Problem 58 - Precalculus with Calculus Previews

Question

For which real numbers \(x\) is (a) \(\sin x \leq \csc x ?\) (b) \(\sin x<\csc x ?\)

Step-by-Step Solution

Answer

(a) True for all \(x\neq k\pi\). (b) True for \(-1 < \sin x < 1\) and \(x\neq k\pi\).

Step 1: Understanding the Relationship

The function \(\csc x\) is the reciprocal of \(\sin x\), so \(\csc x = \frac{1}{\sin x}\). Our task is to compare \(\sin x\) and \(\csc x\): \( \sin x \leq \csc x \) and \( \sin x < \csc x \).

Step 2: Identifying the Domain

The function \(\csc x\) is only defined where \(\sin x eq 0\). Therefore, we exclude integer multiples of \(\pi\), i.e., \(x eq k\pi\) where \(k\) is an integer.

Step 3: Inequality Transformation for (a)

For \(\sin x \leq \csc x\), substitute \(\csc x\) with \(\frac{1}{\sin x}\) to get \(\sin x \leq \frac{1}{\sin x}\). Multiply both sides by \(\sin x\) assuming \(\sin x > 0\): \((\sin x)^2 \leq 1\). This simplifies to \(-1 \leq \sin x \leq 1\). Since \(\sin x\) always satisfies \(-1 \leq \sin x \leq 1\), this condition is automatically met, except for the divisibility condition \(x eq k\pi\).

Step 4: Solution for (a)

The inequality \(\sin x \leq \csc x\) is true for all \(x\) in the domain of sine, except where \(\sin x = 0\) (i.e., \(x eq k\pi\)).

Step 5: Inequality Transformation for (b)

For \(\sin x < \csc x\), we follow the same step as above but look for strict inequality. Convert this to \((\sin x)^2 < 1\), which excludes the boundaries \(\sin x = -1\) and \(\sin x = 1\).

Step 6: Solution for (b)

This inequality \(\sin x < \csc x\) holds true when \(-1 < \sin x < 1\) and \(\sin x eq 0\). Practically, for \(\sin xeq 0\), it is true wherever \(-1 < \sin x < 1\) except integer multiples of half-pi (i.e., \(xeq \frac{(2n+1)\pi}{2}\), \(n\) is an integer).

Key Concepts

Sine Function

The sine function, often denoted as \( \sin x \), is one of the fundamental trigonometric functions in mathematics, which describes the y-coordinate of a point on the unit circle corresponding to an angle \( x \). It is essential in many areas of geometry and physics because it describes how projections change and become cyclical.
The sine function takes input values (angles) in radians or degrees and outputs a real number between -1 and 1. This is because the highest and lowest points on the unit circle correspond to 1 and -1, respectively.

• The sine function is periodic with a period of \( 2\pi \) (or 360 degrees), meaning \( \sin(x) = \sin(x + 2\pi) \).
• The range of the sine function is \(-1 \leq \sin x \leq 1 \).
• The graph of the sine function is wave-like, often referred to as a sine wave.

Understanding the sine function is crucial when dealing with trigonometric inequalities, as it often forms the basis for more complex trigonometric expressions and equations.

Cosecant Function

The cosecant function, denoted as \( \csc x \), is the reciprocal of the sine function. In mathematical terms, this means \( \csc x = \frac{1}{\sin x} \). This implies that the cosecant function is undefined whenever the sine function is zero because division by zero is undefined.
The cosecant function thus inherits the same periodicity of \( 2\pi \) from the sine function.

• For \( \csc x \) to be defined, \( \sin x eq 0 \), meaning \( x eq k\pi \), where \( k \) is an integer.
• The behavior of \( \csc x \) is such that it has vertical asymptotes, lines which it approaches but never intersects, at each point where \( \sin x = 0 \).
• The range of \( \csc x \) is \((-\infty, -1] \cup [1, \infty)\), because the reciprocal function amplifies the extremes of \( \sin x \).

Understanding how \( \csc x \) relates to \( \sin x \) is vital when resolving inequalities involving these functions, as demonstrated in the original exercise.

Trigonometric Domain

In understanding trigonometric functions, it is crucial to consider their domain, which denotes all the permissible input values (angles) for which the functions are defined.
For the sine function, the domain is all real numbers, \( x \in \mathbb{R} \), because \( \sin x \) can take any angle and yield a defined value between -1 and 1.
However, the cosecant function's domain requires greater attention. Since \( \csc x = \frac{1}{\sin x} \), it is only defined where \( \sin x eq 0 \).
This constraint results in excluding angles which are integer multiples of \( \pi \) from the domain.

• Thus, for \( \csc x \), the domain is \( x eq k\pi \) where \( k \) is an integer.
• These exclusions correspond to the points where the sine function crosses zero.

When solving trigonometric inequalities, recognizing and working within these domains is imperative as it avoids undefined expressions and potential errors in solutions.