 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 halfangle 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 halfangle 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 halfangle 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 1cos(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 lefthand side.1179
Lefthand 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 halfangle formulas.1204
In the numerator, I get plus or minus the square root of 1/2.1207
Sin(1/2 x) is 1cos(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 (1cos(x)))/(1/2 (1+cos(x))).1257
The obvious thing to do there is to cancel the (1/2)'s, so you get (1cos(x))/(1+cos(x)).1270
It's not so obvious where to go from here but remember that we've been practicing this rule (ab)×(a+b)=a^{2}b^{2}.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 1cos(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 1cos(x).1322
Multiply top and bottom by 1cos(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
1cos(x), I'll just write that as (1cos(x))^{2}.1342
I don't need to multiply that out.1350
In the bottom, I got 1cos(x) times 1+cos(x), that's the difference of squares formula, that's 1cos^{2}(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 1cos(x), because I had the square root of (1cos(x))^{2}.1382
In the bottom, I still have a square root, 1cos^{2}(x), that's something that should set off some warning bells in your brain.1389
Certainly doesn't mind, because I remember that sin^{2}+cos^{2}=1, the Pythagorean identity.1398
If you move that around, if you see 1cos^{2}, that's equal to sin^{2}.1407
So, 1cos^{2}(x)=sin^{2}(x), that cancels with the square root.1415
I've already got a plus or minus outside, I don't need that another one, (1cos(x))/sin(x).1423
I almost got what I want, I've almost got the righthand side (1cos(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 lefthand side is equal to plus or minus (1cos(x))/sin(x), but we aren't sure if we can remove the plus or minus from the righthand 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
1cos(x) is always greater than 0, the numerator here, the 1cos(x), the numerator 1cos(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 yvalue, 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 xaxis, its ycoordinate 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)=(1cos(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 1cos(x) is always positive, the lefthand and the righthand 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, 1cos(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
1 answer
Last reply by: Dr. William Murray
Tue Aug 13, 2013 5:08 PM
Post by Taylor Wright on July 18, 2013
On example 1, are both answers in simplified form?
1 answer
Last reply by: Dr. William Murray
Mon Apr 29, 2013 9:26 PM
Post by Gabit on April 29, 2013
cosx <= 1, when you plug 0 into x it is still less or equal to 1, cos 0 <=1
1 answer
Last reply by: Dr. William Murray
Sun Apr 28, 2013 10:31 AM
Post by Emily Engle on April 27, 2013
In ex. 3 you say that cos x < 1 ,but my book says that the cos 0 = 1 ( and when you plot cos x it hits one at 0)?
1 answer
Last reply by: Dr. William Murray
Tue Apr 23, 2013 7:42 PM
Post by mickaole walden on April 22, 2013
I know it must have been a lot of work to make the quick notes. I just want to thank you, I really appreciate you. ;>