Problem 111 - Precalculus with Limits

Question

PROOF Prove that \((\cos\ \theta)\mathbf{i} + (\sin\ \theta)\mathbf{j}\) is a unit vector for any value of \(\theta\).

Step-by-Step Solution

Answer

The proof shows that the vector \((\cos\ \theta)\mathbf{i} + (\sin\ \theta)\mathbf{j}\) always has a magnitude of 1, which is the definition of a unit vector. Therefore, it's a unit vector for any value of \(\theta\).

Step 1: Identify the formula for the magnitude of a vector

The magnitude of a vector \(a\mathbf{i} + b\mathbf{j}\) in 2D is given by \(\sqrt{a^2 + b^2}\). So we need to apply this formula to our vector \((\cos\ \theta)\mathbf{i} + (\sin\ \theta)\mathbf{j}\) to find its magnitude.

Step 2: Substitute into the formula

Substitute \(a = \cos \theta\) and \(b = \sin \theta\) into the formula for the magnitude of a vector. So the magnitude is \(\sqrt{(\cos^2 \theta) + (\sin^2 \theta)}\).

Step 3: Simplify the expression using the Pythagorean identity

From trigonometry, we know that \(\cos^2 \theta + \sin^2 \theta = 1\), which is called the Pythagorean identity. Substituting this into our expression for the magnitude, we get \(\sqrt{1}\).

Step 4: Finalize the proof

Since \(\sqrt{1} = 1\), it has been demonstrated that the magnitude of the vector \((\cos\ \theta)\mathbf{i} + (\sin\ \theta)\mathbf{j}\) is 1. This means it's a unit vector for any value of \(\theta\).

Key Concepts

Magnitude of a Vector

Understanding the magnitude of a vector is crucial in this proof. A vector can be thought of as an arrow pointing in a specific direction with a certain length. The length of this arrow is what we call the magnitude. In mathematical terms, for a vector in two dimensions such as \(a\mathbf{i} + b\mathbf{j}\), its magnitude \( \|\mathbf{v}\| \) is calculated using the formula:

• \( \|\mathbf{v}\| = \sqrt{a^2 + b^2} \)

This formula resembles the Pythagorean theorem, as it calculates the hypotenuse of a right triangle formed by the vector components. Using this understanding, when evaluating the vector \( ( \cos \theta ) \mathbf{i} + ( \sin \theta ) \mathbf{j} \), you're essentially determining how long the resulting vector is based on its cosine and sine components. By using the formula, we find that its magnitude plays a substantial role in defining whether the vector is a unit vector – a vector with a magnitude of exactly 1.

Pythagorean Identity

The Pythagorean identity is a fundamental relationship in trigonometry and is pivotal to proving that the given vector is a unit vector. It states:

• \( \cos^2 \theta + \sin^2 \theta = 1 \)

This equation holds true for any angle \(\theta\). It reflects the inherent properties of the sine and cosine functions related to a right triangle or the unit circle. In our proof, when we substitute the sine and cosine components of the vector into the magnitude formula, we end up with \( \sqrt{ \cos^2 \theta + \sin^2 \theta } \).Thanks to the Pythagorean identity, this simplifies to \( \sqrt{1} \), confirming that the magnitude is indeed 1. This step is crucial because it verifies that no matter the angle \(\theta\), the vector maintains a consistent magnitude, affirming its status as a unit vector.

Trigonometric Functions

Trigonometric functions like sine and cosine are core components in understanding vector magnitudes, especially in this context. These functions relate angles to side lengths in right triangles or coordinate positions in the unit circle, serving as the foundation for calculating various properties of vectors.

• \( \cos \theta \): Represents the adjacent side over hypotenuse in a right triangle.
• \( \sin \theta \): Represents the opposite side over hypotenuse in a right triangle.

In our proof, these functions are used as the coefficients of the standard basis vectors \( \mathbf{i} \) and \( \mathbf{j} \), creating a vector that still points in a unit circle path, thanks to its consistent magnitude. By utilizing their properties and understanding their behavior around the unit circle, it becomes evident how they contribute to a vector being a unit vector. This recognition is crucial when determining vector properties and understanding how different values of \(\theta\) influence direction without affecting the overall magnitude.