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

You can figure these out quickly from the addition formulas in the previous lecture, so they shouldn't be hard to memorize if you remember the addition formulas.
Example 1:
Use the doubleangle formulas to find the sine and cosine of (2π /3). Use all three cosine formulas and check that the answers agree. Check that the answers agree with the sine and cosine of (2π /3) derived from the common values.Example 2:
Use the doubleangle formulas to prove the following trigonometric identity:

Example 3:
Use the addition and subtraction formulas to derive a formula for tan2x in terms of tanx. Check the formula on x = (π /6).Example 4:
Use the doubleangle formulas to find the sine and cosine of (4π /3). Use all three cosine formulas and check that the answers agree. Check that the answers agree with the sine and cosine of (2π /3) derived from the common values.Example 5:
Use the doubleangle formulas to prove the following trigonometric identity:

Double Angle Formulas
 Double Angle Formula: sin2θ = 2sinθcosθ
 2θ = π ⇒θ = [(π)/2]
 sinπ = 2sin[(π)/2]cos[(π)/2]
 2(1)(0) = 0
 Double Angle Formula: cos2θ = cos^{2}θ  sin^{2}θ or cos2θ = 2cos^{2}θ  1 or cos2θ = 1  2sin^{2}θ
 2θ = π ⇒θ = [(π)/2]
 cosπ = cos^{2}( [(π)/2] ) − sin^{2}( [(π)/2] )
 0^{2} − (1)^{2}
 Double Angle Formula: sin2θ = 2sinθcosθ
 2θ = [(3π)/2] ⇒θ = [(3π)/4]
 sin[(3π)/2] = 2sin[(3π)/4]cos[(3π)/4]
 2( [(√2 )/2] )( − [(√2 )/2] ) = 2( − [2/4] ) = − 1
 Double Angle Formula: cos2θ = cos^{2}θ  sin^{2}θ or cos2θ = 2cos^{2}θ  1 or cos2θ = 1  2sin^{2}θ
 2θ = [(3π)/2] ⇒θ = ( [(3π)/4] )
 cos[(3π)/2] = 2cos^{2}([(3π)/4]) − 1
 2( [(√2 )/2]^{2} ) − ( 1 ) = 2( [2/4] ) − 1 = 1 − 1 = 0
 Double Angle Formula: sin2θ = 2sinθcosθ
 2θ = 300^{o} ⇒θ = 150^{°}
 sin300^{°} = 2sin150^{°}cos150^{°}
 2( [1/2] ) ( − [(√3 )/2] ) = − [(√3 )/2]
 Double Angle Formula: cos2θ = cos^{2}θ  sin^{2}θ or cos2θ = 2cos^{2}θ  1 or cos2θ = 1  2sin^{2}θ
 2θ = 300^{θ} ⇒ θ = 150^{°}
 cos300^{°} = 1  sin^{2}(150^{°})
 1 − 2( [1/2] )^{2} = 1 − 2( [1/4] ) = 1 − [1/2] = [1/2]
 3(2sinθcosθ), factor out a 3
 2[(2sinxcosx) + 1], factor out a 2
 2[(sin2θ) + 1], Double Angle Formula: sin2θ = 2sinθcosθ
 4(1 − 2sin^{2}θ), factor out a 4
 cos^{2}φ  sin^{2}φ, multiply
*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
Double Angle 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 Formula
 Example 1: Find Sine and Cosine of Angle using Double Angle
 Example 2: Prove Trigonometric Identity using Double Angle
 Example 3: Use Addition and Subtraction Formulas
 Extra Example 1: Find Sine and Cosine of Angle using Double Angle
 Extra Example 2: Prove Trigonometric Identity using Double Angle
 Intro 0:00
 Main Formula 0:07
 How to Remember from Addition Formula
 Two Other Forms
 Example 1: Find Sine and Cosine of Angle using Double Angle 3:16
 Example 2: Prove Trigonometric Identity using Double Angle 9:37
 Example 3: Use Addition and Subtraction Formulas 12:38
 Extra Example 1: Find Sine and Cosine of Angle using Double Angle
 Extra Example 2: Prove Trigonometric Identity using Double Angle
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: Double Angle Formulas
Hi we are trying some examples of the double angle formulas for sin and cos.0000
We are going to try to find the sin and cos of 4pi/3.0005
We are going to use all three cos formulas and check that they agree.0010
We are also going to use our common values to find the sin and cos of 4pi/3 to check our answers.0014
Let me write down the double angle formulas that we are going to be using.0021
We are going to use sin 2X = 2 sin X cos X.0025
These are probably worth remembering, but if you do not remember them, you can work them out from the addition formula for sin and cos.0034
Cos(2X) is cos x ^{ 2}  sin x^{ 2}.0043
Of course, the X here would have to be 2pi/3 because what we are really trying to find is the sin and cos of 4pi/3.0051
So, sin(4pi/3) according to our double angle formula is equal to 2 x sin(2pi/3) x cos(2pi/3).0061
Now, 2pi/3 that is a common value, I remember its sin and cos, its sin is root 3/2 and its cos is 1/2.0078
It is negative because 2pi/3 is in the second quadrant, so its X coordinate is negative.0089
Remember, cos is the X coordinate and this simplifies down to cancel and we will get –root 3/2.0094
cos(4pi/3) is cos(2pi/3)^{2}  sin(2pi/3)^{2}.0106
So, plug those common values in the cos(2pi/3) is 1/2 and sin(2pi/3) is positive root 3/2.0122
I will get ¼ negative goes away because they got squared – root 3 squared is ¾ and so I get 1/2.0137
That was the first of the three formulas for cos(2x).0151
Let me remind you what the other two formulas are.0154
cos(2x) is equal to 2 cos X^{ 2}  1 and the other version we have of that formula is 1 – 2 sin X^{2}.0156
These are all different formulas for cos(2x) and we will try each one now.0177
The first one there is cos(4pi/3) is equal to 2 cos(2pi/3) ^{2}  1.0183
The cos(2pi/3) is 1/2 because it is in the second quadrant 1, that is (2 x ¼  1), which is ½  1, which is – ½.0201
If we use the other version of the formula we will get cos(2pi/3) is (1 2 sin(2pi/3)^{2}) which is (12 x (root 3/2)^{2}).0225
2pi/3 is our common value, I remember its sin, (1 – 2 x root 3 squared is 3), 4 in the denominator so 1 – 3 (1/2), and again we will get 1/2.0243
That is very reassuring because if you look at the three different formulas for cos, we got the same answer for all three of them.0262
That was the first point we wanted to check, but now let us check using our common values, there is 0, pi/2, 3pi/2, 2pi.0272
Now, 4pi/3 is bigger than pi, it is down here.0294
4pi/3 is 2/3 around the unit circle to 2pi.0300
That is one of our common triangles, that is the 30, 60, 90 triangles.0306
I know that the length of the sides there are root 3/2 and 1/2, I can figure out the sin and cos from that.0311
I just have to figure what are the positive or negative.0320
Well, the cos(4pi/3) is negative because the X value is negative, so it is – ½.0323
The sin(4pi/3) is also negative because the Y value is also negative there, it is – root 3/2.0334
Those are the answer we get using the common values on the unit answer.0346
But if you look, that is also the answers we got using the double angle formula breaking 4pi/3 up into ((2 x (2pi/3)).0351
We got sin, was – root 3/2, and cos was – ½.0363
It indeed, in fact, agrees with the values that we got from the unit circle.0367
Finally we are going to use the double angle formulas to prove another trigonometric identity.0000
We are going to prove that sec(2x) is equal to (sec X^{ 2}) / (2 – sec X ^{ 2}).0005
We are going to start with the right side because it looks more complicated.0013
The right hand side and we will try to manipulate it to the left hand side.0017
Let me start with the right hand side, (sec X^{2}) / (2 – sec X ^{ 2}).0022
Again, we do not know what to do with the trigonometric identity, it is often good to start with the more complicated side.0035
Secondly, convert everything to sin and cos.0042
Here I got a lot of sec, I am going to convert it to the definition of sec is (1/cos), this is (1/cos X ^{ 2}).0046
My denominator, I have 2 – 1/ (cos X ^{ 2}).0055
I see a lot of cos^{ 2} in the denominator.0064
I think I’m going to try to clear that by multiplying top and bottom by cos X ^{ 2}.0068
On the top, I will just get 1, on the bottom I get 2 cos X ^{ 2}  1.0077
But look at that, 2 cos X ^{ 2}  1.0090
That is one of the formulas that I remember for cos(2x), this is 1/cos(2x).0094
Now, let us remember by definition, one of our cos is exactly the same as sec.0104
So, this is sec(2x) and that is the left hand side of the identity that we are trying to prove.0110
We proved that we started with the right hand side, we derived the left hand side.0120
The key things to notice in there, the way it worked was, first of all the right hand side is a little more complicated, so we are going to work on that one.0124
When I see a bunch of sec, I try to convert it into sin and cos because I know how to manipulate sin and cos.0131
I got more formulas for them than for sec, tan, cosec, and cot.0138
I converted into sin and cos.0144
I see some cos in the denominator, I decided to multiply by cos X ^{ 2} to clear away those denominators.0147
I'm multiplying that thru and there is really some pattern recognition here knowing your identity formulas.0155
When I see that 2 cos X ^{ 2}  1, a little bell goes of in my head, “wait I have seen that somewhere before, oh yes that is equal to cos(2x)”.0162
Now I got 1/cos(2x), that is by definition sec(2x) and so I converted into the left hand side.0174
That is how you can use the double angle identities to prove more complicated trigonometric identities.0184
That is the end of our lecture on double angle identities.0192
These are the trigonometry lectures for www.educator.com.0196
Hi, welcome back to the trigonometry lectures on educator.com.0000
Today, we're going to learn about the double angle formulas, so here they are.0004
The first one is sin(2x)=2sin(x)×cos(x).0008
You may think there's so many formulas to remember in trigonometry.0014
This one, if you have trouble remembering it, you can work it out from the addition formula.0019
You do have to remember something, but if you can remember the sin(a+b)=sin(a)×cos(b)+cos(a)×sin(b).0024
If you remember that one, then you don't really need to learn anything new here because you can work it out so quickly.0041
Just take a and b, both to be x in the addition formula.0046
If a is x and b is x, then what you get here is sin(2x)=sin(x)×cos(x)+cos(x)×sin(x).0052
What you get is just 2sin(x)×cos(x).0068
If you can remember the addition formula, the double angle formulas are really nothing new to remember here, same goes for the cos(2x) formula.0074
If you remember the addition formula for cosine, you might want to try just plugging in x for each of the a's and b's, and you'll see that what you get is exactly cos^{2}(x)sin^{2}(x).0082
Now, there's two other ways that you often see this formula written as 2cos^{2}(x)1, and 12sin^{2}(x).0096
Those might look different but actually you can figure them out very quickly, or check them very quickly, because 2cos^{2}(x)1 is 2cos^{2}(x) minus, now remember 1 is the same as sin^{2}(x)+cos^{2}(x).0106
If you work with that a little bit, you have 2cos^{2}cos^{2}.0126
That's just a single cos^{2}(x)sin^{2}(x), and so all of a sudden this goes back to the original formula for cos(2x).0132
You could do this, you can check the second formula the exact same way, if you convert the 1 into sin^{2}+cos^{2}, you'll see that it converts back into this original formula for cos(x).0143
Even though it looks like there's 4 new formulas to remember here, really the basic sin(2x) and cos(2x), you can work both of those out from the additional formulas.0158
The other two formulas for cosine, you can just work them out if you remember the original formula for cosine and then the Pythagorean identity, sin^{2}+cos^{2}=1, which certainly any trigonometry student is going to remember the Pythagorean identity.0172
It's really not a lot of new memorization for these formulas.0185
The more interesting question here is how are you going to use them.0190
Let's try them out on some examples.0194
Our first example here is we're just going to get some practice using the sine and cosine of 2x formulas, the double angle formulas.0197
To find the sine and cosine of 2π/3.0206
Even though 2π/3 is a common value, hopefully you can work out the sine and cosine of 2π/3 without using the double angle formulas.0211
We're going to try them out using the double angle formulas, and then we'll just check that the answers we get agree with the values that we know coming from the common values.0219
We'll use that as a check, we won't use that at the beginning.0232
We're also going to use all three of the formulas for cosine and just check and make sure that they all work out, that they all agree with each other.0235
Let's start out by remembering those, actually, four formulas, sin(2x) is 2sin(x)×cos(x), and cos(2x) is cos^{2}(x)sin^{2}(x).0243
Here, we're being asked to find the sine and cosine of 2π/3.0264
We're going to use x=π/3, that way 2x is 2π/3.0267
So, sin(2π/3), using x=π/3, it's 2sin(π/3)×cos(π/3).0277
I remember that the sin(π/3), that's a common value, so the sin(π/3) is root 3 over 2, cos(π/3) is 1/2, the 2 and that in 1/2 cancel, and what we'll get is root 3 over 2.0294
Now, let's try the cosine, cos(2π/3), is cos^{2}(π/3)sin^{2}(π/3) according to our formula, but we're going to check it out and see if it works.0314
Now, cos(π/3) is 1/2, so (1/2)^{2} minus the sin(π/3) is root 3 over 2, we'll square that out.0333
1/2 squared is 1/4, root 3 over 2 squared is, root 3 squared is 3, 2 squared is 4, we get 1/43/4=1/2.0344
Now, there were two other formulas for cos(2x), we want to check out each one of those, cos(2x)=2cos^{2}(x)1.0357
It was also supposed to be equal to 12sin^{2}(x).0370
We're going to check out each one of those.0376
Cos(2π/3), using those other formulas, is equal to 2cos^{2}(π/3)1, which is 2.0379
Now, cos(π/3), that's a common value, that's 1/2, (1/2)^{2}1, which is 2×1/41, which is 1/21, is 1/2.0393
Let's use the other version, 12sin^{2}(π/3), we'll use the last cosine formula there.0413
That's 12, now, sin(π/3), I remember that's a common value, root 3 over 2.0425
We're going to square that out, that's 12 times, root 3 squared is 3, and 2^{2} is a 4.0433
That's 13/2=1/2.0443
The first thing we noticed is that these 3 different formulas for cos(2x) they all gave us the answer 1/2.0452
They do check with each other, that's reassuring.0460
Now, let's work out the sine and cosine of 2π/3 just using the oldfashioned common values.0463
Let me draw my unit circle.0471
There's 0, π/2, π, and 3π/2.0482
2π/3 is 2/3 the way from 0 to π.0490
There it is right there.0493
That's my 306090 triangle, so I know the values there are root 3 over 2 and 1/2.0496
I just have to figure out the sine and cosine, which ones are positive and which ones are negative.0505
I know that the cos(2π/3) because that's the xvalue, and the xvalue is negative, that's 1/2.0512
The sin(2π/3) is the yvalue, which is positive, that's root 3 over 2.0524
We worked those out just looking at the unit circle and remembering the common values but that checks out with the values we got from the formulas there sin(2π/3) and each one of the formulas for cos(π/3).0530
What we're doing there is working out each one of the formulas for sin(2x) and cos(2x) with x=π/3.0545
That separates it out into expressions in terms of sines and cosines of π/3, which I remember so I just plug those in and I get the sine and cosine of 2π/3.0556
All the cosine formulas agree with each other and they all check with the values that I can find just by looking at the unit circle.0567
Our next example is to use the double angle formulas to prove a trigonometric identity.0577
It's not so obvious how to start with this one.0584
We're actually going to start with the righthand side because it looks more complicated.0588
I'm going to start with the righthand side and that's 2tan(x)/1+tan^{2}(x).0592
I'm evaluating the righthand side, I'm going to work with it a bit and hopefully I can simplify it down to the lefthand side, but we'll see how it goes.0605
First thing, I'm going to do is to change everything into sines and cosines.0615
That's a good rule when you're not sure what to do with the trigonometric identity is to change everything into sines and cosines.0619
If you got a tangent or a secant, or a cosecant or a cotangent, convert it into sines or cosines.0626
It will probably make your life easier.0631
I'll write this as 2, tangent, remember is sin/cos, and 1+tan^{2}, that's 1+sin^{2}(x)/cos^{2}(x).0633
Now, I see a lot of cosines in denominators here, I think we're going to try to clear those out.0651
We multiply top and bottom by cos^{2}(x) and see what happens with that.0655
That's multiplying by 1, so that's safe.0662
On the top, I have 2sin(x), now I had a cos(x) in the denominator but I multiplied by cos^{2}, that gives me cos(x) in the numerator.0665
In the bottom, I have cos^{2} times 1+sin^{2}(x) over cos^{2}, that gives me 1×cos^{2} is cos^{2}(x), plus the cos^{2}(x) cancels with the denominator sin^{2}(x).0677
Now, look at this, the top is exactly 2sin(x)×cos(x).0695
I remember that, that's my formula for sin(2x).0701
Now the bottom, that's the Pythagorean identity, so that's just 1, cos^{2}+sin^{2}(x) is 1.0709
This converts into sin(2x), but that's equal to the lefthand side of what we were trying to prove.0716
We started with the righthand side because it looked a little more complicated there.0724
I see a bunch of tangents, I am not so sure what to do with those, I convert them into sines and cosines.0729
I see a lot of cosines in the denominator, so I multiply top and bottom by cos^{2}(x).0736
Then I start noticing some formulas that I recognized, 2sin(x)×cos(x) is a double angle formula, and cos^{2}(x)+sin^{2}(x), that's the Pythagorean identity.0745
It reduces down into the righthand side.0754
Let's try another example here, we're going to use the addition and subtraction formulas to derive a formula for tan(2x) in terms of tan(x).0760
Remember, we have formulas for sin(2x) and cos(2x), we're going to find a formula for tan(2x) just in terms of tan(x).0771
When we get that, we're going to check the formula on a common value π/6, because I know what the tangent of that is, and I know what the tan(2x) is, so we can check whether our formula works.0780
Let me start out with, tan(2x), don't know much about that except that the definition of tan(2x) is sin(2x)/cos(2x).0793
Now, I'm going to use, well, it's the addition and subtraction formulas but it's really the double angle formulas.0811
Of course, those come from the addition and subtraction formulas.0817
Now, sin(2x) is 2sin(x)×cos(x), that's the double angle formula for sine.0820
Of course, you find that out from the addition formula.0829
Cos(2x) is cos^{2}(x)sin^{2}(x), that was the first double angle formula for cosine.0832
Now, it's not totally obvious how to proceed next, but I know that I'm trying to get everything in terms of tan(x).0844
Right now, I've got a bunch of cosines lying around, I'd like to move those down into the denominator.0852
The reason is because tangent is sin/cos, so I would like to be dividing by cosines.0858
What I'm going to do is I'm going to divide the top by cos^{2}(x), and I'll divide the bottom by cos^{2}(x).0866
We're dividing top and bottom by cos^{2}(x), that's dividing by 1, so that's legitimate, we'll see what happens.0876
Now, in the numerator, we get 2sin(x), we had a cos(x) before, we divided by cos^{2}, we get 2sin(x)/cos(x).0882
In the bottom, we're dividing everything by cos^{2}(x), we get 1sin^{2}(x)/cos^{2}(x).0896
That's really nice because now we have sin/cos everywhere and that's tangent.0908
We are asked to find everything in terms of tan(x).0913
What we get here is 2sin/cos is tan(x) over 1sin^{2}(x)/cos^{2}(x) is tan^{2}(x).0918
Our formula, our double angle formula for tangent is tan(2x)=2tan(x)/(1tan^{2}(x)).0931
Now, I didn't list this at the beginning of the lecture as one of the main formulas that you really need to memorize.0944
It kind of depends on your trigonometry class, in some classes they will ask you to memorize this formula, this formula for tan(2x).0949
I don't think it's worth memorizing.0957
In my trigonometry classes, I don't require my students to memorize these formulas for tan(2x).0960
I do require them to memorize sin(2x) and cos(2x) and I figure they can work out the other ones from that.0965
You may have a teacher who requires you to memorize the formula for tan(2x).0973
If so, here it is, here is the formula that you want to remember.0979
Let's check that out on a value that I already know the tangent of, let's try x=π/6.0984
The tan(2π/6), according to this formula, would be 2×tan(π/6)/(1tan^{2}(π/6)).0994
Now, π/6 is a common value, tan(π/6), I remember that, I've got that one memorized, it's root 3 over 3.1011
If you don't have that one memorized, it probably is a good one to memorize, but if you don't have it memorized, you can work it out as long as you remember sine and cosine of π/6.1023
You just divide them together and get the tan(π/6).1034
This is 2 times root 3 over 3, over 1 minus root 3 over 3 squared.1037
Let's do a little over that, that's 2 times root 3 over 3, over 1 minus root 3 over 3 squared, is 3, over 3 squared is 9.1048
That's 3/9 which is 1/3.1063
This is 2 root 3 over 3, divided by 2/3.1067
Remember how you divide fractions, you flip it and multiply, 3/2, that cancels off the 2 and the 3, this whole thing boils down to just a root 3 as tan(2π/6).1076
Of course, 2π/6 is just π/3.1092
π/3 is another common value that I know the tangent of.1101
tan(π/3), I remember, is root 3, that's a common value.1105
Again, if you don't remember that, remember the sine and cosine of π/3, divide them together and you'll get root 3.1112
Look at that, our answers agree.1120
That confirms our formula for tan(2x).1122
To recap the important parts of that problem, we have to figure out tan(2x).1126
We wrote it as sin/cos of 2x.1132
We expanded each one of those using the double angle formulas that we learned at the beginning of the lesson.1135
Then, I was trying to get this in terms of tan(2x).1140
I wanted to get some cosines in the denominator, that's why I divided top and bottom by cos^{2}(x).1144
That converted the thing into something in terms of tan(x).1150
Then we checked that out by plugging in x=π/6, that's something that I know the tangent of, worked through the formula, and we got an answer square root of 3.1156
That checks the common value that I also know tan(π/3) is square root of 3.1170
We'll try some more examples of that later.1176
1 answer
Last reply by: Dr. William Murray
Fri Oct 17, 2014 1:13 PM
Post by Thomas Beguiristain on October 16, 2014
Question about the last example.
According to the double angle identity worked from the addition formula;
sin2x = 2sinXcosX , does that means " 2(sinX) âˆ™ (cosX) " or " 2 (sinX âˆ™ cosX ) " ?
If you work it out from the addition formula indicates the second option, however, in example 3, when deciding
2sinXcosX / cosX cosX , is taked as 2(sinX), since the results is 2sinX/ cosX = 2tan?
How come 2tan forms from 2sin and 1 cos ?
Also, one of the cosX from the denominator formed 1 with the one in the numerator, which is not mentioned. Is that what happened ?
1 answer
Last reply by: Dr. William Murray
Wed Nov 20, 2013 11:52 AM
Post by Sergio Trejo on November 13, 2013
i have a dificult time figuring out how you know what steps to incorporate to start, whats the easiest way to figure that? if any help,
thank you
1 answer
Last reply by: Dr. William Murray
Tue Aug 13, 2013 7:20 PM
Post by Taylor Wright on July 24, 2013
For Sin(3x), could you just turn it into sin(2x+x) and then go from there?
1 answer
Last reply by: Dr. William Murray
Tue Apr 2, 2013 12:57 PM
Post by Anurag Agrawal on February 26, 2013
Thanks! that was really helpful :)
1 answer
Last reply by: Dr. William Murray
Mon Jan 14, 2013 7:06 PM
Post by Jorge Sardinas on January 12, 2013
thanks for the video you made me understand the definition of double angle you are a fantastic and spectacular teacher [the third grader speaking!!!!!]
2 answers
Last reply by: Dr. William Murray
Mon Oct 29, 2012 9:05 PM
Post by Dr. William Murray on October 17, 2012
Ashlee, I'm sorry to hear that. As a teacher, I've been doing this stuff for many years, and sometimes it's hard to know just which parts are difficult for students. If you have some clever tricks to make some of this stuff easier, please post them!
I'm always looking for new ways to understand, remember, and calculate with these ideas, and some of my best tricks over the years cave come from my students. They'll show me something, I'll try it out with my next batch of students, and if it helps them too, I incorporate it into my regular repertoire. It would be great if you can add something too!
Thanks, and take care,
Will Murray
1 answer
Last reply by: Dr. William Murray
Mon Jan 14, 2013 7:21 PM
Post by Ashlee Josephs on October 16, 2012
This doesn't really help. You make things harder than my teacher.
1 answer
Last reply by: Dr. William Murray
Mon Jan 14, 2013 7:30 PM
Post by Judith Gleco on June 6, 2011
This video should include all steps. Not all can see what cancels until you brake it down step by step. I have never had trig before and my instructor at school and this video assume that you just have the double angle or the identities memorizied.
2 answers
Last reply by: Dr. William Murray
Mon Jan 14, 2013 7:30 PM
Post by aloosh aloosh on May 12, 2011
you make things so hard