Problem 30 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-38: Find all solutions of the equation. $$ (\sin t-1) \cos t=0 $$

Step-by-Step Solution

Answer

The solutions are \( t = \frac{\pi}{2} + 2m\pi \) where \( m \) is an integer.

Step 1: Identifying the Zero Product

The equation \[ ( ext{sin } t - 1) ext{ cos } t = 0 \] is a product of two factors set to zero. This means we need to identify when each factor is zero. We'll solve for the conditions where \( ext{sin } t - 1 = 0 \) and \( ext{cos } t = 0 \).

Step 2: Solving \( ext{sin } t - 1 = 0 \)

For \( ext{sin } t - 1 = 0 \), we have:\[ ext{sin } t = 1 \]The sine function equals 1 at angles:\[ t = rac{ ext{π}}{2} + 2k ext{π}, \]where \( k \) is any integer.

Step 3: Solving \( ext{cos } t = 0 \)

For \( ext{cos } t = 0 \), we need:\[ t = rac{ ext{π}}{2} + n ext{π}, \]where \( n \) is any integer. This occurs at odd multiples of \( rac{ ext{π}}{2} \).

Step 4: Combining Solutions

Combine solutions from both conditions, considering both conditions simultaneously:1. From \( ext{sin } t - 1 = 0 \): - \( t = rac{ ext{π}}{2} + 2k ext{π} \)2. From \( ext{cos } t = 0 \): - At \( t = rac{ ext{π}}{2} + n ext{π} \), but this includes odd multiples of \( rac{ ext{π}}{2} \), so they intersect only at: - \( t = rac{ ext{π}}{2} + 2m ext{π}, \)where \( m \) is any integer. Both solutions overlap and represent aligned solutions.

Key Concepts

Sine Function

The sine function is an essential concept in trigonometry. It describes the relationship between the angles and sides of a right triangle. In terms of the unit circle, the sine of an angle \( t \) is the y-coordinate of the point on the circle corresponding to that angle. The function \( \sin t \) ranges between -1 and 1 for all real numbers.

The equation \( \sin t = 1 \) is satisfied when the angle \( t \) corresponds to the maximum value of the sine function. This occurs at angles like \( \frac{\pi}{2} \) and any equivalent angles which can be described as \( \frac{\pi}{2} + 2k\pi \), where \( k \) is an integer. This pattern arises due to the periodic nature of the sine function, repeating every \( 2\pi \).

Understanding where the sine function hits its maximum values gives insight into solving trigonometric equations, such as our original equation involving \( \sin t - 1 = 0 \).

Cosine Function

The cosine function, like the sine function, is a fundamental trigonometric function that expresses the ratio of the adjacent side to the hypotenuse in a right triangle. It is also the x-coordinate of a point on the unit circle at angle \( t \). The cosine function’s range is also between -1 and 1.

In our equation, we had to solve for when \( \cos t = 0 \). This happens when the angle \( t \) points to where the x-coordinate of the circle is zero. These are odd multiples of \( \frac{\pi}{2} \), such as \( \frac{\pi}{2} + n\pi \). The cosine function reaches zero at these positions due to its periodicity with a period of \( \pi \).

Identifying these zeros of the cosine function can help narrow down solutions in trigonometric equations where the zero product property is applied.

Zero Product Property

The zero product property is a base principle encountered in algebra and used in this trigonometric equation as well. It states that if a product of two expressions equals zero, at least one of the expressions must be zero.

In the equation \( (\sin t - 1) \cos t = 0 \), this means either \( \sin t - 1 = 0 \) or \( \cos t = 0 \). Solving these separately gives conditions where the product as whole is zero, allowing us to find solutions easily. This strategy simplifies complex trigonometric equations by breaking them into more manageable pieces.

Use the zero product property with trigonometric functions to systematically solve equations by addressing each factor individually.

Solutions of Trigonometric Equations

Solving trigonometric equations hinges on identifying all values that satisfy an equation within the set functions. Beyond simple algebraic manipulations, solving these requires knowledge of the characteristics of trigonometric functions.

For the original equation \( (\sin t - 1) \cos t = 0 \), solutions involve finding all \( t \) that satisfy either part of the factored equation and understanding where their conditions overlap.

• From \( \sin t = 1 \), solutions are \( t = \frac{\pi}{2} + 2k\pi \).
• From \( \cos t = 0 \), solutions are at odd multiples of \( \frac{\pi}{2} + n\pi \).
• Where these two conditions align is \( t = \frac{\pi}{2} + 2m\pi \), capturing the intersection precisely.

The solutions often need to consider both the periodic nature of the sine and cosine functions and how their cycles relate. The overlap indicates the correct solutions that satisfy the equation, demonstrating the importance of understanding the behavior and periodicity of these functions.

Mastering how to find these solutions involves practice with the properties of sine, cosine, and the zero product property.