Question

Define $$ \begin{array}{l} f(x)=x^{3}-\sin ^{2} x \tan x \\ g(x)=2 x^{2}-\sin ^{2} x-x \tan x . \end{array} $$ Find out, for each of these two functions, whether it is positive or negative for all \(x \in(0, \pi / 2)\), or whether it changes sign. Prove your answer.

Short answer

Based on the given step-by-step solution, it can be concluded that both functions, \(f(x)=x^{3}-\sin ^{2} x \tan x\) and \(g(x)=2 x^{2}-\sin ^{2} x-x \tan x\), are positive on the interval \((0, \pi/2)\).

Step-by-step solution

Simplify the functions

First, we observe that the two functions have some common terms. We can use the trigonometric identity \(\sin^2 x = 1 - \cos^2 x\) to simplify them: $$ \begin{array}{l} f(x)=x^{3}-(1-\cos^{2}x) \tan x \\\ g(x)=2 x^{2}-(1-\cos^{2} x)-x \tan x . \end{array} $$

Analyze the interval of interest

Next, we analyze the interval \((0, \pi/2)\) considering the properties of sine, cosine, and tangent functions. 1. In this interval, the sine function is positive, since \(\sin x\) is positive for \(x \in (0, \pi)\). 2. The cosine function is also positive in this interval, because \(\cos x\) is positive for \(x \in (0, \pi)\). 3. The tangent function is positive as well, since \(\tan x\) is positive for \(x \in (0, \pi/2)\).

Analyze the functions on given interval

Now, let's explore how \(f(x)\) and \(g(x)\) behave over the interval \((0, \pi/2)\). For \(f(x) = x^{3}-(1-\cos^{2}x) \tan x\), we have: 1. \(x^3\) is positive, since \(x \in (0, \pi/2)\). 2. \((1-\cos^{2} x)\) is positive, as it is equal to \(\sin^2 x\). 3. \(\tan x\) is also positive. We observe that all the components of \(f(x)\) are positive, so \(f(x)\) is positive for all \(x \in (0, \pi/2)\). For \(g(x) = 2 x^{2}-(1-\cos^{2} x)-x \tan x\), we have: 1. \(2x^2\) is positive, since \(x \in (0, \pi/2)\). 2. \((1-\cos^{2} x)\) is positive, as it is equal to \(\sin^2 x\). 3. \(x \tan x\) is positive, due to both factors being positive. With all the positive components, \(g(x)\) should also be positive. However, the terms \((1-\cos^{2} x)\) and \(x \tan x\) are subtracted, and the difference could be negative. So it is unclear whether \(g(x)\) is positive, negative, or changes its sign on the interval \((0, \pi/2)\). It will be important to explore this possibility further.

Further exploration of \(g(x)\)

To address the uncertainty of \(g(x) = 2 x^{2}-(1-\cos^{2} x)-x \tan x\), let's rewrite it in another form, using the double-angle identity as clustering together the terms of degree 2: $$ g(x) = 2x^2 - x\tan x - 1 + \cos^2 x $$ Since \(\cos 2x = 1 - 2\sin^2 x\), we can find the double-angle version for cosine: \(\cos^2 x=\dfrac{1+\cos 2x}{2}\). Then, $$ g(x) = 2x^2 - x\tan x - \dfrac{1 - \cos 2x}{2} $$ For \(x \in (0, \pi/2)\), \(\cos 2x\) is positive, thus \(1 - \cos 2x\) is negative. Consequently, the remaining term \(\dfrac{1 - \cos 2x}{2}\) will be negative. If we can show that \(2x^2 - x\tan x\) is always greater than \(\dfrac{1 - \cos 2x}{2}\), it implies that \(g(x)\) is positive. Since \(0<x<\pi/2\), we have \(2x<\pi\), and by the tangent inequality \(x<\tan x\) in the given interval, we can deduce the following: $$ 2x^2 - x\tan x < 2x^2 - x^2 = x^2 $$ On the interval \((0, \pi/2)\), we have \(\cos 2x \leq 1\), and therefore, \(1-\cos 2x \geq 0\). Therefore, $$ \dfrac{1 - \cos 2x}{2} \geq 0 $$ As we can see, \(0\leq \dfrac{1 - \cos 2x}{2} < x^2 < 2x^2 - x\tan x\), so as we have shown, \(g(x)\) remains positive for the interval \((0, \pi/2)\).

Conclusion

In conclusion, function \(f(x)=x^{3}-\sin ^{2} x \tan x\) is positive on the interval \((0, \pi/2)\), and function \(g(x)=2 x^{2}-\sin ^{2} x-x \tan x\) is also positive on the same interval.

Key concepts

Trigonometric Identities

Trigonometric identities are equations that relate the trigonometric functions of angles. These identities are an essential part of solving many problems in mathematics, especially in the fields of algebra, calculus, and real analysis. One of the most used is the Pythagorean identity, which states that for any angle \( x \), \( \sin^2 x + \cos^2 x = 1 \). This relationship lets us express one trigonometric function in terms of another, simplifying equations and computations.

For instance, in our textbook exercise, we utilized the identity \( \sin^2 x = 1 - \cos^2 x \) to simplify the given functions \( f(x) \) and \( g(x) \). This allowed us to more easily analyze their behavior on the interval \( (0, \pi/2) \). \( \cos^2 x \) was then used to convert \( g(x) \) into an alternative form that facilitated further analysis. Understanding and applying these identities is crucial to evaluating trigonometric functions, hereby facilitating the determination of their sign within a specified interval.

Inequality in Trigonometry

Inequalities find frequent application in trigonometry, helping to define the scope within which trigonometric functions will behave under certain constraints. In particular, inequalities in trigonometry are profound when analyzing intervals and assessing function behavior.

For instance, the textbook solution leveraged the fundamental inequalities of trigonometric functions on specific intervals: notably, the sine, cosine, and tangent functions are all positive on the interval \( (0, \pi/2) \). Additionally, inequalities can provide bounds for these functions, such as the fact that for \( x \) in \( (0, \pi/2) \), \( x < \tan x \), which was pivotal in the examination of the function \( g(x) \) within the same interval. This knowledge, together with interval analysis, is essential in determining whether functions remain positive, negative, or change signs within particular bounds.

Interval Analysis

Interval analysis is a technique used to evaluate the behavior of functions over specific ranges of values. In our exercise, we used interval analysis to dissect each term of the functions \( f(x) \) and \( g(x) \) on the interval \( (0, \pi/2) \). Through careful examination, we were able to derive that \( f(x) \) is positive since all components of the function are positive within the given range.

For the function \( g(x) \) however, interval analysis revealed a subtractive term that could potentially result in a negative value. To resolve the ambiguity, the exercise reworked \( g(x) \) using trigonometric identities and analyzed the inequality provided by the interval. By comparing \( 2x^2 - x\tan x \) with \( \dfrac{1 - \cos 2x}{2} \) over the interval, the function \( g(x) \) was similarly concluded to be positive. This systematic approach harnessing interval analysis ensures a detailed understanding of a function's behavior within defined limits, a skill crucial for students studying real analysis.