Problem 17 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 17-20: Use a formula for negatives to find the exact value. (a) \(\sin \left(-90^{\circ}\right)\) (b) \(\cos \left(-\frac{3 \pi}{4}\right)\) (c) \(\tan \left(-45^{\circ}\right)\)

Step-by-Step Solution

Answer

(a) -1, (b) -\frac{\sqrt{2}}{2}, (c) -1

Step 1: Understanding the Problem

We need to find the exact values of trigonometric functions for negative angles using formulas for negative angles. For sine, cosine, and tangent, we'll use these properties: \( \sin(-a) = -\sin(a) \), \( \cos(-a) = \cos(a) \), and \( \tan(-a) = -\tan(a) \).

Step 1: Solving \( \sin(-90^{\circ}) \)

According to the negative angle identity for sine, \( \sin(-a) = -\sin(a) \). Therefore, \( \sin(-90^{\circ}) = -\sin(90^{\circ}) \). Since \( \sin(90^{\circ}) = 1 \), it follows that \( \sin(-90^{\circ}) = -1 \).

Step 2: Solving \( \cos(-\frac{3 \pi}{4}) \)

The cosine function has the property \( \cos(-a) = \cos(a) \). Thus, \( \cos(-\frac{3 \pi}{4}) = \cos(\frac{3 \pi}{4}) \). In the unit circle, \( \cos(\frac{3 \pi}{4}) = -\frac{\sqrt{2}}{2} \). Therefore, \( \cos(-\frac{3 \pi}{4}) = -\frac{\sqrt{2}}{2} \).

Step 3: Solving \( \tan(-45^{\circ}) \)

According to the negative angle identity for tangent, \( \tan(-a) = -\tan(a) \). Therefore, \( \tan(-45^{\circ}) = -\tan(45^{\circ}) \). Since \( \tan(45^{\circ}) = 1 \), it follows that \( \tan(-45^{\circ}) = -1 \).

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental tools in trigonometry, which allow us to understand relationships between different trigonometric functions. These identities are particularly useful when dealing with negative angles. Different trigonometric functions have specific rules regarding negative angles:

• For sine: \( \sin(-a) = -\sin(a) \)
• For cosine: \( \cos(-a) = \cos(a) \)
• For tangent: \( \tan(-a) = -\tan(a) \)

These formulas demonstrate how the trigonometric functions of a negative angle relate to their positive counterparts. The sine and tangent functions change their signs, while cosine remains unchanged. Understanding these identities helps simplify the calculation and analysis of trigonometric expressions, especially when working with angles measured in radians or degrees.

Sine

The sine function, represented as \( \sin(\theta) \), is one of the primary trigonometric functions. It describes the y-coordinate of a point on the unit circle corresponding to an angle \( \theta \). When dealing with negative angles, especially using negative angle identities, it becomes clear how the sine function behaves:

• The identity \( \sin(-a) = -\sin(a) \) indicates that sine takes an opposite sign when the angle is negative.
• For example, to solve \( \sin(-90^{\circ}) \), it can be rewritten as \( -\sin(90^{\circ}) \).

Knowing that \( \sin(90^{\circ}) = 1 \), it follows that \( \sin(-90^{\circ}) = -1 \). This behavior is always consistent, making it easier to evaluate sine for any negative angle using these identities.

Cosine

The cosine function, expressed as \( \cos(\theta) \), refers to the x-coordinate of a point on the unit circle at an angle \( \theta \). Unlike sine and tangent, cosine retains its value for both positive and negative angles:

• The identity \( \cos(-a) = \cos(a) \) shows that cosine of any angle is the same whether the angle is negative or positive.
• For example, to find \( \cos(-\frac{3\pi}{4}) \), we simply use \( \cos(\frac{3\pi}{4}) \).

From the unit circle, \( \cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2} \). Hence, \( \cos(-\frac{3\pi}{4}) \) also results in \( -\frac{\sqrt{2}}{2} \). This consistency in the cosine function greatly simplifies computations involving negative angles.

Tangent

The tangent function, \( \tan(\theta) \), is defined as the ratio of the sine and cosine of a given angle \( \theta \). Just like sine, the tangent function takes on an opposite sign when dealing with negative angles:

• The identity used here is \( \tan(-a) = -\tan(a) \).
• For instance, calculating \( \tan(-45^{\circ}) \) involves re-expressing it as \( -\tan(45^{\circ}) \).

Since \( \tan(45^{\circ}) = 1 \), it means \( \tan(-45^{\circ}) = -1 \). Understanding this identity is crucial as it helps in simplifying expressions and solving problems that involve negative angles in tangent.