Problem 6
Question
Find all solutions of the equation. $$\csc \gamma=\sqrt{2}$$
Step-by-Step Solution
Verified Answer
The solutions are \(\gamma = \frac{\pi}{4} + 2k\pi\) and \(\gamma = \frac{3\pi}{4} + 2k\pi\), where \(k\) is any integer.
1Step 1: Convert Cosecant to Sine
Recognize that the cosecant function is the reciprocal of the sine function. Thus, we have \( ext{csc} \, \gamma = \frac{1}{\sin \gamma}\). So, rewrite the equation in terms of sine: \(\frac{1}{\sin \gamma} = \sqrt{2}\).
2Step 2: Solve for Sine
From \(\frac{1}{\sin \gamma} = \sqrt{2}\), solve for \(\sin \gamma\). Multiply both sides by \(\sin \gamma\) and then divide by \(\sqrt{2}\) to get \(\sin \gamma = \frac{1}{\sqrt{2}}\).
3Step 3: Simplify Sine Expression
Rationalize the denominator of \(\sin \gamma = \frac{1}{\sqrt{2}}\) to get \(\sin \gamma = \frac{\sqrt{2}}{2}\).
4Step 4: Determine General Solutions for Sine
Recall that \(\sin \gamma = \frac{\sqrt{2}}{2}\) at angles \(\gamma = \frac{\pi}{4} + 2k\pi\) and \(\gamma = \frac{3\pi}{4} + 2k\pi\) for any integer \(k\). These represent the principal angles and the periodic nature of the sine function.
Key Concepts
Cosecant FunctionSine FunctionReciprocal Trigonometric IdentitiesGeneral Solutions for Trigonometric Equations
Cosecant Function
The cosecant function is one of the lesser-known trigonometric functions, but it plays a significant role in solving trigonometric equations. In simple terms, cosecant is the reciprocal of the sine function. This means that if you obtain the sine of an angle, the cosecant is the inverse value of that sine. Mathematically, it is represented as:
Converting a problem from cosecant to sine can make it more approachable since sine functions have well-known properties and solutions.
- ext{csc} heta = rac{1}{ ext{sin} heta}
Converting a problem from cosecant to sine can make it more approachable since sine functions have well-known properties and solutions.
Sine Function
The sine function is foundational in trigonometry and common in various applications, from geometry to calculus. It's a periodic function, meaning it repeats its values in regular intervals, specifically every \(2\pi\). The sine of an angle \(\theta\) can be defined as:
In the unit circle context, sine is the \(y\)-coordinate of a point where a line through the origin makes an angle \(\theta\) with the positive \(x\)-axis. Understanding how sine behaves and how it is linked to different angles can help solve trigonometric equations by finding angles that satisfy given conditions.
- \( ext{sin} \, \theta = \frac{ ext{opposite side}}{ ext{hypotenuse}}\)
In the unit circle context, sine is the \(y\)-coordinate of a point where a line through the origin makes an angle \(\theta\) with the positive \(x\)-axis. Understanding how sine behaves and how it is linked to different angles can help solve trigonometric equations by finding angles that satisfy given conditions.
Reciprocal Trigonometric Identities
Reciprocal trigonometric identities are a set of identities that express each of the six trigonometric functions in terms of reciprocal relationships. These identities are crucial when solving trigonometric equations, like when we converted cosecant to sine in the exercise. Here are the fundamental reciprocal identities:
- \( ext{csc} heta = \frac{1}{ ext{sin} heta}\)
- \( ext{sec} heta = \frac{1}{ ext{cos} heta}\)
- \( ext{cot} heta = \frac{1}{ ext{tan} heta}\)
General Solutions for Trigonometric Equations
General solutions in trigonometric equations refer to finding all possible solutions for an equation, given the periodic nature of trigonometric functions. Since trigonometric functions like sine repeat every \(2\pi\), there are typically infinite solutions for a given trigonometric equation, except within specific bounds.
For example, in the problem "\( ext{sin} \, \gamma = \frac{\sqrt{2}}{2}\)", the principal solutions are \(\gamma = \frac{\pi}{4}\) and \(\gamma = \frac{3\pi}{4}\).
However, since sine is periodic with a period of \(2\pi\), the complete solution set includes:
For example, in the problem "\( ext{sin} \, \gamma = \frac{\sqrt{2}}{2}\)", the principal solutions are \(\gamma = \frac{\pi}{4}\) and \(\gamma = \frac{3\pi}{4}\).
However, since sine is periodic with a period of \(2\pi\), the complete solution set includes:
- \(\gamma = \frac{\pi}{4} + 2k\pi\)
- \(\gamma = \frac{3\pi}{4} + 2k\pi\)
Other exercises in this chapter
Problem 6
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