Problem 44 - Algebra 2

Question

\(\begin{array}{l}{\text { How many solutions does the equation } 1=-\sin 2 t \text { have for } 0 \leq t<2 \pi ?} \\ {\text { A. } 1} & {\text { B. } 2} & {\text { C. } 3} & {\text { D. } 4}\end{array}\)

Step-by-Step Solution

Answer

B. 2

Step 1: Analyze the sine function

Firstly, the period of the \(\sin(2t)\) function is actually \(\pi\), because the '2' in front of \(t\) halves the usual period for the sine function, which is \(2\pi\). In one period of the sine function, there is one point where it is equal to -1.

Step 2: Calculate the number of periods

Considering the interval from 0 to \(2\pi\), the \(\sin(2t)\) function completes two full periods. This is because the period of the \(\sin(2t)\) function is \(\pi\), so in the interval from 0 to \(2\pi\), there would be \(2 \div 1 = 2\) periods of the function.

Step 3: Calculate the number of solutions

Since there is one place in every period of the function where \(\sin(2t)\) equals -1, and there are two such periods, there are \(1 * 2 = 2\) solutions.

Key Concepts

Period of Trigonometric Functions

When studying trigonometric functions, the concept of "period" is fundamental. The period of a function is the length of the interval over which the function completes one full cycle. For the basic sine function \(\sin(t)\), the standard period is \(2\pi\). This means that every \(2\pi\) units, the sine function repeats its pattern.

However, if you encounter a function like \(\sin(2t)\), the "2" in front of the \(t\) indicates the frequency has changed. This causes the sine wave to complete its cycle twice as fast. So, the new period becomes half of \(2\pi\), which is \(\pi\).

• Original period of \(\sin(t)\): \(2\pi\)
• Adjusted period for \(\sin(2t)\): \(\pi\)

For any function of the form \(\sin(kt)\): Divide the usual period, \(2\pi\), by \(k\) to find the new period.

Sine Function

The sine function, \(\sin(t)\), is one of the most fundamental building blocks in trigonometry. It represents a smooth, periodic oscillation that varies between -1 and 1. Its graph is a wave that repeats every \(2\pi\) radians, starting at 0, peaking at 1, dropping to -1, and returning to 0.

The equation \(y = -\sin(2t)\) involves a negative sign which flips the graph over the horizontal axis. This inverts the amplitude values, making peaks into troughs. The \(2t\) changes the frequency, as discussed, creating a tighter wave pattern that repeats every \(\pi\).

• Standard sine wave attributes: peaks at 1, troughs at -1
• Inverted sine wave attributes: peaks at -1, troughs at 1

The key point is where the function equals -1, as required in the provided equation.

Number of Solutions

Finding how many solutions an equation like \(1 = -\sin(2t)\) has within a specific interval is all about understanding where the sine function equals a particular value. Here, it involves determining where \(-\sin(2t) = -1\), which simplifies to \(\sin(2t) = 1\).

Given that the period of \(\sin(2t)\) is \(\pi\), we know the function completes two full cycles over \([0, 2\pi)\). In each period of \(\sin(2t)\), it reaches the value of -1 once. By evaluating these points for each cycle:

• 1 solution for \(\sin(2t) = -1\) per period
• 2 periods from \(0 \, ext{to} \, 2\pi\)
• Altogether, the solutions total to 2 times the occurrence per period.
  Understanding intervals and repeating cycles is vital to solving such equations accurately in trigonometry.