Problem 47 - Precalculus with Calculus Previews

Question

Find all angles \(t,\) where \(0 \leq t<\) \(2 \pi,\) that satisfy the given condition. $$ \sin t=0 $$

Step-by-Step Solution

Answer

The angles are \( t = 0 \) and \( t = \pi \).

Step 1: Understanding the Problem

We are asked to find the angles \( t \) in the interval \( 0 \leq t < 2\pi \) for which \( \sin t = 0 \). This problem involves finding when the sine of an angle equals zero.

Step 2: Identifying Key Points

The sine function, \( \sin t \), is zero at integer multiples of \( \pi \). Thus, \( t = n\pi \) for integers \( n \).

Step 3: Considering the Given Range

The problem restricts \( t \) to the interval \( 0 \leq t < 2\pi \). Hence, we must identify all integer multiples of \( \pi \) within this interval.

Step 4: Listing Possible Solutions

Within the interval \( 0 \leq t < 2\pi \), the integer multiples of \( \pi \) are \( 0 \) and \( \pi \). Hence, the solutions are \( t = 0 \) and \( t = \pi \).

Key Concepts

Sine Function

The sine function, represented as \( \sin t \), is a vital component in trigonometry. It is defined based on the coordinates of a point on a unit circle. When you think of the sine function, imagine drawing a right triangle where the angle \( t \) is measured from the origin. Here, sine is the ratio of the length of the opposite side to the hypotenuse.

• It's a periodic function with a cycle of \( 2\pi \), meaning every \( 2\pi \) units, the pattern repeats.
• The sine function oscillates between \(-1\) and \(1\), reaching zero at different points. These zeroes are crucial for solving equations like \( \sin t = 0 \).
• Importantly, sine is symmetric about the origin, which is why it's referred to as an odd function.

For the problem at hand, knowing when \( \sin t = 0 \) reveals the angles, which aren't just theoretical but applicable in waves, physics, and beyond. In the standard interval \( [0, 2\pi) \), these zeroes occur specifically at integer multiples of \( \pi \), such as \( 0 \) and \( \pi \).

Unit Circle

The unit circle is central in understanding trigonometric functions. Imagine a circle with a radius of 1, centered at the origin of a coordinate plane. As you move around this circle, the angle \( t \) dictates your position. Think of this circle as a powerful tool for converting between angles and coordinates.

• Each angle on the unit circle corresponds to a point \((x, y)\) such that \( x^2 + y^2 = 1 \).
• For an angle \( t \), \( x = \cos t \) and \( y = \sin t \), tying the angles to the sine and cosine functions seamlessly.
• Since the circle is based on radius 1, it simplifies complex trigonometric analysis into straightforward geometry.

Understanding the unit circle simplifies many problems, including our own, as you can easily visualize where \( \sin t \) becomes zero. These are aligned perfectly with where the coordinate vertical is 0, which adds clarity to why these angles are multiples of \( \pi \).

Angle Measurement

Angle measurement in trigonometry is often discussed in radians or degrees. A deep grasp of angles helps demystify trigonometric functions and their zeroes.

• Radians are based on the arc length on the unit circle. They are more natural for many mathematical applications compared to degrees.
• One full circle is \(2\pi\) radians, equivalent to 360 degrees.
• Angles such as \(\pi\) and \(2\pi\) are frequently used because they relate directly to half and full rotations.

For our problem, the key interval \(0 \leq t < 2\pi\) uses radian measurement. This is the standard range for one complete circle around the unit circle. Radians provide a direct link between linear and angular understanding, making them indispensable for solving \(\sin t = 0\) with simplicity and insight. Grasping this measurement enables students to confidently work through trigonometric equations and uncover solutions like \(t = 0\) and \(t = \pi\).