Problem 115 - Algebra and Trigonometry Real Mathematics, Real People

Question

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

Step-by-Step Solution

Answer

The given vector \((\cos \theta) \mathbf{i}+(\sin \theta) \mathbf{j}\) is a unit vector for any value of \(\theta\) because its magnitude is 1.

Step 1: Understand what it means for a vector to be a unit vector

A vector is a unit vector if its magnitude or length is 1. The magnitude or length of a vector \(v=a\mathbf{i}+b\mathbf{j}\) in 2D is given by \(\sqrt{a^2 + b^2}\).

Step 2: Substitute the given vector's components into the formula of magnitude

Substituting the given vector's components into the formula, the length of the vector is \(\sqrt{(\cos \theta)^2 + (\sin \theta)^2}\).

Step 3: Simplify the expression inside the square root

The expression \((\cos \theta)^2 + (\sin \theta)^2\) is equal to 1 based on the Pythagorean Identity in trigonometry which states \(\sin^2 \theta + \cos^2 \theta = 1\). Thus, the magnitude of the vector becomes \(\sqrt{1}\).

Step 4: Produce the final output

\(\sqrt{1}\) is 1, and so the magnitude of the given vector is 1. Therefore, the vector \((\cos \theta) \mathbf{i}+(\sin \theta) \mathbf{j}\) is a unit vector for any value of \(\theta\).

Key Concepts

Magnitude of a Vector

To truly grasp why the vector \((\cos \theta) \mathbf{i}+(\sin \theta) \mathbf{j}\) is a unit vector, it's essential to understand the concept of vector magnitude. The magnitude, often called the "length," measures how long the vector is.
In the context of a 2D vector, \((a\mathbf{i} + b\mathbf{j})\), the magnitude is calculated using the formula \(\sqrt{a^2 + b^2}\). This formula is derived from the distance formula in geometry and tells us how far the vector reaches from the origin point (0,0) on the coordinate plane.
So, the vector \((\cos \theta) \mathbf{i}+(\sin \theta) \mathbf{j}\) is a combination of horizontal and vertical movements based on the sine and cosine of \(\theta\) respectively. By calculating its magnitude, we ensure whether the vector truly is of unit length, which means a magnitude of 1.

Pythagorean Identity

A key player in the proof that our vector is a unit vector is the Pythagorean Identity from trigonometry. This identity succinctly states: \(\sin^2 \theta + \cos^2 \theta = 1\).
What this means is that for any angle \(\theta\), when you square the sine of the angle and add it to the square of the cosine of the same angle, you will always get 1.
This identity is crucial for simplifying expressions involving sine and cosine, especially when proving unit vectors. In our scenario, when we calculated the magnitude of the vector \((\cos \theta) \mathbf{i}+(\sin \theta) \mathbf{j}\), the expression inside the square root \(\cos^2 \theta + \sin^2 \theta\) naturally simplified to 1, thanks to this identity. As a result, the magnitude of the vector became \(\sqrt{1} = 1\), confirming that the vector is a unit vector.

Trigonometric Functions

Trigonometric functions, such as \(\sin\) and \(\cos\), are mathematical functions that relate angles of a triangle to the ratios of its sides. These functions are pivotal in the realm of geometry, but they also extend their utility to various fields including physics, engineering, and computer science.
In a unit circle, which is a circle with a radius of one centered at the origin of the coordinate plane, the cosine of an angle \(\theta\) represents the x-coordinate of the point where the terminal side of the angle intersects the circle. Similarly, the sine of the angle \(\theta\) gives the y-coordinate.
Hence, in the vector \((\cos \theta) \mathbf{i}+(\sin \theta) \mathbf{j}\), \(\cos \theta\) and \(\sin \theta\) are directly utilized from their geometric interpretation on the unit circle, emphasizing that these trigonometric functions are fundamentally linked to unit vectors and help describe them.