/*! 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 104 Write each trigonometric express... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 104

Write each trigonometric expression as an algebraic expression (that is, without any trigonometric functions). Assume that x and y are positive and in the domain of the given inverse trigonometric function. $$ \sin \left(\tan ^{-1} x-\sin ^{-1} y\right) $$

Short Answer

Expert verified
The algebraic expression of the given trigonometric equation \( \sin (\tan^{-1}x - \sin^{-1}y) \) is \( \frac{x\sqrt{1-y^2} - y}{\sqrt{1+x^2}} \)

Step by step solution

01

Identify the values

Recognize that the function \( \tan^{-1}x \) is an angle whose tangent is \( x \). Similarly, \( \sin^{-1}y \) is an angle whose sine is \( y \). We will denote these angles as \( A \) and \( B \) respectively.
02

Apply Pythagorean theorem and find cos

For \( A = \tan^{-1}x \), we use the Pythagorean theorem on a right triangle with opposite side \( x \) and adjacent side 1. This gives us \( \cos(A) = \frac{{\text{adjacent side}}}{{\text{hypotenuse}}} = \frac{1}{\sqrt{1+x^2}} \). Similarly, for \( B = \sin^{-1}y \), we find \( \cos(B) = \frac{\sqrt{1-y^2}}{1} = \sqrt{1-y^2} \).
03

Use trigonometric identities for sine of difference of two angles

The expression \( \sin (\tan^{-1}x - \sin^{-1}y) \) is the sine of the difference of two angles. Use the identity \( \sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B) \). This gives \(\sin(\tan^{-1}x)\cos(\sin^{-1}y) - \cos(\tan^{-1}x)\sin(\sin^{-1}y)\). Considering \( \sin(\tan^{-1}x) = x/\sqrt{1+x^2} \) and \( \cos(\sin^{-1}y) = \sqrt{1-y^2} \), and applying these to given equation, which simplifies to \( x\sqrt{1-y^2} - y/\sqrt{1+x^2} \).
04

Simplify the expression

The simplified algebraic expression of given trigonometric equation is \( \frac{x\sqrt{1-y^2} - y}{\sqrt{1+x^2}} \).

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.

Inverse Trigonometric Functions
Inverse trigonometric functions allow us to find angles based on given trigonometric values. They reverse the trigonometric functions (sine, cosine, tangent). For example, \( \tan^{-1}(x) \) finds the angle whose tangent is \( x \). Similarly, \( \sin^{-1}(y) \) finds the angle whose sine is \( y \). These functions are essential when converting trigonometric expressions into algebraic ones.
When using inverse trigonometric functions:
  • \( \tan^{-1}(x) \) tells us that the opposite side is \( x \), and the adjacent side is 1 in a right triangle. The angle we get from this expression is called \( A \).
  • \( \sin^{-1}(y) \) indicates that the opposite side is \( y \) and the hypotenuse is 1, giving us angle \( B \).
These relationships help in translating problems into algebraic forms by understanding the angles and sides involved.
Pythagorean Theorem
The Pythagorean theorem is a vital tool in trigonometry, especially when dealing with inverse functions. The theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. This can be written as \( a^2 + b^2 = c^2 \). When working with inverse trigonometric functions:
  • For \( A = \tan^{-1}(x) \), we form a triangle where the opposite side is \( x \) and the adjacent side is 1. Using the Pythagorean theorem, the hypotenuse becomes \( \sqrt{1 + x^2} \).
  • For \( B = \sin^{-1}(y) \), the opposite side is \( y \) and the hypotenuse is 1. Therefore, the adjacent side is found as \( \sqrt{1 - y^2} \).
This process helps express trigonometric functions in algebraic terms, aiding in solving more complex expressions.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are universally true. They help simplify complex trigonometric expressions. One popular identity is the sine of difference formula used in this exercise:
  • The identity \( \sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B) \) is used when dealing with expressions like \( \sin(\tan^{-1}x - \sin^{-1}y) \).
By substituting angles \( A \) and \( B \) known from inverse trigonometric functions:
  • Find \( \sin(A) \) and \( \cos(A) \) using the triangle from \( \tan^{-1}(x) \).
  • Find \( \sin(B) \) and \( \cos(B) \) using the triangle from \( \sin^{-1}(y) \).
Using these identities, we can break down the problem into simpler, algebraic expressions.
Sine of Difference Formula
The sine of difference formula is a specific trigonometric identity used to find the sine of the difference of two angles. It's particularly handy when inverse trigonometric functions are involved. The formula is given by:
  • \( \sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B) \)
To apply this formula:
  • Identify angles \( A \) and \( B \) using the inverse trigonometric expressions.
  • Calculate \( \sin(A) \), \( \cos(B) \), \( \cos(A) \), and \( \sin(B) \) using the sides of the triangles.
This formula helps in transforming the trigonometric expression into an easier-to-handle algebraic form, offering a new perspective to solve complex trigonometric equations. This process highlights the power of a well-chosen identity in simplifying expressions.

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

Our cycle of normal breathing takes place every 5 seconds. Velocity of air flow, \(y,\) measured in liters per second, after \(x\) seconds is modeled by $$ y=0.6 \sin \frac{2 \pi}{5} x $$Velocity of air flow is positive when we inhale and negative when we exhale. Within each breathing cycle, when are we exhaling at a rate of 0.3 liter per second? Round to the nearest tenth of a second.

When throwing can object, the distance achieved depends on its initial velocity, \(v_{0}\) cand the angle above the horizontal at which the object is thrown, \(\boldsymbol{\theta}\). The distance, \(d,\) in feet, that describes the range covered is given by $$ d=\frac{v_{0}^{2}}{16} \sin \theta \cos \theta $$$$ \text { where } \boldsymbol{\tau}_{0} \text { is measured in feet per second.} $$ You and your friend are throwing a baseball back and forth. If you throw the ball with an initial velocity of \(x_{0}=90\) feet per second, at what angle of elevation, \(\theta\). to the nearest degree, should you direct your throw so that it can be easily caught by your friend located 170 feet away?

The number of hours of daylight in Boston is given by $$ y=3 \sin \left[\frac{2 \pi}{365}(x-79)\right]+12 $$where \(x\) is the number of days after January \(l\). Within a year, when does Boston have 10.5 hours of daylight? Give your answer in days after January 1 and round to the nearest day.

Use a half-angle formula to find the exact value of each expression. $$ \tan 112.5^{\circ} $$

A hot-air balloon is rising vertically. From a point on level ground 120 feet from the point directly under the passenger compartment, the angle of elevation to the balloon changes from \(37.1^{\circ}\) to \(62.4^{\circ} .\) How far, to the nearest tenth of a foot, does the balloon rise during this period?
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

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.