/*! 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 65 Find the solutions of the equati... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the solutions of the equation that are in the interval \([0,2 \pi)\). $$\cot \alpha+\tan \alpha=\csc \alpha \sec \alpha$$

Short Answer

Expert verified
Solutions are all \(\alpha\) in \((0, \frac{\pi}{2})\), \((\frac{\pi}{2}, \pi)\), \((\pi, \frac{3\pi}{2})\), \((\frac{3\pi}{2}, 2\pi)\).\)

Step by step solution

01

Rewrite Trigonometric Identities

We start by expressing all the trigonometric functions in terms of sine and cosine. Recall that \(\cot \alpha = \frac{\cos \alpha}{\sin \alpha}\), \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\), \(\csc \alpha = \frac{1}{\sin \alpha}\), and \(\sec \alpha = \frac{1}{\cos \alpha}\). Substitute these into the equation:\[\frac{\cos \alpha}{\sin \alpha} + \frac{\sin \alpha}{\cos \alpha} = \frac{1}{\sin \alpha \cos \alpha}\]
02

Find a Common Denominator

Combine the left-hand side into a single fraction by finding a common denominator. The common denominator for the terms is \(\sin \alpha \cos \alpha\):\[\frac{\cos^2 \alpha + \sin^2 \alpha}{\sin \alpha \cos \alpha} = \frac{1}{\sin \alpha \cos \alpha}\]Now, remember the Pythagorean identity \(\cos^2 \alpha + \sin^2 \alpha = 1\). Substitute this to get:\[\frac{1}{\sin \alpha \cos \alpha}\]
03

Cross Multiply

Since the equation \(\frac{1}{\sin \alpha \cos \alpha} = \frac{1}{\sin \alpha \cos \alpha}\) holds true identically, cross-multiplying shows that the equation is satisfied for all values where \(\sin \alpha \cos \alpha eq 0\).
04

Determine the Domain

Since the equation holds for all \(\alpha\) such that \(\sin \alpha \cos \alpha eq 0\), we solve for when \(\sin \alpha = 0\) or \(\cos \alpha = 0\). The values excluded are \(\alpha = 0, \pi/2, \pi, 3\pi/2\).
05

List Solutions in Interval

In the interval \([0, 2\pi)\), the solutions are all real numbers except the points where \(\sin \alpha\) and \(\cos \alpha\) are zero. Therefore, the solutions are:\[\alpha \in (0, \frac{\pi}{2}), (\frac{\pi}{2}, \pi), (\pi, \frac{3\pi}{2}), (\frac{3\pi}{2}, 2\pi)\]

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Key Concepts

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

Pythagorean Identity
The Pythagorean identity is one of the fundamental relationships in trigonometry. As explored in this solution, it ties the sine and cosine functions together beautifully. The identity is expressed as:\[ \cos^2 \alpha + \sin^2 \alpha = 1 \]This relationship echoes the Pythagorean Theorem in geometry, demonstrating that the squared hypotenuse of a right-angled triangle equals the sum of the squares of the other two sides. In trigonometric terms, this means the sum of the squares of cosine and sine of an angle always equals one.This identity is useful in many areas of mathematics and physics and helps simplify expressions involving trigonometric functions. For example, in our exercise, recognizing this identity allowed us to substitute it directly, simplifying the equation dramatically. Using this identity can often transform complicated expressions to simpler ones and expose the inherent relationships between different functions.
Interval Notation
Interval notation is a mathematical shorthand for expressing subsets of the real number line. It efficiently represents a range of values within particular bounds. In solving trigonometric equations, this concept is crucial for narrowing down the possible solutions.Consider the interval \([0, 2\pi)\) in our problem. This interval tells us that we're interested in all solutions that start inclusive of 0 but stop just short of including \(2\pi\). The use of brackets and parentheses is key in interval notation:
  • \([\ldots]\) denotes a closed interval, including the endpoint.
  • \((\ldots)\) denotes an open interval, excluding the endpoint.
In finding solutions to the exercise, we exclude certain points within the interval. We are specifically asked to find places where \(\sin \alpha\) and \(\cos \alpha\) are non-zero to satisfy the equation, necessitating exclusions at certain intervals where they equal zero.
Trigonometric Equations
Trigonometric equations are equations that involve trigonometric functions like sine, cosine, tangent, and their reciprocals. They often require you to solve for an angle that satisfies particular conditions. In solving these equations, you typically use known identities to simplify the equation, break it down into more manageable parts, and find all possible solutions. The process can involve:
  • Substituting trigonometric identities.
  • Finding common denominators to combine fractions.
  • Cross-multiplying to simplify expressions.
In our initial problem, we rewrote the functions with their sine and cosine counterparts before simplifying. These kinds of manipulations depend on understanding and applying fundamental identities, allowing one to solve and interpret complex trigonometric relationships effectively.
Sine and Cosine Functions
Sine and cosine functions are the backbone of trigonometry. They describe fundamental wave-like oscillations seen in various natural processes and are vital in mathematics and engineering applications.
  • Sine Function is defined as the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle. Represented as \(\sin \alpha\), it describes periodic oscillations over its range of \([-1, 1]\).
  • Cosine Function is similarly defined, but it is the ratio of the adjacent side to the hypotenuse, denoted as \(\cos \alpha\).
These functions have specific zeros in their cycles. For sine, zeros occur at multiples of \(\pi\) (0, \(\pi\), \(2\pi\), etc.), and for cosine, they occur at odd multiples of \(\frac{\pi}{2}\). Recognizing these basics helps identify when they contribute as zeros in trigonometric equations, as seen in our problem's solution. Understanding their behavior is key to effectively solving equations that incorporate them.

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Most popular questions from this chapter

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