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# 1.25: Pythagorean Identities

Difficulty Level: At Grade Created by: CK-12
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Practice Pythagorean Identities
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What if you were working on a problem using the unit circle and had the value of one trig function (such as sine), but wanted instead to find the value of another trig function (such as cosine)? Is this possible?

Try it with sinθ=12\begin{align*}\sin \theta = \frac{1}{2}\end{align*}

### Pythagorean Identities

One set of identities are called the Pythagorean Identities because they rely on the Pythagorean Theorem. In other Concepts we used the Pythagorean Theorem to find the sides of right triangles.

Consider the way that the trig functions are defined. Let’s look at the unit circle:

The legs of the right triangle are x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*}. The hypotenuse is 1. Therefore the following equation is true for all x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} on the unit circle:

x2+y2=1\begin{align*}x^2 + y^2 = 1\end{align*}

Now remember that on the unit circle, cosθ=x\begin{align*}\cos \theta = x\end{align*} and sinθ=y\begin{align*}\sin \theta = y\end{align*}. Therefore the following equation is an identity:

cos2θ+sin2θ=1\begin{align*}\cos^2 \theta + \sin^2 \theta = 1\end{align*}

Note: Writing the exponent 2 after the cos\begin{align*}cos\end{align*} and sin\begin{align*}sin\end{align*} is the standard way of writing exponents. Just keeping mind that cos2θ\begin{align*}\cos^2 \theta\end{align*} means (cosθ)2\begin{align*}(\cos \theta)^2\end{align*} and sin2θ\begin{align*}\sin ^2 \theta\end{align*} means (sinθ)2\begin{align*}(\sin \theta)^2\end{align*}.

We can use this identity to find the value of the sine function, given the value of the cosine, and vice versa. We can also use it to find other identities.

#### What is the value of sinθ\begin{align*}\sin \theta\end{align*}?

If cosθ=14\begin{align*}\cos \theta = \frac{1}{4}\end{align*} what is the value of sinθ\begin{align*}\sin \theta\end{align*}? Assume that θ\begin{align*}\theta\end{align*} is an angle in the first quadrant.

sinθ=154\begin{align*}\sin \theta = \frac{\sqrt{15}}{4}\end{align*}

cos2θ+sin2θ(14)2+sin2θ116+sin2θsin2θsin2θsinθsinθ=1=1=1=1116=1516=±1516=±154\begin{align*}\cos^2 \theta + \sin^2 \theta & = 1\\ \left ( \frac{1}{4} \right )^2 + \sin ^2 \theta & = 1\\ \frac{1}{16} + \sin^2 \theta & = 1\\ \sin^2 \theta & = 1 -\frac{1}{16}\\ \sin^2 \theta & = \frac{15}{16}\\ \sin \theta & = \pm \sqrt{\frac{15}{16}}\\ \sin \theta & = \pm \frac{\sqrt{15}}{4}\end{align*}

Remember that it was given that θ\begin{align*}\theta\end{align*} is an angle in the first quadrant. Therefore the sine value is positive, so sinθ=154\begin{align*}\sin \theta = \frac{\sqrt{15}}{4}\end{align*}.

#### Use the identity cos2θ+sin2θ=1\begin{align*}\cos^2\theta + \sin^2\theta = 1\end{align*} to show that cot2θ+1=csc2θ\begin{align*}\cot^2 \theta + 1 = \csc^2 \theta\end{align*}

cos2θ+sin2θcos2θ+sin2θsin2θcos2θsin2θ+sin2θsin2θcos2θsin2θ+1cosθsinθ×cosθsinθ+1cotθ×cotθ+1cot2θ+1=1=1sin2θ=1sin2θ=1sin2θ=1sinθ×1sinθ=cscθ×cscθ=csc2θDivide both sides bysin2θ.sin2θsin2θ=1Write the squared functions in termsof their factors.Use the quotient and reciprocalidentities.Write the functions as squaredfunctions.\begin{align*}\cos^2\theta + \sin^2\theta & = 1 && \text{Divide both sides by} \sin^2 \theta.\\ \frac{\cos^2\theta + \sin^2\theta}{\sin^2\theta} & = \frac{1}{\sin^2 \theta}\\ \frac{\cos^2\theta}{\sin^2\theta} + \frac{\sin^2\theta}{\sin^2\theta} & = \frac{1}{\sin^2\theta} && \frac{\sin^2\theta}{\sin^2\theta} = 1\\ \frac{\cos^2\theta}{\sin^2\theta} + 1 & = \frac{1}{\sin^2\theta}\\ \frac{\cos \theta}{\sin \theta} \times \frac{\cos \theta}{\sin \theta} + 1 & = \frac{1}{\sin \theta} \times \frac{1}{\sin \theta} && \text{Write the squared functions in terms}\\ &&& \text{of their factors.}\\ \cot \theta \times \cot \theta + 1 & = \csc \theta \times \csc \theta && \text{Use the quotient and reciprocal}\\ &&& \text{identities.}\\ \cot^2\theta + 1 & = \csc^2 \theta && \text{Write the functions as squared}\\ &&& \text{functions.}\end{align*}

#### What is the value of cosθ\begin{align*}\cos \theta\end{align*}?

If sinθ=12\begin{align*}\sin \theta = \frac{1}{2}\end{align*} what is the value of cosθ\begin{align*}\cos \theta\end{align*}? Assume that θ\begin{align*}\theta\end{align*} is an angle in the first quadrant.

cosθ=34\begin{align*}\cos \theta = \sqrt{\frac{3}{4}}\end{align*}

sin2θ+cos2θ(12)2+cos2θ14+cos2θcos2θcos2θcosθ=1=1=1=114=34=±34\begin{align*}\sin^2 \theta + \cos^2 \theta & = 1\\ \left ( \frac{1}{2} \right )^2 + \cos ^2 \theta & = 1\\ \frac{1}{4} + \cos^2 \theta & = 1\\ \cos^2 \theta & = 1 -\frac{1}{4}\\ \cos^2 \theta & = \frac{3}{4}\\ \cos \theta & = \pm \sqrt{\frac{3}{4}}\\ \end{align*}

Remember that it was given that \begin{align*}\theta\end{align*} is an angle in the first quadrant. Therefore the cosine value is positive, so \begin{align*}\cos \theta = \sqrt{\frac{3}{4}}\end{align*}.

### Examples

#### Example 1

Earlier, you were given a problem about trig function.

Since we now know that:

\begin{align*}\sin^2 \theta + \cos^2 \theta = 1\end{align*}

we can use this to help us compute the cosine of the angle from the problem at the beginning of this Concept. It was given at the beginning of this Concept that:

\begin{align*}\sin \theta = \frac{1}{2}\end{align*}

Therefore, \begin{align*}\sin^2 \theta = \frac{1}{4}\end{align*}

If we use this to solve for cosine:

\begin{align*} \sin^2 \theta + \cos^2 \theta = 1\\ \cos^2 \theta = 1 - \sin^2 \theta\\ \cos^2 \theta = 1 - \frac{1}{4}\\ \cos^2 \theta = \frac{3}{4}\\ \cos \theta = \frac{\sqrt{3}}{2}\\ \end{align*}

#### Example 2

If \begin{align*}\cos \theta = \frac{1}{2}\end{align*} what is the value of \begin{align*}\sin \theta\end{align*}? Assume that \begin{align*}\theta\end{align*} is an angle in the first quadrant.

The solution is \begin{align*}\sin \theta = \sqrt{\frac{3}{4}}\end{align*}. We can see this from the Pythagorean Identity:

\begin{align*}\cos^2 \theta + \sin^2 \theta & = 1\\ \left ( \frac{1}{2} \right )^2 + \sin ^2 \theta & = 1\\ \frac{1}{4} + \sin^2 \theta & = 1\\ \sin^2 \theta & = 1 -\frac{1}{4}\\ \sin^2 \theta & = \frac{3}{4}\\ \sin \theta & = \pm \sqrt{\frac{3}{4}}\\ \end{align*}

#### Example 3

If \begin{align*}\sin \theta = \frac{1}{8}\end{align*} what is the value of \begin{align*}\cos \theta\end{align*}? Assume that \begin{align*}\theta\end{align*} is an angle in the first quadrant.

The solution is \begin{align*}\cos \theta = \sqrt{\frac{63}{64}}\end{align*}. We can see this from the Pythagorean Identity:

\begin{align*}\cos^2 \theta + \sin^2 \theta & = 1\\ \left ( \frac{1}{8} \right )^2 + \cos ^2 \theta & = 1\\ \frac{1}{64} + \cos^2 \theta & = 1\\ \cos^2 \theta & = 1 -\frac{1}{64}\\ \cos^2 \theta & = \frac{63}{64}\\ \cos \theta & = \pm \sqrt{\frac{63}{64}}\\ \end{align*}

#### Example 4

If \begin{align*}\sin \theta = \frac{1}{3}\end{align*} what is the value of \begin{align*}\cos \theta\end{align*}? Assume that \begin{align*}\theta\end{align*} is an angle in the first quadrant.

The solution is \begin{align*}\cos \theta = \sqrt{\frac{8}{9}}\end{align*}. We can see this from the Pythagorean Identity:

\begin{align*}\sin^2 \theta + \cos^2 \theta & = 1\\ \left ( \frac{1}{3} \right )^2 + \cos ^2 \theta & = 1\\ \frac{1}{9} + \cos^2 \theta & = 1\\ \cos^2 \theta & = 1 -\frac{1}{9}\\ \cos^2 \theta & = \frac{8}{9}\\ \cos \theta & = \pm \sqrt{\frac{8}{9}}\\ \end{align*}

### Review

1. If you know \begin{align*}\sin \theta\end{align*}, what other trigonometric value can you determine using a Pythagorean Identity?
2. If you know \begin{align*}\sec \theta\end{align*}, what other trigonometric value can you determine using a Pythagorean Identity?
3. If you know \begin{align*}\cot \theta\end{align*}, what other trigonometric value can you determine using a Pythagorean Identity?
4. If you know \begin{align*}\tan \theta\end{align*}, what other trigonometric value can you determine using a Pythagorean Identity?

For questions 5-14, assume all angles are in the first quadrant.

1. If \begin{align*}\sin \theta = \frac{1}{2}\end{align*}, what is the value of \begin{align*}\cos \theta\end{align*}?
2. If \begin{align*}\cos \theta = \frac{\sqrt{2}}{2}\end{align*}, what is the value of \begin{align*}\sin \theta\end{align*}?
3. If \begin{align*}\tan \theta = 1\end{align*}, what is the value of \begin{align*}\sec \theta\end{align*}?
4. If \begin{align*}\csc \theta = \sqrt{2}\end{align*}, what is the value of \begin{align*}\cot \theta\end{align*}?
5. If \begin{align*}\sec \theta = 2\end{align*}, what is the value of \begin{align*}\tan \theta\end{align*}?
6. If \begin{align*}\cot \theta = \sqrt{3}\end{align*}, what is the value of \begin{align*}\csc \theta\end{align*}?
7. If \begin{align*}\cos \theta = \frac{1}{4}\end{align*}, what is the value of \begin{align*}\sin \theta\end{align*}?
8. If \begin{align*}\sec \theta = 3\end{align*}, what is the value of \begin{align*}\tan \theta\end{align*}?
9. If \begin{align*}\sin \theta = \frac{1}{5}\end{align*}, what is the value of \begin{align*}\cos \theta\end{align*}?
10. If \begin{align*}\tan \theta = \frac{\sqrt{3}}{3}\end{align*}, what is the value of \begin{align*}\sec \theta\end{align*}?
11. Use the identity \begin{align*}\sin^2\theta + \cos^2\theta = 1\end{align*} to show that \begin{align*}\tan^2 \theta + 1 = \sec^2 \theta\end{align*}

### Vocabulary Language: English

Pythagorean Identity

Pythagorean Identity

The Pythagorean identity is a relationship showing that the sine of an angle squared plus the cosine of an angle squared is equal to one.

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Date Created:
Sep 26, 2012