Problem 70 - Contemporary Precalculus

Question

Here is proof that the sine function has period \(2 \pi .\) We saw in the text that \(\sin (t+2 \pi)=\sin t\) for every \(t .\) We must show that there is no positive number smaller than \(2 \pi\) with this property. Do this as follows: (a) Find a number \(t\) such that \(\sin (t+\pi) \neq \sin t\) (b) Find all numbers \(k\) such that \(0

Step-by-Step Solution

Answer

Question: Prove that the sine function has a period of 2π. Answer: Using the step-by-step solution, we showed that there is no positive number k less than 2π with the property that sin(t+k)=sin(t) for every number t. As a result, we can conclude that the sine function has a period of 2π.

Step 1: (a) Find a number t such that \(\sin(t+\pi) \neq \sin t\)

We can choose any number t within the range of 0 to \(2\pi\). For example, let us choose \(t=\frac{\pi}{2}.\) Therefore, we have: \(\sin(\frac{\pi}{2} + \pi) = \sin\frac{3\pi}{2} = -1\) and \(\sin(\frac{\pi}{2}) = 1\) Since \(\sin(t+\pi) \neq \sin t\), we have successfully found a number t such that these values are not equal.

Step 2: (b) Find all numbers k where \(0<k<2\pi\) and \(\sin(k)=0\)

By using the definition of sine and drawing a unit circle, we can determine when the sine function is equal to 0. This occurs when the y-coordinate of the point on the unit circle is equal to 0: - at \(k = \pi\) - at \(k = 2\pi\) - and any integer multiple of pi (n\(\pi\)) However, we are looking for values of k between 0 and \(2\pi\), so the only valid value for k in this range is \(\pi\).

Step 3: (c) Suppose k is a number such that \(\sin(t+k)=\sin(t)\) for every number \(t\). Show that \(\sin(k)=0\)

Let's consider \(t=0\) for the sine function property \(\sin(t+k)=\sin(t)\). We obtain: \(\sin(0+k) = \sin(0)\) \(\sin(k) = 0\) This shows that the sine of k must be equal to 0, as suggested.

Step 4: (d) Use parts (a)-(c) to show that there is no positive number k less than \(2\pi\) with the property that \(\sin(t+k)=\sin t\) for every number t

In part (b), we found that the only value of k where the sine is 0 and k is between 0 and \(2\pi\) is \(k=\pi\). However, in part (a), we showed that there is at least one number t such that \(\sin(t+\pi)\neq\sin(t)\). This means that our initial assumption that there is a positive number k less than \(2\pi\) with the property that \(\sin(t+k)=\sin(t)\) for every number t is false. Since we have shown that there is no positive number k less than \(2\pi\) with the property that \(\sin(t+k)=\sin(t)\) for every number t, we can conclude that the sine function has a period of \(2\pi\).

Key Concepts

Sine Function

The sine function, or \(\sin(t)\), is a fundamental trigonometric function. It oscillates between -1 and 1. When you look at the sine of a number \(t\), it helps describe the y-coordinate of a point on the unit circle. This function is periodic, meaning it repeats its values in regular intervals.
In practical terms, whenever you increment the angle \(t\) by full rotations of a circle (e.g., by \(2\pi\)), the sine value remains unchanged. This mirrors the repeating circular path traced on the unit circle.

Key aspects of the sine function include:

• The function is continuous and smooth.
• Defined for all real numbers.
• Mainly used in wave patterns and harmonic oscillations.

Understanding the sine function's behavior is crucial for solving trigonometric equations and modeling cyclical phenomena.

Unit Circle

The unit circle is a perfect circle with a radius of one unit centered at the origin of a coordinate plane. It's a vital tool in trigonometry for defining the sine, cosine, and other trigonometric functions. By "unit circle," we emphasize the use of a circle with radius 1 to simplify trigonometric calculations and relationships.
Each point \((x, y)\) on the unit circle corresponds to an angle \(t\) from the positive x-axis, with \(x = \cos(t)\) and \(y = \sin(t)\). This foundational concept allows easy identification of sine values as the y-coordinates.

Some critical points on the unit circle are:

• At \(t = 0\) or \(2\pi\), the point is \((1,0)\).
• At \(t = \frac{\pi}{2}\), the point is \((0,1)\).
• At \(t = \pi\), the point is \((-1,0)\).
• At \(t = \frac{3\pi}{2}\), the point is \((0,-1)\).

Using the unit circle, we can visualize how the sine function is structured, alongside understanding its symmetries and periodicity.

Trigonometric Identities

Trigonometric identities are equations that hold true for every value of the variable involved. They play a significant role when working with trigonometric functions like sine, helping simplify expressions and solve equations quickly.

For the sine function, these identities prove useful:

• \(\sin(-t) = -\sin(t)\): This shows sine is an odd function.
• \(\sin(t + 2\pi) = \sin(t)\): This identity confirms the periodicity of sine.
• \(\sin^2(t) + \cos^2(t) = 1\): This is Pythagorean identity, highlighting the relationship with cosine.

To solve the problems involving periodicity or symmetry, these identities offer a solid foundation. By applying these trigonometric identities, we effectively reduce complex equations to a more manageable form.

Period of Sine

The period of the sine function is the interval over which the function completes one full cycle before starting again. For the sine function, this period is \(2\pi\). This means after every \(2\pi\) interval, the sine function values will repeat.
In the original exercise solution, determining if \(\sin(t+2\pi) = \sin(t)\) holds was central. This confirmed that \(t+2\pi\) reached the same point on the unit circle, proving \(2\pi\) is the minimal repeating duration.

Important observations about sine's period include:

• No positive number smaller than \(2\pi\) satisfies \(\sin(t+k) = \sin(t)\) for all \(t\).
• The function shows symmetry about its period, reflecting its wave-like nature.

Understanding the sine function’s period helps in modeling continuous periodic functions, assisting in applications ranging from physics to engineering.