/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 Prove the following identities. ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 68

Prove the following identities. $$\tan \theta=\frac{\sin \theta}{\cos \theta}$$

Short Answer

Expert verified
Identity: $$ \frac{\sin \theta}{\cos \theta} = \tan \theta $$ Proof: 1. Define trigonometric functions in right-angled triangles: $$ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}},\; \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}},\; \tan \theta = \frac{\text{opposite}}{\text{adjacent}} $$ 2. Label the side lengths as O, A, and H: $$ \sin \theta = \frac{O}{H},\; \cos \theta = \frac{A}{H},\; \tan \theta = \frac{O}{A} $$ 3. Divide the sine function by the cosine function and simplify: $$ \frac{\sin \theta}{\cos \theta} = \frac{\frac{O}{H}}{\frac{A}{H}} = \frac{O \cdot H}{A \cdot H} = \frac{O}{A} $$ 4. Compare the result with the tangent function definition: $$ \frac{\sin \theta}{\cos \theta} = \frac{O}{A} = \tan \theta $$ Therefore, we have proved the given trigonometric identity: $$ \frac{\sin \theta}{\cos \theta} = \tan \theta $$

Step by step solution

01

Define trigonometric functions in right-angled triangles

In a right-angled triangle, we can define sine, cosine, and tangent functions as follows: $$ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}},\; \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}},\; \tan \theta = \frac{\text{opposite}}{\text{adjacent}} $$ Here, 胃 is an angle in the triangle other than the 90-degree angle, the opposite side is the side opposite to angle 胃, the adjacent side is the side adjacent to the angle, and the hypotenuse is the longest side.
02

Define the opposite, adjacent, and hypotenuse sides

Let O be the length of the opposite side, A be the length of the adjacent side, and H be the length of the hypotenuse. We can rewrite the trigonometric functions using these side lengths: $$ \sin \theta = \frac{O}{H},\; \cos \theta = \frac{A}{H},\; \tan \theta = \frac{O}{A} $$
03

Find the relation between sine and cosine functions

Now, let's divide the sine function by the cosine function: $$ \frac{\sin \theta}{\cos \theta} = \frac{\frac{O}{H}}{\frac{A}{H}} $$ Since both fractions have the same denominator (H), we can simplify the expression by multiplying the numerator of the first fraction by the denominator of the second fraction and vice versa: $$ \frac{\sin \theta}{\cos \theta} = \frac{O \cdot H}{A \cdot H} $$ Since H (the hypotenuse) is in both the numerator and the denominator, we can cancel them out: $$ \frac{\sin \theta}{\cos \theta} = \frac{O}{A} $$
04

Relate the result to the tangent function

Now, we know that the tangent function is defined as the ratio of the opposite and adjacent sides: $$ \tan \theta = \frac{O}{A} $$ Comparing this with the result we obtained in Step 3, we can see that: $$ \frac{\sin \theta}{\cos \theta} = \tan \theta $$ Thus, we have proved the given trigonometric identity.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Right-Angled Triangle
A right-angled triangle is a triangle in which one of the angles is exactly 90 degrees. The fundamental properties of such triangles allow us to explore and define the trigonometric functions. These functions relate the angles of the triangle to its sides.
The side opposite the right angle is called the hypotenuse and it is always the longest side. The side opposite to a given angle (other than the right angle) is known as the opposite side, and the side adjacent to the angle is called the adjacent side.
  • Hypotenuse: long side opposite the right angle.
  • Opposite: side opposite the angle of interest.
  • Adjacent: side next to the angle of interest.
These definitions are crucial as they set the basis for understanding the trigonometric functions that we use to solve various geometric problems.
Trigonometric Functions
Trigonometric functions are mathematical relations that are derived from the ratios of the sides of a right-angled triangle. They are vital in bridging the gap between geometry and algebra by translating geometric properties into algebraic equations. The primary trigonometric functions are sine (\( \sin \theta \)), cosine (\( \cos \theta \)), and tangent (\( \tan \theta \)).
Each function associates an angle in a right-angled triangle with a ratio of two sides:
  • Sine (\( \sin \theta \)): ratio of the length of the opposite side to the hypotenuse.
  • Cosine (\( \cos \theta \)): ratio of the length of the adjacent side to the hypotenuse.
  • Tangent (\( \tan \theta \)): ratio of the length of the opposite side to the adjacent side.
These functions form the backbone of trigonometry, enabling us to describe and calculate angles and distances in triangles and more complex shapes.
Tangent Function
The tangent function in trigonometry is expressed as the ratio of the opposite side to the adjacent side in a right-angled triangle. It is denoted as \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \).
This function is particularly useful because it relates the angle directly to the vertical height relative to the horizontal base of the triangle. The tangent function is crucial for solving triangles, especially when dealing with slopes or angles of elevation and depression. It provides a direct connection between the angle and the gradient of the slope, which is often required in real-world applications.
Sine Function
The sine function, indicated by \( \sin \theta \), represents the ratio of the opposite side to the hypotenuse in a right-angled triangle. This can be formulated as \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \).
The sine function is highly significant in trigonometry because it links an angle to the peak height it rises to when projected along the hypotenuse. Its application is extensive in fields like engineering, physics, and even music theory, where wave patterns are examined.
Cosine Function
Cosine (\( \cos \theta \)) is another fundamental trigonometric function. It provides a vital relationship by linking an angle in a right-angled triangle to the ratio of the adjacent side over the hypotenuse: \( \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \).
This function makes it possible to find one side of the triangle if the angle and the hypotenuse are known. Its principal use extends from classical mechanics to modern computer graphics, where the cosine function assists in transforming angles to linear dimensions, most notably in calculating projections and optimising models.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Transformations of \(f(x)=x^{2}\) Use shifts and scalings to transform the graph of \(f(x)=x^{2}\) into the graph of \(g .\) Use a graphing utility to check your work. a. \(g(x)=f(x-3)\) b. \(g(x)=f(2 x-4)\) c. \(g(x)=-3 f(x-2)+4\) d. \(g(x)=6 f\left(\frac{x-2}{3}\right)+1\)

Shifting and scaling Use shifts and scalings to graph the given functions. Then check your work with a graphing utility. Be sure to identify an original function on which the shifts and scalings are performed. \(f(x)=x^{2}-2 x+3\) (Hint: Complete the square first.)

Inverse of composite functions a. Let \(g(x)=2 x+3\) and \(h(x)=x^{3} .\) Consider the composite function \(f(x)=g(h(x)) .\) Find \(f^{-1}\) directly and then express it in terms of \(g^{-1}\) and \(h^{-1}\). b. Let \(g(x)=x^{2}+1\) and \(h(x)=\sqrt{x}\). Consider the composite function \(f(x)=g(h(x)) .\) Find \(f^{-1}\) directly and then express it in terms of \(g^{-1}\) and \(h^{-1}\) c. Explain why if \(g\) and \(h\) are one-to-one, the inverse of \(f(x)=g(h(x))\) exists.

Evaluate the following expressions. $$\tan ^{-1}(\tan (\pi / 4))$$

Shifting and scaling Use shifts and scalings to graph the given functions. Then check your work with a graphing utility. Be sure to identify an original function on which the shifts and scalings are performed. $$g(x)=-3 x^{2}$$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.