/*! 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 26 \(\csc (x-y)+\sec (x+y)=x\)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\csc (x-y)+\sec (x+y)=x\)

Short Answer

Expert verified
There is no general solution for this complex equation. Specific values, such as x=0 or x=1 might work under specific conditions.

Step by step solution

01

Understand the given equation

The given trigonometric equation is \( \csc (x-y)+\sec (x+y)=x \). We need to find the solutions for this equation.
02

Break down the equation using trigonometric identities

Rewrite the trigonometric functions \( \csc \) and \( \sec \) in terms of sine and cosine: \ \csc (x-y) = \frac{1}{\sin (x-y)} \ and \ \sec (x+y) = \frac{1}{\cos (x+y)} \.
03

Substitute the identities into the equation

Substitute the identities into the equation: \[ \frac{1}{\sin (x-y)} + \frac{1}{\cos (x+y)} = x \].
04

Find a common denominator and simplify

The common denominator for \( \sin (x-y) \) and \( \cos (x+y) \) is \( \sin (x-y) \cos (x+y) \). Write the equation with the common denominator: \[ \frac{\cos (x+y) + \sin (x-y)}{\sin (x-y) \cos (x+y)} = x \].
05

Simplify the numerator

Simplify the numerator \ \cos (x+y) + \sin (x-y) \ to identify potential values of \( x \): \[ \cos (x+y) + \sin (x-y) = x \sin (x-y) \cos (x+y) \].
06

Isolate the variable

Identify ways to isolate \( x \) through algebraic manipulation or further use of trigonometric identities: \[ \cos (x+y) + \sin (x-y) = x \sin (x-y) \cos (x+y) \]. This complex approach suggests a symmetry or specific solutions, typically x=0 or x=1 can be considered.

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 91Ó°ÊÓ!

Key Concepts

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

trigonometric identities
Trigonometric identities are crucial tools in solving trigonometric equations. They help us express trigonometric functions in a simpler or alternative form. In our exercise, we used two fundamental identities: \(\csc\) and \(\sec\). The cosecant function, \(\csc(\theta)\), is the reciprocal of sine, which means \(\csc(\theta) = \frac{1}{\sin(\theta)}\).\

Similarly, the secant function, \(\sec(\theta)\), is the reciprocal of cosine, giving us \(\sec(\theta) = \frac{1}{\cos(\theta)}\).\

By rewriting our trigonometric equation \(\csc(x-y) + \sec(x+y) = x\) using these identities, we obtain: \(\frac{1}{\sin(x-y)} + \frac{1}{\cos(x+y)} = x\).\ This step simplifies future algebraic manipulations, making it easier to isolate variables.
substitution method
The substitution method is a powerful technique for solving complex equations. It involves replacing one part of an equation with another equivalent expression to simplify the problem. In our case: \(\frac{1}{\sin(x-y)} + \frac{1}{\cos(x+y)} = x\). \

We can introduce new variables, like \(\text('u') = \frac{1}{\sin(x-y)} \) and \(\text('v') = \frac{1}{\cos(x+y)}\) to make things simpler: \(\text('u') + \text('v') = x\).\

Now, substitute back to get: \(\frac{1}{u} = \sin(x-y)\) and \(\frac{1}{v} = \cos(x+y)\). This approach temporarily removes the trigonometric identities and lets us work with simpler algebraic expressions.
algebraic manipulation
After substituting the identities, the next step is algebraic manipulation. This includes combining fractions, finding common denominators, and simplifying expressions. Given: \(\frac{1}{\sin(x-y)} + \frac{1}{\cos(x+y)} = x\),\

first find a common denominator: \(\sin(x-y) \cos(x+y)\). Rewrite the equation as: \(\frac{\cos(x+y) + \sin(x-y)}{\sin(x-y) \cos(x+y)} = x\).\

Now simplify the numerator: \(\cos(x+y) + \sin(x-y) = x \sin(x-y) \cos(x+y)\).\

This simplification may lead us to potential values of \(\text{'x'}\), highlighting the importance of skillful algebraic techniques in solving and simplifying trigonometric equations. This might suggest specific solutions like \(\text{'x'=0}\) or \(\text{'x'=1}\) can be considered.

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

\(f(x)=\left|\csc \frac{1}{2} x\right|\)

\(h(x)=\ln \left|\cot \frac{1}{2} x\right|\)

A particle moving in a straight line is said to have simple harmonic motion if the measure of its acceleration is always proportional to the measure of its displacement from a fixed point on the line and its acceleration and displacement are oppositely directed. So if at \(t \mathrm{sec}, s \mathrm{ft}\) is the directed distance of the particle from the origin and \(v \mathrm{ft} / \mathrm{sec}\) is the velocity of the particle, then a differential equation for simple harmonic motion is $$ \frac{d v}{d t}=-k^{2} s $$ where \(k^{2}\) is the constant of proportionality and the minus sign indicates that the acceleration is opposite in direction from the displacement. Because \(d v / d t=(d v / d s)(d s / d t)=v(d v / d s)\), Eq. (7) may be written as $$ v \frac{d v}{d s}=-k^{2} \mathrm{~s} $$ (a) Solve Eq. (8) for \(v\) to get \(v=\pm k \sqrt{a^{2}-s^{2}} .\) Note: Take \(a^{2} k^{2}\) as the arbitrary constant of integration and justify this choice. (b) Letting \(v=d s / d t\) in the solution of part (a), we obtain the differential equation $$ \frac{d s}{d t}=\pm k \sqrt{a^{2}-s^{2}} $$ Taking \(t=0\) at the instant when \(v=0\) (and hence \(s=a\) ), solve Eq. (9) to obtain $$ s=a \cos k t $$

Find, to the nearest degree, the measurements of the interior angles of the triangle having vertices at \((1,0),(-3,2)\), and \((2,3)\)

A steel girder 27 ft long is moved horizontally along a passageway \(8 \mathrm{ft}\) wide and into a corridor at right angles to the passageway. How wide must the corridor be in order for the girder to go around the corner? Neglect the horizontal width of the girder.

See all solutions

Recommended explanations on Math Textbooks

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.