Problem 39 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 39-42: Prove that the statement is true for every positive integer \(n\). \(\sin (\theta+n \pi)=(-1)^{n} \sin \theta\)

Step-by-Step Solution

Answer

The identity is proven: \( \sin (\theta+n \pi)=(-1)^{n} \sin \theta \) for all positive integers \( n \).

Step 1: Understand the Problem

We need to prove the identity \( \sin (\theta+n \pi)=(-1)^{n} \sin \theta \) for all positive integers \( n \). This means for each \( n \), the left side of the equation should equal the right side.

Step 2: Use Sine Addition Formula

Recall that \( \sin(a+b) = \sin a \cos b + \cos a \sin b \). Apply this formula to \( \sin (\theta + n\pi) \), where \( a = \theta \) and \( b = n\pi \), gives us \( \sin (\theta + n\pi) = \sin \theta \cos(n\pi) + \cos \theta \sin(n\pi) \).

Step 3: Evaluate Cosine and Sine Terms

Since \( \cos(n\pi) = (-1)^n \) and \( \sin(n\pi) = 0 \) for any integer \( n \), we substitute these values into the formula: \( \sin \theta \cos(n\pi) = \sin \theta (-1)^n \) and \( \cos \theta \sin(n\pi) = \cos \theta (0) = 0 \). Therefore, \( \sin(\theta + n\pi) = \sin \theta (-1)^n \).

Step 4: Conclude the Proof

Simplifying the expression, we get \( \sin (\theta + n\pi) = (-1)^n \sin \theta \), which matches our original statement. Thus, the identity is proven for all positive integers \( n \).

Key Concepts

Understanding the Sine Function

The sine function is a fundamental concept in trigonometry. It represents the ratio of the side of a right triangle opposite the angle to the hypotenuse. The sine function is commonly denoted as \( \sin \theta \), where \( \theta \) represents the angle.
The sine function is periodic, meaning it repeats its values in a regular cycle. The period of the sine function is \( 2\pi \), meaning that the function repeats its pattern every \( 2\pi \) radians.
This periodic property is crucial in proving trigonometric identities and allows us to understand relationships between angles and their associated sine values.

Developing the Mathematical Proof

A mathematical proof involves convincing logic and reasoning to show that a particular statement or identity holds true. In this case, we are required to prove the identity \( \sin (\theta+n \pi)=(-1)^{n} \sin \theta \) for all positive integer values of \( n \).
We approach this proof by using the sine addition formula: \( \sin(a+b) = \sin a \cos b + \cos a \sin b \). By applying this formula to \( \sin (\theta+n \pi) \), we rewrite it in terms of known functions of \( n\pi \): \( \sin(\theta + n\pi) = \sin \theta \cos(n\pi) + \cos \theta \sin(n\pi) \).
Recognizing that \( \cos(n\pi) = (-1)^n \) and \( \sin(n\pi) = 0 \) for any integer \( n \), the expression reduces to \( \sin(\theta + n\pi) = \sin \theta (-1)^n \), thus proving our original identity. This simplification confirms the identity, showing that our proof holds for any positive integer \( n \).

Exploring Integer Values in Trigonometry

When working with trigonometric identities, integer values, particularly those involving multiples of \( \pi \), play a significant role. For an angle \( n\pi \), where \( n \) is an integer, specific trigonometric functions like cosine and sine take on consistent, predictable values.
The cosine of \( n\pi \), \( \cos(n\pi) \), results in either \( 1 \) or \( -1 \) depending on whether \( n \) is even or odd. This aligns with the property: \( \cos(n\pi) = (-1)^n \).
On the other hand, the sine of \( n\pi \), \( \sin(n\pi) \), is always \( 0 \) for any integer \( n \). These predictable outcomes are what allow certain identities, like our proven statement \( \sin(\theta + n\pi) = (-1)^n \sin \theta \), to be simplified and validated. Understanding these consistent integer-based results is critical in solving and proving trigonometric equations.