Problem 47 - Algebra 2

Question

For what value(s) of \(t\) in the interval from 0 to 2\(\pi\) is the value of \(f(t)=3 \cos 2 t\) the least?

Step-by-Step Solution

Answer

The minimum value of the function occurs at \(t = \pi/2\).

Step 1: Determine The Value of Function

We know that \(3 \cos 2t\) will take its minimum value of -3 because \(\cos 2t\) varies between -1 and 1 and we're multiplying it by 3. Thus, we set \(3 \cos 2t\) equal to -3 and solve for \(\cos 2t\). This gives us \(\cos 2t = -1\).

Step 2: Solve for t

We now need to solve the equation \(\cos 2t = -1\) for \(t\) in the interval from 0 to \(2\pi\). The solutions of \(\cos \theta = -1\) in this interval occur at \(\pi, 2\pi\). Hence \(2t = \pi, 2\pi\) and therefore, \(t = \pi/2, \pi\). However, according to our predefined interval, we only consider \(t = \pi/2\).

Key Concepts

Cosine Function

The cosine function is a crucial part of trigonometry, often represented as \( \cos(\theta) \). This function gives us the horizontal coordinate of a point on the unit circle at an angle \( \theta \). It's a periodic function, meaning it repeats its values in regular intervals. The basic properties of the cosine function include:

• Domain: All real numbers
• Range: From -1 to 1
• Periodicity: Repeats every \(2\pi\)

In the exercise, we examine a function \( f(t) = 3 \cos 2t \). Here, the 2 inside the cosine argument indicates that the usual period of the cosine function is halved. As \( \cos(2t) \) oscillates, it takes values from -1 to 1. Multiplying by 3 stretches this range to -3 to 3. Understanding these properties helps us easily identify when the function achieves its minimum or maximum values.

Interval Analysis

Interval analysis involves studying functions over a specific domain. In this exercise, we focus on the interval from 0 to \(2\pi\). Looking at a function on a defined interval helps us determine where particular function values occur.

For \( f(t) = 3 \cos 2t \), we focus on when this function reaches its least value. Since \( \cos 2t = -1 \) minimizes \( \cos 2t \), we determine the specific points within our interval where this occurs. For cosine, \( \cos \theta = -1 \) results at \( \pi, 3\pi, \ldots \). However, with a doubled argument in \( 2t \), we see solutions at \( 2t = \pi, 3\pi \), or \( t = \pi/2, \pi \).

Only values fitting our interval are considered, refining to \( t = \pi/2 \). This solution highlights the necessity of interval analysis in trigonometry.

Function Minimization

Function minimization is about finding the smallest value a function can take. This involves both understanding the function itself and analyzing its behavior over a certain range.

In our example, \( 3 \cos 2t \), minimization occurs with \( \cos 2t = -1 \). This yields \( 3 \times (-1) = -3 \). Hence, we aim to solve for when this occurs.

• Solve for \( \cos 2t = -1 \).
• The solutions are \(2t = \pi, 3\pi\).
• For \(t\), this means \(t = \pi/2, \pi\).

By limiting our scope to the interval from 0 to \(2\pi\), we find \( t = \pi/2 \) where the function is minimized. This showcases the analytical value of understanding and applying trigonometric properties to simplify complex problems.