Pythagorean Trigonometric Identities

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Lesson Priority: VIP Knowledge

Pre-Calculus $\longrightarrow$
Trigonometry $\longrightarrow$
  • Learn the most fundamental Pythagorean trigonometric identity, which shows up a lot in advanced mathematics
  • Derive two other equivalent Pythagorean identities from the first one
  • Use these identities to answer questions about unknown quantities
Lesson Description

One of the most common identities of trigonometry is derived from applying the Pythagorean Theorem to our coordinate plane triangle scheme. This lesson covers the derivation of Pythagorean trig identities, and showcases ways to use them to answer questions

Practice Problems

Practice problems and worksheet coming soon!


The Most Important Identity

As you learn trig, the word "identity" will mean a couple of things to you. Generally speaking, identity refers to two things that are equivalent.Without further ado, let's see the most important identity you'll (almost 100% certainly) be expected to know from memory.
The Pythagorean Trig IdentityFor any angle $\theta$, the following relationship exists.$$\sin^2(\theta) + \cos^2(\theta) = 1$$
Most of us will get by just fine taking this information as a fact to memorize and running with it. For those who need / want to see where it comes from, the next few paragraphs will show just that (that is to say, skip to the next header » if you don't care).When we study the definitions of trig functions in the coordinate plane, we defined the sine and cosine of the standard position angle that goes through a point as follows:$$\sin(\theta) = \frac{y}{r}$$$$\cos(\theta) = \frac{x}{r}$$This definition comes from the SOH-CAH-TOA idea applied to the diagram of whatever point in the plane that we're looking at:In words, $x$ is the $x$ direction distance of the point from the origin, $y$ is the $y$ direction distance, and $r$ is the hypotenuse of the resulting right triangle.From the figure, it is also clear that, by the Pythagorean Theorem, the relationship$$x^2 + y^2 = r^2$$exists. We can combine these facts by first dividing the Pythagorean Theorem relationship on both sides by $r^2$, and then using substitution.$$\frac{x^2 + y^2}{r^2} = \frac{r^2}{r^2}$$$$\longrightarrow \frac{x^2}{r^2} + \frac{y^2}{r^2} = 1$$$$\longrightarrow \left(\frac{x}{r}\right)^2 + \left(\frac{y}{r}\right)^2 = 1$$Now comes the substitution step. Replace $x/r$ with $\cos(\theta)$, and $y/r$ with $\sin(\theta)$.$$\sin^2(\theta) + \cos^2(\theta) = 1$$Voila! This relationship is true, regardless of what angle $\theta$ is. And because this relationship is derived from the Pythagorean relationship between $x$, $y$, and $r$, this identity and the two equivalent forms that follow are referred to as "Pythagorean Identities" by teachers and textbooks.

Two Equivalent Forms

Because of the way in which all six trig functions » can be defined in terms of $\sin$ and $\cos$, we can quickly re-write this master identity in two other ways (and we'll commonly need to do so).Take the core identity and divide through by $\sin^2(\theta)$:$$\sin^2(\theta) + \cos^2(\theta) = 1$$$$\frac{\sin^2{\theta}}{\sin^2{\theta}} + \frac{\cos^2{\theta}}{\sin^2{\theta}} = \frac{1}{\sin^2{\theta}}$$Recalling that $\cos(\theta) / \sin(\theta) = \cot(\theta)$, and that $1 / \sin(\theta) = \csc(\theta)$, the equation simplifies to$$1 + \cot^2(\theta) = \csc^2(\theta)$$This is the first of the two spin-off forms of the most important identity. You can probably make a good guess as to how we'll get the other!Dividing the original most important identity by $\cos^2(\theta)$ yields$$\sin^2(\theta) + \cos^2(\theta) = 1$$$$\frac{\sin^2{\theta}}{\cos^2{\theta}} + \frac{\cos^2{\theta}}{\cos^2{\theta}} = \frac{1}{\cos^2{\theta}}$$$$\tan^2(\theta) + 1 = \sec^2(\theta)$$Thus, in total, the three Pythagorean trigonometric identities, with the first being the "base" or "master" one, are$$\sin^2(\theta) + \cos^2(\theta) = 1$$$$1 + \cot^2(\theta) = \csc^2(\theta)$$$$\tan^2(\theta) + 1 = \sec^2(\theta)$$
Pro Tip
Under exam pressure, it is a very common mistake to see students "mess up" the other two Pythagorean identities by mixing up terms. The division concept I used to derive these other two identities isn't just something I'm showing for fun - it's what I actually do when I need them. It's super fast and accurate to start with the foundation identity $\sin^2(\theta) + \cos^2(\theta) = 1$ and divide out by either $\sin^2$ or $\cos^2$ to get each of the other two versions. I highly recommend you do the same!

Using Pythag Trig IDs

Now that we have three shiny new facts to memorize, it's fair to ask what we're going to do with them.For starters, when we do need to pull out these formulas, be prepared to re-write them in any equivalent algebra form. For example, sometimes we'll need to know that $\sin^2(\theta) + \cos^2(\theta)$ is $1$, but sometimes we'll need to know that $1 - \cos^2(\theta)$ is $\sin^2(\theta)$.Typically these are most useful for a task we call proving trig identities ». This is a skill akin to puzzle solving that will require a combination of these types of base identities and strong algebra manipulation skills, which we will study in its own lesson very soon.Another relatively quick and common task that we can use these Pythagorean trig identities for is algebraically finding all six trig function values when we're only given one of the six values. We are able to do this with a picture and SOH-CAH-TOA, but these identities allow us to work purely algebraically.
Example 1Given $\sin(\theta) = 2/7$ and $\tan(\theta) \lt 0$, find the values of all six trig functions for the same angle $\theta$.$\blacktriangleright$ Knowing the value of the sine, we can immediately use the core Pythagorean relationship.$$\sin^2(\theta) + \cos^2(\theta) = 1$$$$\Rightarrow \left(\frac{2}{7}\right)^2 + \cos^2(\theta) = 1$$$$\cos^2(\theta) = 1 - \left(\frac{2}{7}\right)^2$$$$\cos^2(\theta) = \frac{45}{49}$$$$\cos(\theta) = \pm \frac{3\sqrt{5}}{7}$$Now, we can look at the other piece of given information - the fact that $\tan(\theta)$ is negative. Since the value of the sine is positive, the unknown angle $\theta$ must be a quadrant 2 angle for the tangent value to be negative. Additionally, if $\theta$ is a quadrant 2 angle, then the cosine value must also be negative.This deductive work is very important. Since the algebra left us with a $\pm$ sign, it is up to us to figure out which sign the answer should have based on whatever other information we have.As we said, this angle is in Quadrant 2. Therefore, we know that $\cos(\theta) = -3\sqrt{5}/7$.With the sine and cosine values in hand, we can use the classic trig function relationships to get the four remaining trig values of the angle $\theta$.Immediately, we know that$$\sin(\theta) = \frac{2}{7} \Rightarrow \csc(\theta) = \frac{7}{2}$$$$\cos(\theta) = \frac{-3\sqrt{5}}{7} \Rightarrow \sec(\theta) = \frac{-7}{3\sqrt{5}}$$$$= \frac{-7\sqrt{5}}{15}$$To get tangent and cotangent, use their definitions in terms of sines and cosines:$$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$$$$=\frac{\frac{2}{7}}{\frac{-3\sqrt{5}}{7}} = \frac{-2\sqrt{5}}{15}$$$$\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$$$$=\frac{\frac{-3\sqrt{5}}{7}}{\frac{2}{7}} = \frac{-3\sqrt{5}}{2}$$And there we have it!

Put It To The Test

Example 2List all three Pythagorean Trig Identities from memory.
Show solution
$\blacktriangleright$ While you need to straight-up memorize the so-called "core" identity, remember that division helps us derive the other two on-the-fly from the core equation.$$\sin^2(x) + \cos^2(x) = 1$$And the other two, are$$1 + \cot^2(x) = \csc^2(x)$$$$\tan^2(x) + 1 = \sec^2(x)$$It doesn't matter what you use for the variable unless you were told to use a particular one. We usually use $x$ or $\theta$.
Example 3Under all possibilities, algebraically determine all six trig values of the angle $\theta$ given that $\sec(\theta) = 4$.
Show solution
$\blacktriangleright$ Let's start with the trig identity that involves the secant:$$\tan^2(\theta) + 1 = \sec^2(\theta)$$Plugging in what we know:$$\tan^2(\theta) + 1 = (4)^2$$$$\longrightarrow \tan^2(\theta) = 15$$$$\tan(\theta) = \pm \sqrt{15}$$Since we were asked to answer the question under all possibilities, and since we do not have any additional information, we have to consider both cases.If $\tan(\theta) = \sqrt{15}$, then the angle $\theta$ is a Quadrant 1 angle, because that's where both secant and tangent values are positive. If $\tan(\theta) = -\sqrt{15}$, then $\theta$ is a Quadrant 4 angle, which is the quadrant in which the secant is positive but the tangent is negative.Let's do each scenario's results, starting with assuming $\theta$ is a Q1 angle.Given $\tan(\theta)$ is $\sqrt{15}$, we know by the reciprocal relationships that $\cot(\theta)$ is $1/\sqrt{15}$, which means that $\cot^2(\theta) is $1/15$. Put this into the Pythag relationship that uses cotangent:$$1 + \cot^2(\theta) = \csc^2(\theta)$$$$1 + \frac{1}{15} = \csc^2(\theta)$$$$\longrightarrow \frac{16}{15} = \csc^2(\theta)$$$$\csc(\theta) = \pm \frac{4}{\sqrt{15}} = \pm \frac{4\sqrt{15}}{15}$$In this scenario we decided that the angle has to be in Quadrant 1, so the value of cosecant is positive.$$\csc(\theta) = \frac{4\sqrt{15}}{15}$$Finally, using simple reciprocal relationships, let's write down the last two trig functions that we haven't stated yet: sine and cosine.$$\sec(\theta) = 4 \longrightarrow \cos(\theta) = \frac{1}{4}$$$$\csc(\theta) = \frac{4\sqrt{15}}{15} \longrightarrow \sin(\theta) = \frac{\sqrt{15}}{4}$$Now let's look at the scenario where $\theta$ is a Q4 angle. The only thing that will be different is the plus and minus signs, and since we know the numeric parts of the end results will be same, we can just use those numbers and choose the correct signs based on the fact that we're looking at a Q4 angle.In Quadrant 4, the cosine and secant are positive, and everything else is negative. Therefore we will have:$$\sin(\theta) = -\frac{\sqrt{15}}{4}$$$$\cos(\theta) = \frac{1}{4}$$$$\tan(\theta) = -\sqrt{15}$$$$\cot(\theta) = -\frac{\sqrt{15}}{15}$$$$\sec(\theta) = 4$$$$\csc(\theta) = -\frac{4\sqrt{15}}{15}$$
Example 4Reduce the following expression to one that uses as few trig functions as possible.$$\frac{\cos(x)}{1 + \sin(x)} + \frac{1+\sin(x)}{\cos(x)}$$
Show solution
$\blacktriangleright$ Let's start with getting these to be a common denominator. That means multiplying the first fraction by $\cos(x)$ top and bottom, and multiplying the second fraction by $(1+\sin(x))$ top and bottom.$$\frac{\cos(x)}{1 + \sin(x)} \cdot \frac{\cos(x)}{\cos(x)} + \frac{1+\sin(x)}{\cos(x)} \cdot \frac{(1+\sin(x))}{(1 + \sin(x))}$$$$ \longrightarrow \frac{\cos^2(x) + (1 + \sin(x))^2}{\cos(x)(1 + \sin(x))}$$Let's expand out the squared binomial on top. Although we often seek to find and keep factors, the squaring operation is a hint to expand because we expect pythag trig identities to appear when we work with squared trig functions.$$ \longrightarrow \frac{\cos^2(x) + (1 + 2\sin(x) + \sin^2(x))}{\cos(x)(1 + \sin(x))}$$$$ = \frac{\sin^2(x) + \cos^2(x) + 1 + 2\sin(x)}{\cos(x)(1 + \sin(x))}$$Now on top, recognize that we have the sum of $\sin^2(x)$ and $\cos^2(x)$, which we know is equal to $1$.$$ \longrightarrow \frac{1 + 1 + 2\sin(x)}{\cos(x)(1 + \sin(x))}$$$$ = \frac{2 + 2\sin(x)}{\cos(x)(1 + \sin(x))}$$$$ = \frac{2\cancel{(1 + \sin(x))}}{\cos(x)\cancel{(1 + \sin(x))}}$$$$ = \frac{2}{\cos(x)}$$$$ = 2\sec(x)$$
Lesson Takeaways
  • Memorize the "core" Pythagorean Trig identity $\sin^2(x) + \cos^2(x) = 1$
  • Be able to derive the other two Pythagorean trig identities that use $\tan^2(x)$, $\cot^2(x)$, $\sec^2(x)$, and $\csc^2(x)$
  • Use these equations to algebraically determine all six trig functions of an unknown angle
  • Reduce complicated expressions to simpler forms by using these identities

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