Question

Two trigonometric inequalities Consider the angle \(\theta\) in standard position in a unit circle where \(0 \leq \theta<\pi / 2\) or \(-\pi / 2<\theta<0\) (use both figures). Figure cannot copy a. Show that \(|A C|=|\sin \theta|,\) for \(-\pi / 2<\theta<\pi / 2 .\) (Hint: Consider the cases \(0<\theta<\pi / 2\) and \(-\pi / 2<\theta<0\) separately.) b. Show that \(|\sin \theta|<|\theta|,\) for \(-\pi / 2<\theta<\pi / 2 .\) (Hint: The length of arc \(A B\) is \(\theta\) if \(0 \leq \theta<\pi / 2,\) and is \(-\theta\) if \(-\pi / 2<\theta<0 .)\) c. Conclude that \(-|\theta| \leq \sin \theta \leq|\theta|,\) for \(-\pi / 2<\theta<\pi / 2\). d. Show that \(0 \leq 1-\cos \theta \leq|\theta|,\) for \(-\pi / 2<\theta<\pi / 2\). e. Show that \(1-|\theta| \leq \cos \theta \leq 1,\) for \(-\pi / 2<\theta<\pi / 2\).

Short answer

Question: Show that the inequality \(|\sin \theta| < |\theta|\) holds for \(-π/2 < \theta < π/2\). Answer: For \(-\pi / 2<\theta<0\), the arc \(AB\) is of length \(-\theta\), which is greater than the chord \(|AC|\) since an arc in a unit circle has a larger length than the corresponding chord between the two endpoints. From part (a), we know that \(|AC|=|\sin \theta|\), so \(|\sin \theta| < |-θ|=|\theta|\) for \(-\pi / 2<\theta<0\). A similar argument applies for \(0<\theta<\pi / 2\). Therefore, the inequality \(|\sin \theta| < |\theta|\) holds for \(-π/2 < \theta < π/2\).

Step-by-step solution

a. Show that \(|A C|=|\sin \theta|\) for \(-π/2 < \theta < π/2\)

Consider the value of \(\sin \theta\) in the first quadrant (\(0 < \theta < π/2\)). In this case, \(AC\) represents the vertical opposite side to angle \(\theta\) in a right triangle inscribed in the unit circle. Thus, the length \(|A C|\) is equal to \(\sin \theta\) in this range of values. Now consider the negative values of \(\theta\) in the fourth quadrant (\(-π/2 < \theta < 0\)). In this case, \(AC\) represents the vertical leg of a right triangle inscribed in the unit circle, but now in the direction opposite to angle \(\theta\). We know that \(\sin (-\theta)=-\sin \theta\), and in this scenario, \(|A C|\) is equal to the absolute value of \(\sin \theta\). Therefore, in both cases, \(|AC|\) is equal to \(|\sin \theta|\) for \(-π/2 < \theta < π/2\).

b. Show that \(|\sin \theta| < |\theta|\) for \(-π/2 < \theta < π/2\)

For \(-\pi / 2<\theta<0\), we observe that the arc \(AB\) is of length \(-\theta\). Since an arc in a unit circle has a larger length than the corresponding chord between the two endpoints, we can say that \(|AB| > |AC|\). As we know from part (a): \(|AC|=|\sin \theta|\). We also have: \(|AB|=|-θ|=|\theta|\). Therefore, \(|\sin \theta| < |\theta|\) for \(-\pi / 2<\theta<0\). Similar logic applies for \(0<\theta<\pi / 2\). In this case, the arc \(AB\) is of length \(\theta\). Therefore, \(|AB| > |AC|\), leading to \(|\sin \theta| < |\theta|\) for \(0<\theta<\pi / 2\). Since these inequalities hold in both quadrants, the inequality \(|\sin \theta| < |\theta|\) holds for \(-\pi / 2<\theta<\pi / 2\).

c. Conclude that \(-|\theta| \leq \sin \theta \leq |\theta|\) for \(-π/2 < \theta < π/2\)

Since \(|\sin \theta| < |\theta|\), it directly implies that \(-\vert\theta\vert \leq \sin \theta \leq |\theta|\) for \(-\pi / 2<\theta<\pi / 2\).

d. Show that \(0 \le 1-\cos \theta \le |\theta|\) for \(-π/2 < \theta < π/2\)

Observe that the angle \(\theta\) is subtended by the arc \(AB\) in the unit circle. It's known that the length of the arc \(AB\) of a unit circle is equal to the central angle \(\theta\). Moreover, \(AD\) is the horizontal component, so the length \(|AD|\) is \(\cos \theta\). Notice that lengths \(AB\) and \(BD\) are both positive. Now consider the triangle \(ABD\). By the triangle inequality, it follows that \(AB+BD>AD\). Let us write this inequality in terms of the trigonometric functions and angle \(\theta\). We get \(|\theta| + 1 > \cos\theta\). Rearrange the inequality and we get \(0 < 1 - \cos\theta \le |\theta|\) for \(-\pi / 2<\theta<\pi / 2\).

e. Show that \(1 - |\theta| \le \cos \theta \le 1\) for \(-π/2 < \theta < π/2\)

From part (d), we have \(0 \le 1-\cos \theta \le |\theta|\). Subtract 1 from both sides of the right inequality to get: \(-1 \le -\cos \theta \le -|\theta|\). Now multiply every side of the inequality by -1, which switches the inequality directions. Hence, we get \(1 \ge \cos \theta \ge |\theta|\). We know that \(\cos \theta \le 1\) for all \(\theta\). Thus, we can combine these inequalities and finally conclude that \(1 - |\theta| \le \cos \theta \le 1\) for \(-π/2 < \theta < π/2\).

Key concepts

Unit Circle

The unit circle is a circle with a radius of 1, centered at the origin of a coordinate plane. It's a foundational concept in trigonometry.

By considering angles in the unit circle, we can easily determine trigonometric functions like sine, cosine, and tangent. Key angles and their corresponding coordinates on the unit circle help us understand these functions better. For instance, at an angle \(\theta\) in the first quadrant, the vertical distance from the x-axis is equal to \(\sin \theta\).

When working with trigonometric inequalities, the unit circle is especially helpful for visualizing how different trigonometric values relate to the angle \(\theta\) and understanding the symmetries of sine and cosine in different quadrants.

Trigonometric Functions

Trigonometric functions like sine and cosine relate the angles of a triangle to the lengths of its sides. These functions are mainly defined based on the unit circle.

The sine function, \(\sin \theta\), represents the y-coordinate, while the cosine function, \(\cos \theta\), represents the x-coordinate of a point on the unit circle. For any angle \(\theta\), the values of these functions vary between -1 and 1.

Understanding these functions is crucial when tackling inequalities, as they allow us to convert geometric properties into algebraic expressions. We're able to show relationships, like \(|\sin \theta| < |\theta|\), using these functions, demonstrating how the vertical distance (sine) compares with the arc length (\theta).

Inequalities

Trigonometric inequalities compare the size of trigonometric expressions. They often use properties of trigonometric functions and angles to establish these relationships.

For instance, the inequality \(|\sin \theta| < |\theta|\) suggests that the vertical projection in the unit circle is smaller than the arc's length. This understanding stems from comparing straight-line distances (sine) with curved distances (arc length).

Inequalities are powerful tools to predict how angles relate to trigonometric values. They're key in other areas of mathematics, ensuring balanced expressions and providing solutions within specific ranges.

Right Triangle

A right triangle in trigonometry has one 90-degree angle. It's a fundamental shape for defining sine, cosine, and other trigonometric functions.

When placed in a unit circle, the hypotenuse equals the circle's radius (1). This positions one leg along the horizontal (cosine) and another vertical (sine), making calculations direct and intuitive.

Using right triangles, we explore inequalities by interpreting sides' lengths for angles in different quadrants. This helps us see, for example, why \(|AC| = |\sin \theta|\) as derived in step 1, extending understanding from a geometric to an algebraic perspective.

Quadrants

The coordinate plane divides revolution into four quadrants. Each quadrant affects the sign of trigonometric functions.

- **Quadrant I**: Both sine and cosine are positive.
- **Quadrant II**: Sine is positive, cosine is negative.
- **Quadrant III**: Both sine and cosine are negative.
- **Quadrant IV**: Sine is negative, cosine is positive.

Understanding quadrants is crucial because it directly impacts the outcomes of trigonometric inequalities. For example, when \(-\pi/2 < \theta < 0\), we're in Quadrant IV, where sine is negative. This must be taken into account when proving inequalities like \(-|\theta| \leq \sin \theta \leq |\theta|\). Knowing the quadrant tells us the expected direction and sign of our functions, crucial for validating solutions.