**Pythagorean identities**are important trigonometric identities that allow us to simplify trigonometric expressions, derive other trigonometric identities, and solve equations. Understanding these identities is essential when building a strong foundation to master trigonometric concepts and learn more advanced math topics.

**In this article, weâ€™ll break down**

*The Pythagorean identities are derived from the Pythagorean theorem. We use these identities to simplify processes involving trigonometric expressions, equations and identities.***the proof of these three Pythagorean identities,**show key applications of these identities, and provide ample examples to help you master this topic.

## What Are the Pythagorean Identities?

The Pythagorean identities are**the three most-used trigonometric identities that have been derived from the Pythagorean theorem**, hence its name. Here are the three Pythagorean identities that weâ€™ll learn and apply throughout our discussion. \begin{aligned}\color{DarkOrange}\textbf{Pythagorean}\,\,\color{DarkOrange}\textbf{Iden}&\color{DarkOrange}\textbf{tities}\\\\\sin^2\theta + \cos^2 \theta = &1\\\tan^2 \theta +1= \sec^2 &\theta\\1+ \cot^2 \theta = \csc^2 &\theta\end{aligned} The first Pythagorean identity is

**the most fundamental**since it will be easier for us to derive the two remaining Pythagorean identities with this. From the first equation, the Pythagorean states that the sum of squares of $\sin \theta$ and $\cos \theta$ will always be equal to $1$. \begin{aligned}\sin^2 45^{\circ} + \cos^2 45^{\circ} &= 1\\\sin^2 \left(\dfrac{2\pi}{3}\right) + \cos^2 \left(\dfrac{2\pi}{3}\right)&= 1\end{aligned} Why donâ€™t we

**evaluate the left-hand side of the equations**to confirm that the Pythagorean identity $\sin^2 \theta + \cos^2\theta =1$ remain true for these two equations?

\begin{aligned}\boldsymbol{\sin^2 45^{\circ} + \cos^2 45^{\circ}} &= \boldsymbol{1}\end{aligned} | \begin{aligned}\boldsymbol{\sin^2 \dfrac{2\pi}{3}+ \cos^2 \dfrac{2\pi}{3}}&= \boldsymbol{1}\end{aligned} |

\begin{aligned}\sin^2 45^{\circ} + \cos^245^{\circ} &=1\\\left(\dfrac{1}{\sqrt{2}}\right)^2+ \left(\dfrac{1}{\sqrt{2}}\right)^2&= 1\\\dfrac{1}{2}+ \dfrac{1}{2}&=1\\1&=1 \checkmark\end{aligned} | \begin{aligned}\sin^2 \left(\dfrac{2\pi}{3}\right) + \cos^2\left(\dfrac{2\pi}{3}\right)&=1\\\left(\dfrac{\sqrt{3}}{2}\right)^2+ \left(-\dfrac{1}{2}\right)^2&= 1\\\dfrac{3}{4}+ \dfrac{1}{4}&=1\\1&=1 \checkmark\end{aligned} |

**will remain true for all angle measures**. This is what makes these identities helpful â€“ we can simplify complex trigonometric expressions and use them to rewrite and prove identities. For us to appreciate the Pythagorean identities, itâ€™s important that we

**understand their origin and derivation first**.

### Pythagorean Identity Definition and Proof

Given an angle, $\theta$, the Pythagorean identities allow us to**show the relationship between the squares of the trigonometric ratios**. Letâ€™s place our focus on the first Pythagorean identity. \begin{aligned}\sin^2 \theta + \cos^2 \theta &= 1\end{aligned} Itâ€™s most crucial to remember this Pythagorean identity â€“ thatâ€™s because once we know this by heart, the two remaining Pythagorean identities

**will be easy to remember and derive**. For now, letâ€™s understand that we can apply the Pythagorean Theorem to derive the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$. Suppose that

**we have a unit circle**. Observe the relationship between the sides of the right triangle formed inside the first quadrant of the unit circle as shown below.

**the side adjacent to**$\theta$

**is equal to**$\cos \theta$

**and the side opposite**$\theta$ is $\sin \theta$. Apply the Pythagorean theorem to relate the sides of the right triangle formed.

**the side adjacent to**$\theta$

**is equal to**$\cos \theta$

**and the side opposite**$\theta$ is $\sin \theta$. Apply the Pythagorean theorem to relate the sides of the right triangle formed. This proves our first Pythagorean identity, $\sin^2\theta + \cos^2 \theta = 1$. To prove that $\sec^2 \theta- \tan^2 \theta = 1$ is true,

**divide both sides of the equation by**$\cos^2 \theta$. Apply the basic trigonometric identities $\sec \theta =\dfrac{1}{\cos\theta}$ and $\tan \theta =\dfrac{\sin \theta}{\cos \theta}$. \begin{aligned}\sin^2\theta+\cos^2\theta &=1\\\dfrac{\sin^2\theta}{\color{DarkOrange}\cos^2\theta}+\dfrac{\cos^2\theta}{\color{DarkOrange}\cos^2\theta}&=\dfrac{1}{\color{DarkOrange}\cos^2\theta}\\\left(\dfrac{\sin\theta}{\cos\theta}\right)^2+1&=\left(\dfrac{1}{\cos\theta}\right)^2\\\color{DarkOrange}\boldsymbol{\tan^2 \theta + 1} &\color{DarkOrange}\boldsymbol{=\sec^2\theta}\end{aligned} Derive the third Pythagorean identity by applying a similar process. This time,

**divide both sides of**$\sin^2\theta + \cos^2\theta =1$

**by**$\sin^2\theta$. Use the trigonometric identities $\csc \theta =\dfrac{1}{\sin\theta}$ and $\cot \theta =\dfrac{\cos \theta}{\sin \theta}$ to simplify the identity. \begin{aligned}\sin^2\theta + \cos^2 \theta &=1\\\dfrac{\sin^2\theta}{\color{DarkOrange}\sin^2\theta} +\dfrac{\cos^2\theta}{\color{DarkOrange}\sin^2\theta} &=\dfrac{1}{\color{DarkOrange}\sin^2\theta}\\1+ \left(\dfrac{\cos\theta}{\sin\theta}\right)^2&= \left(\dfrac{1}{\sin\theta}\right)^2\\\color{DarkOrange}\boldsymbol{1 + \cot^2 \theta} &\color{DarkOrange}\boldsymbol{=\csc^2\theta}\end{aligned} Now that weâ€™ve shown you

**how the identities were derived**, it’s time for us to learn how to apply them in solving problems and proving other trigonometric identities.

## How To Use the Pythagorean Identity?

The Pythagorean identity can be used to**solve equations, evaluate expressions, and prove identities**by rewriting trigonometric expressions using the three identities. This is how to use the Pythagorean identities. \begin{aligned}\sin^2\theta + \cos^2 \theta = &1\\\tan^2 \theta +1= \sec^2 &\theta\\1+ \cot^2 \theta = \csc^2 &\theta\end{aligned}

### Evaluating Expressions Using Pythagorean Identities

When using the Pythagorean identity to evaluate expressions,*we can:*

- Identify which of the three identities will be the most helpful.
- Use the given values into the chosen Pythagorean identity, then solve for the unknown value.

**weâ€™re working with sine and cosine**, letâ€™s use the first Pythagorean identity. \begin{aligned}\sin^2\theta + \cos^2\theta = 1\end{aligned} Substitute $\sin \theta = \dfrac{12}{13}$ into the Pythagorean identity. Simplify the equation to find the exact value of $\cos \theta$. \begin{aligned}\sin^2\theta+ \cos^2 \theta &= 1\\\left({\color{DarkOrange}\dfrac{12}{13}}\right)^2 +\cos^2 \theta &= 1\\\dfrac{144}{169}+\cos^2 \theta &= 1\\\cos^2\theta&= \dfrac{25}{169}\\\cos \theta &= \pm \dfrac{5}{13}\end{aligned} The angle, $\theta$, lies on the first quadrant, so $\cos \theta$ is positive. Hence, $\cos \theta = \dfrac{5}{13}$. Apply a similar process when

**asked to find the exact values of other trigonometric expressions**. For now, letâ€™s take a look at how we can use the Pythagorean identities when solving trigonometric equations.

### Solving Equations Using Pythagorean Identities

When given a trigonometric equation, see whether we can rewrite any of the terms using the Pythagorean identities. These terms are normally those that**contain the terms from the three Pythagorean identities**.

- When either $\sin \theta$ and $\cos \theta$ are part of the equation and at least one of them is squared
- Similarly, when $\sec \theta$ and $\tan \theta$ are present as well as $\csc \theta$ and $\cot \theta$
- To simplify the equation, rewrite one of the trigonometric expression in terms of the other

**the equation contains**$\sec^2 \theta$ and $\tan \theta$,

**so rewrite**$\sec^2 \theta$

**using the Pythagorean identity**$\tan^2 \theta +1 = \sec^2 \theta$. \begin{aligned}1 – \sec^2\theta &= \tan \theta\\1 – {\color{DarkOrange}(\tan^2 \theta +1 )} &= \tan \theta\\1 -\tan^2\theta -1&= \tan\theta\\\tan^2\theta +\tan\theta&=0\end{aligned} We now have a quadratic equation with only $\tan \theta$ and $\tan^2{\theta}$ to worry about.

**Apply appropriate algebraic techniques**to find $\tan \theta$ and $\theta$.

\begin{aligned}\tan \theta(\tan\theta +1)&=0\\\tan \theta = 0,\tan \theta &+ 1=0 \end{aligned} | |

\begin{aligned}\tan \theta&= 0\\\theta &=\pi \end{aligned} | \begin{aligned}\tan \theta + 1&= 0\\\tan \theta &= -1\\\theta &= \dfrac{3\pi}{4} \end{aligned} |

**now easier to simplify and solve**.

### Proving Trigonometric Identities Using Pythagorean Identities

The reason why Pythagorean identities are important is that**they lead to a wide range of other trigonometric identities and properties**. Knowing how to simplify, derive and even prove identities using Pythagorean identities is essential, especially when advancing to other trigonometry and math topics. \begin{aligned}\cos^2\theta &= (1 – \sin \theta)(1 +\sin\theta)\end{aligned}

**Simplify the right-hand side**of the equation by applying algebraic techniques learned in the past. \begin{aligned}\cos^2\theta&= (1 – \sin \theta)(1 +\sin\theta)\\&= 1^2 – (\sin \theta)^2\\&= 1 – \sin^2 \theta\end{aligned}

*Does the right-hand side of the equation now look familiar?*If we rewrite the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$, we can show that $1 – \sin^2\theta = \cos^2\theta$. \begin{aligned}\cos^2\theta &= 1 – \sin^2\\&= \cos^2\theta \end{aligned} This shows how important Pythagorean identities are

**when simplifying and proving trigonometric expressions and identities**. When youâ€™re ready, head on over to the next section to solve more problems!

### Example 1

Suppose that $\sec \theta = -\dfrac{29}{20}$, what is the exact value of $\tan \theta$ if it is also negative?__Solution__

We want to find $\tan \theta$â€™s value given the value of $\sec\theta$. Use the Pythagorean identity $\tan^2\theta + 1= \sec^2\theta$ and the fact that $\sec \theta = -\dfrac{29}{20}$.
\begin{aligned}\tan^2\theta + 1= \sec^2\theta\\ \tan^2\theta + 1&= {\color{DarkOrange}\left(-\dfrac{29}{20}\right)}^2\\\tan^2\theta +1 &= \dfrac{841}{400}\\\tan^2\theta &=\dfrac{441}{400}\\\tan \theta &= \pm \dfrac{21}{20}\end{aligned}
Since we know that $\tan \theta$ is negative, we let go of the positive solution. This means that we have $\tan \theta=-\dfrac{21}{20}$.
### Example 2

If $\csc \thetaÂ – \cot \theta = -4$, what is the value of $\csc \theta + \cot \theta$?__Solution__

Since weâ€™re working with cosecant and cotangent functions, itâ€™s best to focus on the third Pythagorean identity, $1+ \cot^2\theta = \csc^2\theta$. Rewrite this identity so that we can isolate $1$ on the right-hand side of the equation.
\begin{aligned}1+ \cot^2\theta &= \csc^2\theta\\\csc^2\theta – \cot^2\theta &= 1\\(\csc \theta – \cot \theta)(\csc \theta + \cot \theta) &= 1\end{aligned}
Notice anything familiar on the left-hand side of the resulting equation? We now have the expression thatâ€™s given in the problem and we have the expression we need to find as well.
\begin{aligned}(\csc \theta – \cot \theta)(\csc \theta + \cot \theta) &= 1\\({\color{DarkOrange}-4})(\csc \theta + \cot \theta)&= 1\\\csc \theta + \cot \theta &= – \dfrac{1}{4}\end{aligned}
This means that $\csc \theta + \cot \theta$ is equal to $-\dfrac{1}{4}$.
### Example 3

Show that the trigonometric identity $\tan\theta -\tan\theta\sec^2\theta = \tan^3 \theta$ is true.__Solution__

First, letâ€™s factor our $\tan \theta$ from each of the terms on the left-hand side of the equation.
\begin{aligned}\tan\theta -\tan\theta\sec^2\theta = \tan^3 \theta\\\tan\theta(1- \sec^2\theta )= \tan^3 \theta\end{aligned}
Weâ€™re working with $\sec^2 \theta$ and $\tan \theta$, so the best Pythagorean identity to use is $\tan^2 \theta +1 = \sec^2\theta$. Rewrite $1 – \sec^2\theta$ in terms of $\tan \theta$ to simplify the left-hand side of the equation.
\begin{aligned}\tan\theta({\color{DarkOrange}\tan^2\theta})&= \tan^3 \theta\\\tan^3\theta &= \tan^3\theta \,\checkmark\end{aligned}
This confirms that $\tan\theta -\tan\theta\sec^2\theta = \tan^3 \theta$ is true.
*Practice Questions*

1. If $\sin \theta\cos\theta = \dfrac{1}{4}$, what is the value of $\sin \theta – \cos \theta$?
A. $\dfrac{\sqrt{2}}{2}$
B. $\dfrac{\sqrt{3}}{2}$
C. $\dfrac{1}{2}$
D. $\dfrac{3}{2}$
2. Suppose that $\cos \theta = \dfrac{3}{7}$ and $\cot^2 \theta = \dfrac{a}{b}$, what is the value of $a + b$?
A. $31$
B. $40$
C. $49$
D. $98$
3. Which of the following is equivalent to $\dfrac{\cos \theta}{1 + \sin \theta}$?
A. $-\dfrac{1}{\sin \theta \cot \theta}$
B. $\dfrac{1 – \sin \theta}{\sin \theta \cot \theta}$
C. $\dfrac{1 + \sin \theta}{\sin \theta \cot \theta}$
D. $\dfrac{1}{\sin \theta \cot \theta}$
*Answer Key*

1. A
2. C
3. B
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