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Chapter 5: Problem 3

$$ \text { Verify that } \cos x<\left(\frac{\sin x}{r}\right)^{3} \text { for } 0<|x|<\frac{\pi}{2} \text {. } $$

Short Answer

Expert verified
Question: Verify if the inequality \(\cos x < \left(\frac{\sin x}{r}\right)^3\) holds for \(0 < |x| < \frac{\pi}{2}\) when \(r > 1\). Answer: Yes, the inequality \(\cos x < \left(\frac{\sin x}{r}\right)^3\) holds for \(0 < |x| < \frac{\pi}{2}\) when \(r > 1\).

Step by step solution

01

Observe the ranges of the functions

First, let's observe the ranges in which the functions can take values: \(\cos x\) can take values between \(0\) and \(1\) for \(0 < |x| < \frac{\pi}{2}\) because the cosine function has a range of \([-1, 1]\) and is positive in the given range of x. \(\left(\frac{\sin x}{r}\right)^3\), since \(\sin x\) can take values between \(0\) and \(1\) and \(r\) is a positive constant, can also take values between \(0\) and \(1\).
02

Analyze the behavior of the functions

Now let's analyze the behavior of both functions in the interval \(0 < |x| < \frac{\pi}{2}\): When \(0 < |x| < \frac{\pi}{2}\), \(\sin x\) is an increasing function which means that the function \(\left(\frac{\sin x}{r}\right)^3\) will also be increasing. In the same interval, \(\cos x\) is a decreasing function since it starts at a maximum at \(|x|=0\) and reaches a minimum at \(|x|=\frac{\pi}{2}\).
03

Apply trigonometric identities

To further evaluate the inequality, we can apply the following trigonometric identity: \(\cos^{2} x + \sin^{2} x = 1\). Since \(0 < |x| < \frac{\pi}{2}\), both \(\cos x\) and \(\sin x\) are positive. We can rewrite the inequality in terms of \(\sin x\) and \(\cos x\) using the trigonometric identity: $$\cos x < \left(\frac{\sin x}{r}\right)^3 \text { can be re-written as }1 - \sin^2 x < \left(\frac{\sqrt{1-\cos^2 x}}{r}\right)^3$$ Since both functions are positive, we can square the inequality: $$\cos^2 x < \frac{1 - \cos^2 x}{r^2}$$
04

Verify the inequality

Given the behavior of the functions, it is clear that the value of \(\cos x\) decreases as the value of \(|x|\) increases, while the value of \(\left(\frac{\sin x}{r}\right)^3\) increases as the value of \(|x|\) increases. As both functions have ranges between \(0\) and \(1\), the inequality holds as follows: $$\cos^2 x < \frac{1 - \cos^2 x}{r^2}$$ We can simplify this inequality and find a condition for \(r\): $$r^2 \cos^2 x < (1 - \cos^2 x)$$ As long as \(r > 1\), this inequality will hold for \(0 < |x| < \frac{\pi}{2}\). Thus, if \(r > 1\), then we can verify that \(\cos x < \left(\frac{\sin x}{r}\right)^3\) for \(0 < |x| < \frac{\pi}{2}\).

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Key Concepts

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

Cosine Function
The cosine function, denoted as \( \cos x \), is a fundamental piece of trigonometry. It describes the x-coordinate of a point on a unit circle at a given angle \( x \). For angles \( 0 < |x| < \frac{\pi}{2} \), the cosine function takes values between 0 and 1, behaving as a decreasing function. At \( x = 0 \), \( \cos x \) reaches its maximum of 1, gradually decreasing as \( x \) approaches \( \frac{\pi}{2} \), where \( \cos x \) becomes 0. This decreasing behavior is crucial for comparing trigonometric functions within specific intervals.
Sine Function
The sine function, represented by \( \sin x \), corresponds to the y-coordinate of a point on the unit circle. Within the interval \( 0 < |x| < \frac{\pi}{2} \), sine values range from 0 to 1. Unlike cosine, \( \sin x \) is an increasing function within this range, beginning at 0 and reaching 1 as \( x \) approaches \( \frac{\pi}{2} \). By transforming \( \sin x \) with a constant scaling of \( r \), we assess inequality involving \( \left( \frac{\sin x}{r} \right)^3 \). Keeping \( r \) positive ensures that \( \left( \frac{\sin x}{r} \right)^3 \) is a valid function for comparison.
Trigonometric Identities
Trigonometric identities are essential tools for manipulating and simplifying expressions. The identity \( \cos^2 x + \sin^2 x = 1 \) is particularly useful. In our scenario, both \( \cos x \) and \( \sin x \) are positive, allowing us to express one function in terms of the other. Using the identity, the inequality \( \cos x < \left( \frac{\sin x}{r} \right)^3 \) is rewritten as \( 1 - \sin^2 x < \left( \frac{\sqrt{1-\cos^2 x}}{r} \right)^3 \). Squares and square roots transform these functions to enable a direct comparison suitable for inequality verification.
Inequality Verification
Verifying inequalities involves evaluating the behavior of each function within the specified range. For \( 0 < |x| < \frac{\pi}{2} \), \( \cos x \) decreases, while \( \left( \frac{\sin x}{r} \right)^3 \) increases, setting the stage for the given inequality. By squaring both sides, we obtain \( \cos^2 x < \frac{1 - \cos^2 x}{r^2} \). Simplifying further, \( r^2 \cos^2 x < 1 - \cos^2 x \) indicates that the inequality is satisfied if \( r > 1 \). This analysis shows that the initial inequality holds as long as \( r \) is chosen appropriately.

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