For more information, please see full course syllabus of Trigonometry
For more information, please see full course syllabus of Trigonometry
Pythagorean Identity
Main formulas:
 The Pythagorean theorem: The side lengths of a right triangle satisfy a^{2} + b^{2} = c^{2}.
 The Pythagorean identity: For any angle x, we have sin^{2} x + cos^{2} x = 1.
Example 1:
Use the Pythagorean theorem to prove the Pythagorean identity.Example 2:
If cosθ = 0.47 and θ is in the fourth quadrant, find sinθ .Example 3:
Verify the following trigonometric identity :

Example 4:
Use the Pythagorean identity to prove the Pythagorean theorem.Example 5:
If sinθ = − [5/13] and θ is in the third quadrant, find cosθ .Pythagorean Identity
 Use the Pythagorean Identity to solve for sinθ: sin^{2}θ + cos^{2}θ = 1
 sin^{2}θ + (0.32)^{2} = 1 ⇒ sin^{2}θ + (0.1024) = 1 ⇒ sin^{2}θ = 1  0.1024 ⇒ sin^{2}θ = 0.8974 ⇒ sinθ = ±√{0.8974}
 sinθ = ± 0.9473 Remember the mnemonic ASTC. This tells which quadrant sine values will be positive. Sine is only positive in quadrants I and II
 Use the Pythagorean Identity to solve for cosθ: sin^{2}θ + cos^{2}θ = 1
 (0.47)^{2} + cos^{2}θ = 1 ⇒ (0.2209) + cos^{2}θ = 1 ⇒ cos^{2}θ = 1  0.2209 ⇒ cos^{2}θ = 0.7791 ⇒ cosθ = ±√{0.7791}
 cosθ = ± 0.8827 Remember the mnemonic ASTC. This tells which quadrant cosine values will be positive. Cosine is only positive in quadrants I and IV
 Use the Pythagorean Identity to solve for sinθ: sin^{2}θ + cos^{2}θ = 1
 sin^{2}θ + ([12/13])^{2} = 1 ⇒ sin^{2}θ + ([144/169]) = 1 ⇒ sin^{2}θ = [169/169] − [144/169] ⇒ sin^{2}θ = [25/169] ⇒ sinθ = ±√{[25/169]}
 sinθ = ±[5/13] Remember the mnemonic ASTC. This tells which quadrant sine values will be positive. Sine is only positive in quadrants I and II
 Use the Pythagorean Identity to solve for cosθ: sin^{2}θ + cos^{2}θ = 1
 ([3/5])^{2} + cos^{2}θ = 1 ⇒ ([9/25]) + cos^{2}θ = 1 ⇒ cos^{2}θ = [25/25]  [9/25] ⇒ cos^{2}θ = [16/25] ⇒ cosθ = ±√{[16/25]}
 cosθ = ±[4/5] Remember the mnemonic ASTC. This tells which quadrant cosine values will be positive. Cosine is only positive in quadrants I and IV
 Try to get the left hand side to look like the right hand side because it is the more complicated side
 1  cos^{2}θ = sin^{2}θ by multiplying
 Try to get the left hand side to look like the right hand side because it is the more complicated side
 [(sinα)/(cosα)] · cosα = sinα because tanα = [(sinα)/(cosα)]
 [(sinαcosα)/(cosα)] = sinα by multiplying
 Try to get the left hand side to look like the right hand side because it is the more complicated side
 1  sin^{2}θ = cos^{2}θ by multiplying
 Try to get the left hand side to look like the right hand side because it is the more complicated side
 sin^{2}θ  (1  sin^{2}θ) = 2sin^{2}θ  1 by substituting the Pythagorean Identity for cos^{2}θ
 sin^{2}θ  1 + sin^{2}θ = 2sin^{2}θ  1 by multiplying
 Try to get the left hand side to look like the right hand side because it is the more complicated side
 [(sinθsinθ+ cosθcosθ)/(cosθsinθ)] = cscθsecθ by adding fractions
 [(sin^{2}θ+ cos^{2}θ)/(cosθsinθ)] = cscθsecθ by multiplying
 [1/(cosθsinθ)] = cscθsecθ by the Pythagorean Identity: sin^{2}θ + cos^{2}θ = 1
 [1/(cosθ)] · [1/(sinθ)] = cscθsecθ by separating fractions
 [(1 + sinθ+ 1 − sinθ)/((1 − sinθ)(1 + sinθ))] = 2sec^{2}θ by adding fractions
 [2/(1 − sin^{2}θ)] = 2sec^{2}θ by simplifying
 [2/(cos^{2}θ)] = 2sec^{2}θ by the Pythagorean Identity
*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
Pythagorean Identity
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 0:00
 Pythagorean Identity 0:17
 Pythagorean Triangle
 Pythagorean Identity
 Example 1: Use Pythagorean Theorem to Prove Pythagorean Identity 1:14
 Example 2: Find Angle Given Cosine and Quadrant 4:18
 Example 3: Verify Trigonometric Identity 8:00
 Extra Example 1: Use Pythagorean Identity to Prove Pythagorean Theorem
 Extra Example 2: Find Angle Given Cosine and Quadrant
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: Pythagorean Identity
We are learning about the Pythagorean identity, what we are going to try now is to start with the Pythagorean identity and prove the Pythagorean theorem.0000
Remember what we did in the earlier example, was we started with the Pythagorean theorem and we proved the Pythagorean identity.0009
The point of this is to show that you can get from one to the other or from the other back to the first one.0015
And so that the two factor are equivalent even though one seems like geometric fact and one seems like a trigonometric fact.0022
Let us do that, remember that the Pythagorean identity says that sin^{2}(x) + cos^{2}(x) = 1.0033
The Pythagorean theorem, that is the theorem about right triangles, so let me set up a right triangle here.0045
I will label the sides a, b, and c, that is Pythagorean identity, now I’m going to use my SOHCAHTOA.0055
SOHCAHTOA tells us that if we have an angle here, I will this angle (x) in the corner here, the sin(x) is equal to the opposite/hypotenuse.0065
The opposite side here is b/c, the cos(x) is equal to a/c, the adjacent/hypotenuse.0090
I’m going to plug those in the Pythagorean identity because remember we are allowed to use the Pythagorean identity here.0105
That says the sin^{2}(x) + cos^{2}(x) = 1, if I plug those in I will get b/c^{2} + a/c^{2}= 1.0112
Now I’m just going to do a little bit of algebraic manipulation b^{2}/c^{2} + a^{2}/c^{2} is equal to 1.0127
I’m going to multiply both sides there by c^{2} and that will clear my denominators on the left I got b^{2} + a ^{2} and on the right I have c^{2}.0137
If I switch around the two terms here a^{2} + b^{2} is equal to c^{2}.0151
That is the very familiar equation we approved the Pythagorean theorem.0160
We started out with the Pythagorean identity from trigonometry and we ended up proving the Pythagorean theorem from geometry a^{2} + b^{2}=c^{2}.0177
It is just a matter of writing down the right angle in one corner of the triangle and then working through the algebra, you end up with the Pythagorean theorem from geometry.0190
That shows that the Pythagorean identity and the Pythagorean theorem really are equivalent to each other.0200
You ought to be able to start with either one and prove the other one.0206
Let us try one more example, we are given that sin(theta) is 5/13, and (theta) is in the third quadrant and we want to find cos(theta).0000
Let me try graphing out what that might be.0010
Ok (theta) is in the third quadrant, that is down here and so its sum angle down there, I do not know exactly where it is but I will draw it down there.0022
What I’m given is that sin(theta) is 5/13 and I want to find cos(theta).0035
Well I have this Pythagorean identity that says sin^{2}(theta) + cos^{2}(theta) = 1.0040
I will plug in sin(theta ) that is 5/13^{2} + cos^{2}(theta) =10051
5/13 when you square, the negative goes away so we get 25/169 + cos^{2}(theta) is equal to0062
Well I’m going to have to subtract the 25/169 so I will write 1 as 169/169 then I will subtract 25/169 from both sides.0074
I got cos^{2}(theta) is 144/169, then if I take the square root of both sides to solve for cos(theta), I get cos(theta) is equal to + or – square root of 144 is 12, square root of 169 is 13.0087
I know the my cos(theta) is equal to either positive or negative 12/13.0110
That is all I can get from the Pythagorean identity because it only told me what cos^{2}(theta) is, I can not figure out from that whether cos(theta) is positive or negative.0116
But the problem gave us a little extra information, it says that (theta) is in the quadrant.0127
Knowing that (theta) is in the third quadrant, I looked down there and I remember that cos is equal to the x coordinate of my angle.0133
Cos is the x coordinate, remember all students take calculus, down there in the third quadrant tan are positive but nothing else is positive.0145
That means that cos is not positive, it is negative.0158
The cos(theta) is equal to 12/13 and must be the negative value because it is down in the third quadrant, that is where the x coordinate is negative.0162
The key to this problem is remembering the Pythagorean identity, sin^{2}(theta) + cos^{2}(theta)=1.0181
Then you plug the value you are given into the Pythagorean identity and you try to solve for cos(theta).0189
Once you work through the arithmetic you get the value for cos(theta) but you do not know if it is positive or negative.0197
Then you go over and look whether what quadrant the angle is in, it is in the quadrant and then you either remember the all students take calculus.0203
That tells you the plus or minus on the different functions or you just remember that in the third quadrant the x values are negative so the cos value has to be negative.0212
Either way you end up with cos(theta) is equal to 12/13.0222
That is the end of our set on the Pythagorean identity, this is www.educator.com.0229
Hello, this is the trigonometry lectures for educator.com and today we're going to learn about probably the single most important identity in all trigonometry which is the Pythagorean identity.0000
It says that sin^{2}x + cos^{2}x = 1.0011
This is known as the Pythagorean identity.0017
It takes its name from the Pythagorean theorem which you probably already heard of.0019
The Pythagorean theorem says that if you have a right triangle, very important that one of the angles be a right angle, then the side lengths satisfy a^{2} + b^{2} = c^{2}.0024
You probably heard that already.0040
The new fact for trigonometry class is that sin^{2}x + cos^{2}x = 1.0042
What we're going to learn is we work through the exercises for these lectures.0050
Is if these are really two different sides of the same coin, you should think of this as being sort of facts that come out of each other.0055
In fact, we're going to use each one of these facts to prove the other one.0064
These are really equivalent to each other.0069
Let's go ahead and start doing that.0071
In our first example, we are going to start with the Pythagorean theorem, remember that's a^{2} + b^{2} = c^{2}.0074
We're going to try to prove the Pythagorean identity sin^{2}x + cos^{2}x = 1.0084
The way we'll do that is let x be an angle.0095
Let's draw x on the unit circle.0107
The reason I'm drawing it on the unit circle is because remember the definition of sine and cosine is the x and y coordinates of that angle.0112
If we draw x on the unit circle, the hypotenuse has length 1 and the xcoordinate of that point, remember, is the cos(x), and the ycoordinate is the sin(x).0124
Now, what we have here is a right triangle and we're allowed to use the Pythagorean theorem, we're given that and we're going to use that and try to prove the Pythagorean identity.0148
The Pythagorean theorem says that in a right triangle, by the Pythagorean theorem...0160
Let me draw my right triangle a little bigger, there's x, there's 1, this is cosx, this is sinx.0174
By the Pythagorean theorem, one side squared, let me write that first of all as cosine x squared plus the other side squared is equal to 1^{2}.0186
That's the length of the hypotenuse.0205
If we just do a little semantic cleaning up here, 1^{2}, of course, is just 1, cosine x squared, the common notation for that is cos^{2}x + sin^{2}x = 1.0207
We just derived an equation, and look this is the Pythagorean identity.0228
What we've done is we started by assuming the Pythagorean theorem and then we used the Pythagorean theorem to derive the Pythagorean identity.0246
Let's see an application of that in the next example.0256
We're given that θ is an angle whose cosine is 0.47, and θ is in the fourth quadrant.0260
We have to find sinθ.0266
Let me draw θ, θ is somewhere down there in the fourth quadrant.0272
I don't know exactly where it is but θ looks like that.0279
Here is what I know, by the Pythagorean identity, sin^{2}θ + cos^{2}θ = 1.0284
I'm going to fill in the one that I know, cosθ, cosθ is 0.47.0295
This is 0.47^{2} = 1 + sin^{2}θ.0302
Now, 0.47, that's not something I can easily find the square of, so I'll do that on my calculator.0310
0.47^{2} = 0.2209, so that's +0.2209, sin^{2}θ +0.2209 = 1, sin^{2}θ = 1  0.2209, which is 0.7791.0317
Sinθ, if we take the square root of both sides, sinθ is equal to plus or minus the square root of 0.7791, which is approximately equal to 0.8827.0360
Now, it's plus or minus because I know that sine squared is this positive number, but I don't know whether this sine is a positive or negative.0382
We're given more information in the problem, θ is in the fourth quadrant.0390
Remember, sine is the ycoordinate, so the sine in the fourth quadrant is going to be negative because the ycoordinate is negative.0395
Because θ is in quadrant 4, sinθ is going to be negative, so we take the negative value, sinθ is approximately equal to 0.8827.0414
The whole key to doing this problem was to start with the Pythagorean identity sin^{2}θ + cos^{2}θ = 1.0446
Once you're given sine or cosine, you could plug those in and figure out the other one except that you can't figure out whether they're positive or negative.0457
Their identity doesn't tell you that so we had to get this little extra information about θ being in the fourth quadrant, that totals that the sinθ is negative and we were able to figure out that it was 0.8827.0463
Let's try another example of that.0480
We're going to verify a trigonometric identity.0483
This is a very common problem in trigonometry classes as you'll be given some kind of identity involving the trigonometric functions and you have to verify it.0486
For this one, what I want to do is start with the right hand side.0496
I'm going to label this RHS.0503
RHS stands for righthand side.0504
The righthand side here is equal to sinθ/(1  cosθ).0510
Now, I'm going to do a little trick here which is very common when you have something plus something in the denominator, or something minus something in the denominator.0519
The trick is to multiply the conjugate of that thing.0529
Here I have 1  cosθ in the denominator, I'm going to multiply by 1 + cosθ, and then, of course, I have to multiply the numerator by the same thing, 1 + cosθ).0533
The reason you do that, this is really an algebraic trick so you probably have learned about this in the algebra lectures.0547
The reason you do that is you want to take advantage of this formula, (a + b) × (a  b^{2}.0554
That's often the way of simplifying things using that algebraic formula.0566
What we get here in the numerator is (sinθ) × (1 + cosθ), in the denominator, using this (a^{2} + b^{2}) formula, we get (1  cos^{2}θ).0570
Now, let's remember the Pythagorean identity.0586
Pythagorean identity says sin^{2}θ + cos^{2}θ = 1.0590
That means 1  cos^{2}θ = sin^{2}θ.0596
We can substitute that in into our work here, sinθ×(1 + cosθ).0603
The denominator, by the Pythagorean identity, turns into sin^{2}θ.0613
We get some cancellation going on, the sine in the numerator cancels with one of the sines in the denominator leaving us just with (1 + cosθ)/sinθ in the denominator.0623
That's the same as the lefthand side that we started with.0639
We started with the righthand side and we're able to work it all the way down and end up with the lefthand side verifying the trigonometric identity.0644
There were sort of two key steps there.0654
One was in looking at the denominator and recognizing that it was a good candidate to invoke this algebraic trick where you multiply by the conjugate.0657
If you have (a + b), you multiply by (a  b).0669
If you have (a  b), you multiply by (a + b).0672
Either way, you get to invoke this identity.0674
Here, we had (a  b), we multiplied by (a + b) and then we got to invoke the identity and get something nice on the bottom.0679
The second trick there was to remember the Pythagorean identity and notice that (1  cos^{2}θ) converts into sin^{2}θ.0686
Once we did that, it was pretty to simplify it down to the lefthand side of the original identity.0697
We'll try some more examples later.0703
1 answer
Last reply by: Dr. William Murray
Tue Aug 5, 2014 3:52 PM
Post by Farhat Muruwat on August 1, 2014
This is really fundamental compared to what they teach in school. I need help with trig identities and examples!
3 answers
Last reply by: Dr. William Murray
Mon Dec 22, 2014 9:15 PM
Post by Rand Mahmood on December 28, 2013
When you check the answer on the second question (practice questions) it gives you .9473, while the actual answer is .8827. Can you take a look at it?
1 answer
Last reply by: Dr. William Murray
Fri Jul 5, 2013 9:29 AM
Post by Manfred Berger on June 28, 2013
I have encountered a bit of a bump in the road with example 3. Is there a way to navigate around the fact that the denominators vanish for multiples of pi?
1 answer
Last reply by: Dr. William Murray
Sat Nov 3, 2012 6:16 PM
Post by Daisy soto on November 1, 2012
I am having trouble with the video...it will not load to the 1st example
1 answer
Last reply by: Dr. William Murray
Tue Apr 2, 2013 12:51 PM
Post by Dr. William Murray on October 30, 2012
Hi Carri,
The idea here is that you can read the example problems in the video and then stop the video and try to work them out on your own. Then you can restart the video and see if your answers agree with mine. So that's five practice problems.
You're probably taking a trig class in high school or college right now. (These Educator videos aren't meant to be standalone courses; they're meant to be supplements to regular courses.) So you probably have a textbook with lots of practice problems, and hopefully, the answers to the odd problems are in the back.
For specific homework questions, you can try posting on here. If they're relevant to the video lectures and I think other people would benefit, I'll answer them as I have time. If you're just looking for help with your own stuff, I recommend talking to your teacher or getting a local tutor  it's easier to work specific questions out face to face.
Thanks for taking trigonometry!
Will Murray
1 answer
Last reply by: Dr. William Murray
Tue Apr 2, 2013 12:48 PM
Post by carri campbell on October 30, 2012
I'm disappointed in this already. Where are the practice problems? I have specific homework questions I would like to ask, is that a possibility?
1 answer
Last reply by: Dr. William Murray
Wed Aug 22, 2012 1:25 PM
Post by jim kwon on August 17, 2012
what if it is sin^2 40 degree + cos^2 40 degree?
1 answer
Last reply by: Sandra Bruce
Mon Jun 4, 2012 2:17 PM
Post by David Burns on August 17, 2011
Again, the quick notes section on this page does not correspond to the lecture  it's one off. Please fix this, it's kind of annoying.
1 answer
Last reply by: Marco Zendejo
Tue Jun 21, 2011 10:48 PM
Post by Donald Bada on June 10, 2011
why is the x place between the sin and x instead of after the x?
1 answer
Last reply by: Dr. William Murray
Tue Apr 2, 2013 12:50 PM
Post by Greg Banta on July 14, 2010
Can't here sound on this one :(
I heard sound on others