Problem 61 - Single Variable Calculus

Question

Prove that \(2 x / \pi<\sin x

Step-by-Step Solution

Answer

First, find the derivative of \(f(x) = \frac{\sin x}{x}\) using the quotient rule, so \(f'(x) = \frac{(\cos x)(x) - (\sin x)(1)}{x^2}\). Next, examine the sign of \(f'(x)\) on the interval \(\left(0, \frac{\pi}{2}\right)\). Since \(\sin x < x\) and \(\cos x > 0\), the numerator is positive. Thus, \(f'(x)\) is positive on this interval, meaning the function is increasing. Use this result to prove the inequality: \(\frac{2}{\pi} < \frac{\sin x}{x} < 1\), which when multiplied by x, gives the desired result: \(2x/\pi < \sin x < x\) for \(0 < x < \frac{\pi}{2}\).

Step 1: Find the derivative of the function f(x)

First, we need to find the derivative of the function \(f(x) = \frac{\sin x}{x}\). Using the quotient rule, the derivative of a function \(\frac{u(x)}{v(x)}\) is given by \(\frac{u'(x)v(x) - u(x)v'(x)}{v^2(x)}\). Therefore, the derivative of \(f(x)\) is: \( f'(x) = \frac{(\cos x)(x) - (\sin x)(1)}{x^2} \)

Step 2: Examine the sign of the derivative f'(x)

Now, we want to determine the sign of the derivative \(f'(x)\) on the interval \(\left(0, \frac{\pi}{2}\right)\). This will tell us if the function is decreasing. We know that \(\cos x\) is positive and \(\sin x\) is increasing for \(0 < x < \frac{\pi}{2}\). Additionally, \(x\) is positive on the given interval, thus both terms in the numerator are positive. However, since \(\sin x < x\) on this interval (due to the fact that \(\sin x\) is always smaller than the arc itself), the second term in the numerator will be smaller than the first one. As a result, the overall numerator is positive: \(\cos x \cdot x - \sin x > 0 \) And since \(x^2\) is always positive, the derivative \(f'(x)\) is positive over the interval \(\left(0, \frac{\pi}{2}\right)\).

Step 3: Interpret result of the sign of f'(x)

Since we have established that the derivative, \(f'(x)\), is positive over the interval \(\left(0, \frac{\pi}{2}\right)\), we can now conclude that the function \(f(x) = \frac{\sin x}{x}\) is increasing on this interval.

Step 4: Combine results to prove the inequality

Now we know that the function \(f(x) = \frac{\sin x}{x}\) is increasing on the interval \(\left(0, \frac{\pi}{2}\right)\). Therefore, for any \(0 < x < \frac{\pi}{2}\), we have: \(\frac{2}{\pi} < \frac{\sin x}{x} < 1\) By multiplying both sides of the inequality by x, we get: \(2x/\pi < \sin x < x\) Hence, we have proved that \(2x/\pi <\sin x

Key Concepts

Quotient Rule

The quotient rule is an essential tool in calculus used to find the derivative of a function expressed as a quotient of two other functions. If we have a function of the form \( f(x) = \frac{u(x)}{v(x)} \), the quotient rule states that its derivative is given by:

• \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \)

This rule is particularly useful when dealing with complex fractions, allowing us to differentiate the numerator and denominator separately.
In the given problem, where \( f(x) = \frac{\sin x}{x} \), we apply the quotient rule by setting \( u(x) = \sin x \) and \( v(x) = x \). Here, the derivatives \( u'(x) = \cos x \) and \( v'(x) = 1 \) are used to find the derivative \( f'(x) \). This gives us:

• \( f'(x) = \frac{(\cos x)x - \sin x \cdot 1}{x^2} \)

Understanding how to utilize the quotient rule simplifies the process of finding derivatives for rational functions.

Derivative Sign Analysis

Once the derivative of a function is known, analyzing its sign over a particular interval provides insights into the function's behavior. The sign of the derivative indicates whether the function is increasing or decreasing:

• A positive derivative means the function is increasing.
• A negative derivative signifies the function is decreasing.

In the given context, determining the sign of \( f'(x) \) helps to show that \( \frac{\sin x}{x} \) is decreasing.
We observe the interval \( (0, \frac{\pi}{2}) \), where \( \cos x \), and \( x \) are both positive, making \( x \cdot \cos x \) positive. Since \( \sin x < x \) over this range, the term \( \cos x \cdot x - \sin x \) remains positive. Thus:

• For \( 0 < x < \frac{\pi}{2} \), \( f'(x) > 0 \).

This result indicates that the function is increasing over the specified interval, a foundation for the given inequality.

Trigonometric Inequalities

Trigonometric inequalities involve inequalities containing trigonometric functions such as sine, cosine, or tangent. In this exercise, the core inequality is \( 2x/\pi < \sin x < x \) for \( 0 < x < \frac{\pi}{2} \).
Here, understanding the behavior of \( \sin x \) is crucial. It's known that \( \sin x \) grows from 0 to 1 as \( x \) increases from 0 to \( \frac{\pi}{2} \). This means:

• \( \sin x < x \) because the curve of \( \sin x \) lies below the line \( y = x \) in this interval.
• \( 2x/\pi < \sin x \) arises as a result of comparing the rate at which \( \sin x \) approaches 1 compared to the linear function \( 2x/\pi \).

Applying derivative sign analysis to show that the function \( \frac{\sin x}{x} \) is increasing establishes that \( \sin x \) stays between these bounding lines over \( 0 < x < \frac{\pi}{2} \).

Increasing Functions

An increasing function is one where the function value increases as the input increases. To mathematically confirm whether a function is increasing, its derivative must be evaluated:

• If \( f'(x) > 0 \) for all \( x \) in a given interval, the function is increasing over that interval.

In this problem, \( f(x) = \frac{\sin x}{x} \) was shown to have a positive derivative over \( (0, \frac{\pi}{2}) \). Thus, \( f(x) \) is increasing.
As a consequence, as \( x \) approaches \( \frac{\pi}{2} \), \( \frac{\sin x}{x} \) approaches its maximum within this interval. Establishing that the function is increasing was key in proving the given trigonometric inequality, bridging the connection between increasing functions and trigonometric bounds. This reinforces the understanding of how functions behave over specific intervals and how they relate to inequalities.

Previous

Problem 61

Next

Problem 61

Other exercises in this chapter

Problem 61

Flight Path of a Plane The function \(f(x)=\left\\{\begin{array}{ll}0 & \text { if } 0 \leq x

View solution

Problem 61

Chemical Pollution As a result of an abandoned chemical dump leaching chemicals into the water, the main well of a town has been contaminated with trichloroethy

View solution

Problem 61

Effect of Smoking Bans The sales (in billions of dollars) in restaurants and bars in California from the beginning of \(1993(t=0)\) to the beginning of \(2000(t

View solution

Problem 61

Maximizing Profit The total daily profit in dollars realized by the TKK Corporation in the manufacture and sale of \(x\) dozen recordable DVDs is given by the t

View solution

Key Concepts

Quotient Rule

The quotient rule is an essential tool in calculus used to find the derivative of a function expressed as a quotient of two other functions. If we have a function of the form \( f(x) = \frac{u(x)}{v(x)} \), the quotient rule states that its derivative is given by:

• \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \)

This rule is particularly useful when dealing with complex fractions, allowing us to differentiate the numerator and denominator separately.
In the given problem, where \( f(x) = \frac{\sin x}{x} \), we apply the quotient rule by setting \( u(x) = \sin x \) and \( v(x) = x \). Here, the derivatives \( u'(x) = \cos x \) and \( v'(x) = 1 \) are used to find the derivative \( f'(x) \). This gives us:

• \( f'(x) = \frac{(\cos x)x - \sin x \cdot 1}{x^2} \)

Understanding how to utilize the quotient rule simplifies the process of finding derivatives for rational functions.

Derivative Sign Analysis

Once the derivative of a function is known, analyzing its sign over a particular interval provides insights into the function's behavior. The sign of the derivative indicates whether the function is increasing or decreasing:

• A positive derivative means the function is increasing.
• A negative derivative signifies the function is decreasing.

In the given context, determining the sign of \( f'(x) \) helps to show that \( \frac{\sin x}{x} \) is decreasing.
We observe the interval \( (0, \frac{\pi}{2}) \), where \( \cos x \), and \( x \) are both positive, making \( x \cdot \cos x \) positive. Since \( \sin x < x \) over this range, the term \( \cos x \cdot x - \sin x \) remains positive. Thus:

• For \( 0 < x < \frac{\pi}{2} \), \( f'(x) > 0 \).

This result indicates that the function is increasing over the specified interval, a foundation for the given inequality.

Trigonometric Inequalities

Trigonometric inequalities involve inequalities containing trigonometric functions such as sine, cosine, or tangent. In this exercise, the core inequality is \( 2x/\pi < \sin x < x \) for \( 0 < x < \frac{\pi}{2} \).
Here, understanding the behavior of \( \sin x \) is crucial. It's known that \( \sin x \) grows from 0 to 1 as \( x \) increases from 0 to \( \frac{\pi}{2} \). This means:

• \( \sin x < x \) because the curve of \( \sin x \) lies below the line \( y = x \) in this interval.
• \( 2x/\pi < \sin x \) arises as a result of comparing the rate at which \( \sin x \) approaches 1 compared to the linear function \( 2x/\pi \).

Applying derivative sign analysis to show that the function \( \frac{\sin x}{x} \) is increasing establishes that \( \sin x \) stays between these bounding lines over \( 0 < x < \frac{\pi}{2} \).

Increasing Functions

An increasing function is one where the function value increases as the input increases. To mathematically confirm whether a function is increasing, its derivative must be evaluated:

• If \( f'(x) > 0 \) for all \( x \) in a given interval, the function is increasing over that interval.

In this problem, \( f(x) = \frac{\sin x}{x} \) was shown to have a positive derivative over \( (0, \frac{\pi}{2}) \). Thus, \( f(x) \) is increasing.
As a consequence, as \( x \) approaches \( \frac{\pi}{2} \), \( \frac{\sin x}{x} \) approaches its maximum within this interval. Establishing that the function is increasing was key in proving the given trigonometric inequality, bridging the connection between increasing functions and trigonometric bounds. This reinforces the understanding of how functions behave over specific intervals and how they relate to inequalities.