# Nakafa Learning Content
> For AI agents: use [llms.txt](https://nakafa.com/llms.txt) for the site index. Markdown versions are available by appending `.md` to content URLs or sending `Accept: text/markdown`.
URL: https://nakafa.com/en/subjects/mathematics/function-modeling/trigonometric-identity
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/material/lesson/mathematics/function-modeling/trigonometric-identity/en.mdx
Learn trigonometric identities through proofs of Pythagorean, reciprocal, and quotient forms, then use them to simplify expressions.
---
## Understanding Trigonometric Identities
Have you ever noticed that some mathematical equations are always true for any value of their variables? For example, $$(a + b)^2 = a^2 + 2ab + b^2$$ is always true for any values of $$a$$ and $$b$$. Equations like this are called identities.
Visible text: Have you ever noticed that some mathematical equations are always true for any value of their variables? For example, is always true for any values of and . Equations like this are called identities.
In trigonometry, we also have equations that are always true for any angle value. These are called **trigonometric identities**. These identities are very useful for simplifying trigonometric expressions and solving equations.
## Basic Trigonometric Identities
### Pythagorean Identity
Let's start with the most fundamental identity. Consider a unit circle with point $$P(x, y)$$ that forms angle $$\theta$$ with the positive $$x$$-axis.
Visible text: Let's start with the most fundamental identity. Consider a unit circle with point that forms angle with the positive -axis.
Component: UnitCircle
Props:
- title: Unit Circle and Pythagorean Identity
- description: Notice how the coordinates of point $$P$$ change as the
angle changes. These coordinates are the values of{" "}
$$\cos\theta$$ and $$\sin\theta$$.
Visible text: Notice how the coordinates of point change as the
angle changes. These coordinates are the values of{" "}
and .
- angle: 45
On the unit circle:
- Radius $$= 1$$
- $$x$$-coordinate is $$\cos \theta$$
- $$y$$-coordinate is $$\sin \theta$$
Visible text: - Radius
- -coordinate is
- -coordinate is
Using the Pythagorean theorem for point $$P$$:
Visible text: Using the Pythagorean theorem for point :
```math
x^2 + y^2 = 1^2
```
Substituting the values of $$x$$ and $$y$$:
Visible text: Substituting the values of and :
```math
(\cos \theta)^2 + (\sin \theta)^2 = 1
```
Or can be written as:
```math
\sin^2 \theta + \cos^2 \theta = 1
```
This is the **Pythagorean identity**, the most fundamental identity in trigonometry.
**Other Forms of Pythagorean Identity:**
From the basic identity above, we can derive two other forms:
**Second form:** Divide both sides by $$\cos^2 \theta$$ (for $$\cos \theta \neq 0$$)
Visible text: **Second form:** Divide both sides by (for )
Component: MathContainer
Children:
```math
\frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta}
```
```math
\tan^2 \theta + 1 = \sec^2 \theta
```
**Third form:** Divide both sides by $$\sin^2 \theta$$ (for $$\sin \theta \neq 0$$)
Visible text: **Third form:** Divide both sides by (for )
Component: MathContainer
Children:
```math
\frac{\sin^2 \theta}{\sin^2 \theta} + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta}
```
```math
1 + \cot^2 \theta = \csc^2 \theta
```
### Reciprocal Identities
Each trigonometric function has its reciprocal. This relationship forms reciprocal identities:
Component: MathContainer
Children:
```math
\sin \theta = \frac{1}{\csc \theta}
```
```math
\cos \theta = \frac{1}{\sec \theta}
```
```math
\tan \theta = \frac{1}{\cot \theta}
```
Or in the opposite form:
Component: MathContainer
Children:
```math
\csc \theta = \frac{1}{\sin \theta}
```
```math
\sec \theta = \frac{1}{\cos \theta}
```
```math
\cot \theta = \frac{1}{\tan \theta}
```
### Quotient Identities
Quotient identities relate tangent and cotangent to sine and cosine:
Component: ContentStack
Children:
```math
\tan \theta = \frac{\sin \theta}{\cos \theta}
```
```math
\cot \theta = \frac{\cos \theta}{\sin \theta}
```
Component: Triangle
Props:
- title: Trigonometric Function Visualization
- description: Observe how the ratios of triangle sides give $$\sin$$, $$\cos$$, and $$\tan$$ values. Also notice how these values change as the angle changes.
Visible text: Observe how the ratios of triangle sides give , , and values. Also notice how these values change as the angle changes.
- angle: 30
- labels: {
opposite: "Opposite Side",
adjacent: "Adjacent Side",
hypotenuse: "Hypotenuse",
}
Both identities can be proven directly from the definition of trigonometric functions on the unit circle.
### Even and Odd Function Identities
When angles are negative, trigonometric functions have special properties:
**Even function (symmetry about $$y$$-axis):**
Visible text: **Even function (symmetry about -axis):**
```math
\cos(-\theta) = \cos \theta
```
**Odd functions (symmetry about origin):**
Component: ContentStack
Children:
```math
\sin(-\theta) = -\sin \theta
```
```math
\tan(-\theta) = -\tan \theta
```
Component: UnitCircle
Props:
- title: Exploring Even and Odd Properties
- description: Try moving the slider to negative and positive values. Notice how $$\cos$$, $$\sin$$, and $$\tan$$ values change for opposite angles.
Visible text: Try moving the slider to negative and positive values. Notice how , , and values change for opposite angles.
- angle: 60
## Using Identities in Proofs
Let's see how trigonometric identities are used to prove other equations.
### Simplifying Expressions
Simplify $$\frac{\sin^2 \theta}{\cos \theta} + \cos \theta$$
Visible text: Simplify
**Solution:**
Component: MathContainer
Children:
```math
\frac{\sin^2 \theta}{\cos \theta} + \cos \theta = \frac{\sin^2 \theta}{\cos \theta} + \frac{\cos^2 \theta}{\cos \theta}
```
```math
= \frac{\sin^2 \theta + \cos^2 \theta}{\cos \theta}
```
```math
= \frac{1}{\cos \theta} \quad \text{(using Pythagorean identity)}
```
```math
= \sec \theta
```
### Proving Identities
Prove that $$\frac{1 + \tan^2 \theta}{\sec \theta} = \sec \theta$$
Visible text: Prove that
**Solution:**
We start from the left side:
Component: MathContainer
Children:
```math
\frac{1 + \tan^2 \theta}{\sec \theta} = \frac{\sec^2 \theta}{\sec \theta} \quad \text{(using } 1 + \tan^2 \theta = \sec^2 \theta \text{)}
```
```math
= \sec \theta
```
It is proven that the left side equals the right side.
## Determining Trigonometric Function Values
Trigonometric identities are very useful for determining the values of all trigonometric functions when one of them is known.
### Identity Applications
If $$\sin \theta = \frac{3}{5}$$ and $$90^\circ < \theta < 180^\circ$$ (quadrant II), determine the values of other trigonometric functions.
Visible text: If and (quadrant II), determine the values of other trigonometric functions.
**Solution:**
Use the Pythagorean identity to find $$\cos \theta$$:
Visible text: Use the Pythagorean identity to find :
Component: MathContainer
Children:
```math
\sin^2 \theta + \cos^2 \theta = 1
```
```math
\left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1
```
```math
\frac{9}{25} + \cos^2 \theta = 1
```
```math
\cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}
```
```math
\cos \theta = \pm \frac{4}{5}
```
Since $$\theta$$ is in quadrant II, then $$\cos \theta < 0$$.
Therefore, $$\cos \theta = -\frac{4}{5}$$
Visible text: Since is in quadrant II, then .
Therefore,
Next, calculate the other trigonometric functions:
Component: MathContainer
Children:
```math
\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/5}{-4/5} = -\frac{3}{4}
```
```math
\csc \theta = \frac{1}{\sin \theta} = \frac{5}{3}
```
```math
\sec \theta = \frac{1}{\cos \theta} = -\frac{5}{4}
```
```math
\cot \theta = \frac{1}{\tan \theta} = -\frac{4}{3}
```
## Exercises
1. Simplify the expression $$\frac{\tan \theta \cdot \cos \theta}{\sin \theta}$$
2. Prove the identity $$\frac{\sin \theta}{1 + \cos \theta} = \frac{1 - \cos \theta}{\sin \theta}$$
3. If $$\cos \theta = \frac{5}{13}$$ and $$270^\circ < \theta < 360^\circ$$, determine the values of all trigonometric functions.
4. Simplify $$\sin^4 \theta - \cos^4 \theta$$
5. If $$\tan \theta = \frac{4}{3}$$ and $$\sin \theta < 0$$, determine the values of $$\sin \theta$$ and $$\cos \theta$$.
Visible text: 1. Simplify the expression
2. Prove the identity
3. If and , determine the values of all trigonometric functions.
4. Simplify
5. If and , determine the values of and .
### Answer Key
1. Let's simplify step by step:
```math
\frac{\tan \theta \cdot \cos \theta}{\sin \theta} = \frac{\frac{\sin \theta}{\cos \theta} \cdot \cos \theta}{\sin \theta}
```
```math
= \frac{\sin \theta}{\sin \theta} = 1
```
2. To prove the identity, we will transform the left side:
```math
\frac{\sin \theta}{1 + \cos \theta} = \frac{\sin \theta}{1 + \cos \theta} \cdot \frac{1 - \cos \theta}{1 - \cos \theta}
```
```math
= \frac{\sin \theta (1 - \cos \theta)}{(1 + \cos \theta)(1 - \cos \theta)}
```
```math
= \frac{\sin \theta (1 - \cos \theta)}{1 - \cos^2 \theta}
```
```math
= \frac{\sin \theta (1 - \cos \theta)}{\sin^2 \theta}
```
```math
= \frac{1 - \cos \theta}{\sin \theta}
```
3. Given $$\cos \theta = \frac{5}{13}$$ in quadrant IV.
Finding $$\sin \theta$$:
```math
\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{25}{169} = \frac{144}{169}
```
```math
\sin \theta = -\frac{12}{13} \text{ (negative in quadrant IV)}
```
Other trigonometric functions:
```math
\tan \theta = \frac{-12/13}{5/13} = -\frac{12}{5}
```
```math
\csc \theta = -\frac{13}{12}
```
```math
\sec \theta = \frac{13}{5}
```
```math
\cot \theta = -\frac{5}{12}
```
4. Use difference of squares factoring:
```math
\sin^4 \theta - \cos^4 \theta = (\sin^2 \theta)^2 - (\cos^2 \theta)^2
```
```math
= (\sin^2 \theta + \cos^2 \theta)(\sin^2 \theta - \cos^2 \theta)
```
```math
= 1 \cdot (\sin^2 \theta - \cos^2 \theta)
```
```math
= \sin^2 \theta - \cos^2 \theta
```
5. Given $$\tan \theta = \frac{4}{3}$$ and $$\sin \theta < 0$$.
Since $$\tan \theta > 0$$ and $$\sin \theta < 0$$, then $$\cos \theta < 0$$ (quadrant III).
Use the identity $$1 + \tan^2 \theta = \sec^2 \theta$$:
```math
1 + \frac{16}{9} = \sec^2 \theta
```
```math
\sec^2 \theta = \frac{25}{9}
```
```math
\sec \theta = -\frac{5}{3} \text{ (negative in quadrant III)}
```
```math
\cos \theta = -\frac{3}{5}
```
For $$\sin \theta$$:
```math
\tan \theta = \frac{\sin \theta}{\cos \theta}
```
```math
\frac{4}{3} = \frac{\sin \theta}{-3/5}
```
```math
\sin \theta = \frac{4}{3} \cdot \left(-\frac{3}{5}\right) = -\frac{4}{5}
```
Visible text: 1. Let's simplify step by step:
2. To prove the identity, we will transform the left side:
3. Given in quadrant IV.
Finding :
Other trigonometric functions:
4. Use difference of squares factoring:
5. Given and .
Since and , then (quadrant III).
Use the identity :
For :