Understanding the Range of Trigonometric Functions

Oct 13, 2025 by thesweetdesigns.com 32 views

Introduction

Trigonometric equations are a fundamental concept in mathematics, and solving them requires a deep understanding of trigonometric functions and their properties. In this article, we will explore how to solve trigonometric equations involving cosine and tangent functions. We will use the given values of cosine and tangent to find the exact value of a trigonometric expression.

Given Values

We are given two trigonometric equations:

$\cos \theta = -\frac{\sqrt{5}}{6}$ , where $\pi \leq \theta \leq \frac{3\pi}{2}$
$\tan \beta = \frac{4}{3}$ , where $0 \leq \beta \leq \frac{\pi}{2}$

Understanding the Range of Trigonometric Functions

Before we proceed to solve the trigonometric equations, it's essential to understand the range of trigonometric functions. The range of a trigonometric function is the set of all possible output values it can produce.

The range of the cosine function is $[-1, 1]$ .
The range of the tangent function is all real numbers.

Solving the Cosine Equation

We are given the equation $\cos \theta = -\frac{\sqrt{5}}{6}$ , where $\pi \leq \theta \leq \frac{3\pi}{2}$ . To solve this equation, we need to find the value of $\theta$ that satisfies the equation.

Since the cosine function is negative in the second and third quadrants, we can use the reference angle to find the value of $\theta$ .

Finding the Reference Angle

The reference angle is the acute angle between the terminal side of the angle and the x-axis. In this case, the reference angle is the angle whose cosine is $\frac{\sqrt{5}}{6}$ .

We can find the reference angle using the inverse cosine function:

$\cos^{-1} \left(\frac{\sqrt{5}}{6}\right) = \alpha$

Finding the Value of $\theta$

Since the cosine function is negative in the second and third quadrants, we can use the reference angle to find the value of $\theta$ .

In the second quadrant, $\theta = \pi + \alpha$
In the third quadrant, $\theta = 2\pi - \alpha$

We can substitute the value of $\alpha$ into these equations to find the value of $\theta$ .

Solving the Tangent Equation

We are given the equation $\tan \beta = \frac{4}{3}$ , where $0 \leq \beta \leq \frac{\pi}{2}$ . To solve this equation, we need to find the value of $\beta$ that satisfies the equation.

Since the tangent function is positive in the first and third quadrants, we can use the inverse tangent function to find the value of $\beta$ .

$\tan^{-1} \left(\frac{4}{3}\right) = \beta$

Finding the Value of $\beta$

We can use the inverse tangent function to find the value of $\beta$ .

$\beta = \tan^{-1} \left(\frac{4}{3}\right)$

Finding the Exact Value of the Trigonometric Expression

We are given the trigonometric expression $\sin \theta \cos \beta + \cos \theta \sin \beta$ . To find the exact value of this expression, we can use the given values of cosine and tangent.

We can substitute the value of $\cos \theta$ and $\tan \beta$ into the expression:

$\sin \theta \cos \beta + \cos \theta \sin \beta = \sin \left(-\frac{\sqrt{5}}{6}\right) \cos \beta + \cos \left(-\frac{\sqrt{5}}{6}\right) \sin \beta$

We can use the trigonometric identity $\sin (-\theta) = -\sin \theta$ and $\cos (-\theta) = \cos \theta$ to simplify the expression:

$\sin \left(-\frac{\sqrt{5}}{6}\right) \cos \beta + \cos \left(-\frac{\sqrt{5}}{6}\right) \sin \beta = -\sin \left(\frac{\sqrt{5}}{6}\right) \cos \beta + \cos \left(\frac{\sqrt{5}}{6}\right) \sin \beta$

We can use the trigonometric identity $\sin \alpha \cos \beta + \cos \alpha \sin \beta = \sin (\alpha + \beta)$ to simplify the expression:

$-\sin \left(\frac{\sqrt{5}}{6}\right) \cos \beta + \cos \left(\frac{\sqrt{5}}{6}\right) \sin \beta = \sin \left(\frac{\sqrt{5}}{6} - \beta\right)$

We can substitute the value of $\beta$ into the expression:

$\sin \left(\frac{\sqrt{5}}{6} - \beta\right) = \sin \left(\frac{\sqrt{5}}{6} - \tan^{-1} \left(\frac{4}{3}\right)\right)$

We can use the trigonometric identity $\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$ to simplify the expression:

$\sin \left(\frac{\sqrt{5}}{6} - \tan^{-1} \left(\frac{4}{3}\right)\right) = \sin \left(\frac{\sqrt{5}}{6}\right) \cos \left(\tan^{-1} \left(\frac{4}{3}\right)\right) - \cos \left(\frac{\sqrt{5}}{6}\right) \sin \left(\tan^{-1} \left(\frac{4}{3}\right)\right)$

We can use the trigonometric identity $\cos \alpha = \frac{1}{\sec \alpha}$ and $\sin \alpha = \frac{\tan \alpha}{\sec \alpha}$ to simplify the expression:

$\sin \left(\frac{\sqrt{5}}{6}\right) \cos \left(\tan^{-1} \left(\frac{4}{3}\right)\right) - \cos \left(\frac{\sqrt{5}}{6}\right) \sin \left(\tan^{-1} \left(\frac{4}{3}\right)\right) = \sin \left(\frac{\sqrt{5}}{6}\right) \frac{1}{\sec \left(\tan^{-1} \left(\frac{4}{3}\right)\right)} - \cos \left(\frac{\sqrt{5}}{6}\right) \frac{\tan \left(\tan^{-1} \left(\frac{4}{3}\right)\right)}{\sec \left(\tan^{-1} \left(\frac{4}{3}\right)\right)}$

We can use the trigonometric identity $\sec \alpha = \frac{1}{\cos \alpha}$ and $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$ to simplify the expression:

$\sin \left(\frac{\sqrt{5}}{6}\right) \frac{1}{\sec \left(\tan^{-1} \left(\frac{4}{3}\right)\right)} - \cos \left(\frac{\sqrt{5}}{6}\right) \frac{\tan \left(\tan^{-1} \left(\frac{4}{3}\right)\right)}{\sec \left(\tan^{-1} \left(\frac{4}{3}\right)\right)} = \sin \left(\frac{\sqrt{5}}{6}\right) \frac{1}{\frac{1}{\cos \left(\tan^{-1} \left(\frac{4}{3}\right)\right)}} - \cos \left(\frac{\sqrt{5}}{6}\right) \frac{\frac{\sin \left(\tan^{-1} \left(\frac{4}{3}\right)\right)}{\cos \left(\tan^{-1} \left(\frac{4}{3}\right)\right)}}{\frac{1}{\cos \left(\tan^{-1} \left(\frac{4}{3}\right)\right)}}$

We can simplify the expression:

$$\sin \left(\frac{\sqrt{5}}{6}\right) \frac{1}{\frac{1}{\cos \left(\tan^{-1} \left(\frac{4}{3}\right)\right)}} - \cos \left(\frac{\sqrt{5}}{6}\right) \frac{\frac{\sin \left(\tan^{-1} \left(\frac{4}{3}\right)\right)}{\cos \left(\tan^{-1} \left(\frac{4}{3}\right)\right)}}{\frac{1}{\cos \left(\tan^{-1} \left(\frac{4}{3}\right)\right)}} = \sin \left(\frac{\sqrt{5}}{6}\right) \cos \left(\tan^{-1} \left(\frac{4}{3}\right)\right) - \cos \left(\frac{\sqrt{5}}{6}\right) \sin \left(\tan^{-1} \left(\frac{<br/> **Q&A: Solving Trigonometric Equations**$

Q: What is the range of the cosine function?

A: The range of the cosine function is $[-1, 1]$ .

Q: How do you find the reference angle for a given angle?

A: To find the reference angle, you can use the inverse cosine function: $\cos^{-1} (x)$ , where $x$ is the cosine of the angle.

Q: What is the difference between the sine and cosine functions?

A: The sine function is the ratio of the opposite side to the hypotenuse in a right triangle, while the cosine function is the ratio of the adjacent side to the hypotenuse.

Q: How do you solve a trigonometric equation involving the tangent function?

A: To solve a trigonometric equation involving the tangent function, you can use the inverse tangent function: $\tan^{-1} (x)$ , where $x$ is the tangent of the angle.

Q: What is the range of the tangent function?

A: The range of the tangent function is all real numbers.

Q: How do you simplify a trigonometric expression involving the sine and cosine functions?

A: To simplify a trigonometric expression involving the sine and cosine functions, you can use the trigonometric identities:

$\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$
$\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

Q: What is the difference between the sine and cosine functions in the context of trigonometric equations?

A: In the context of trigonometric equations, the sine and cosine functions are used to represent the ratios of the sides of a right triangle. The sine function is the ratio of the opposite side to the hypotenuse, while the cosine function is the ratio of the adjacent side to the hypotenuse.

Q: How do you use the given values of cosine and tangent to find the exact value of a trigonometric expression?

A: To use the given values of cosine and tangent to find the exact value of a trigonometric expression, you can substitute the values into the expression and simplify using the trigonometric identities.

Q: What is the importance of understanding the range of trigonometric functions?

A: Understanding the range of trigonometric functions is important because it helps you to determine the possible values of the function and to solve trigonometric equations.

Q: How do you use the inverse cosine and inverse tangent functions to solve trigonometric equations?

A: To use the inverse cosine and inverse tangent functions to solve trigonometric equations, you can substitute the values into the functions and simplify using the trigonometric identities.

Q: What is the difference between the sine and cosine functions in the context of trigonometric identities?

A: In the context of trigonometric identities, the sine and cosine functions are used to represent the ratios of the sides of a right triangle. The sine function is the ratio of the opposite side to the hypotenuse, while the cosine function is the ratio of the adjacent side to the hypotenuse.

Q: How do you simplify a trigonometric expression involving the sine and cosine functions using trigonometric identities?

A: To simplify a trigonometric expression involving the sine and cosine functions using trigonometric identities, you can use the following identities:

$\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$
$\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

Q: What is the importance of understanding the trigonometric identities?

A: Understanding the trigonometric identities is important because it helps you to simplify trigonometric expressions and to solve trigonometric equations.

Q: How do you use the given values of cosine and tangent to find the exact value of a trigonometric expression using trigonometric identities?

A: To use the given values of cosine and tangent to find the exact value of a trigonometric expression using trigonometric identities, you can substitute the values into the expression and simplify using the trigonometric identities.

Q: What is the difference between the sine and cosine functions in the context of trigonometric equations and identities?

A: In the context of trigonometric equations and identities, the sine and cosine functions are used to represent the ratios of the sides of a right triangle. The sine function is the ratio of the opposite side to the hypotenuse, while the cosine function is the ratio of the adjacent side to the hypotenuse.