Problem 37 - College Trigonometry

Question

For the given vector \(\vec{v}\), find the magnitude \(\|\vec{v}\|\) and an angle \(\theta\) with \(0 \leq \theta<360^{\circ}\) so that \(\vec{v}=\|\vec{v}\|\langle\cos (\theta), \sin (\theta)\rangle\) (See Definition 11.8.) Round approximations to two decimal places. $$ \vec{v}=\left\langle-\frac{1}{2},-\frac{\sqrt{3}}{2}\right\rangle $$

Step-by-Step Solution

Answer

The magnitude is 1 and \( \theta = 240^\circ \).

Step 1: Find the magnitude of the vector

The magnitude of a vector \( \vec{v} = \langle a, b \rangle \) is given by \[\|\vec{v}\| = \sqrt{a^2 + b^2}\]Here, \( a = -\frac{1}{2} \) and \( b = -\frac{\sqrt{3}}{2} \). Substitute these values:\[\|\vec{v}\| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1\]Hence, the magnitude \( \|\vec{v}\| \) is 1.

Step 2: Write vector relation

Now, we consider the standard form for the vector using magnitude and trigonometric components: \[\vec{v} = \|\vec{v}\| \langle \cos(\theta), \sin(\theta) \rangle\]Which becomes:\[\langle -\frac{1}{2}, -\frac{\sqrt{3}}{2} \rangle = 1 \cdot \langle \cos(\theta), \sin(\theta) \rangle\]

Step 3: Determine \(\theta\) from \(\cos(\theta)\) and \(\sin(\theta)\)

From:\[\cos(\theta) = -\frac{1}{2}\]and \[\sin(\theta) = -\frac{\sqrt{3}}{2}\]These conditions correspond to the trigonometric values on the unit circle for \(\theta = 240^\circ\) in standard position, as this is in the third quadrant where both cosine and sine are negative. Thus, \(\theta = 240^\circ\).

Key Concepts

Trigonometric Functions

Trigonometric functions are fundamental in vector analysis. They connect angles with ratios derived from right-angled triangles. A primary operation is determining the cosine and sine of an angle. These represent the horizontal and vertical components of a unit circle point, respectively.
For a given angle \( \theta \),

• \( \cos(\theta) \) gives the x-coordinate of a point on the unit circle.
• \( \sin(\theta) \) gives the y-coordinate.

In our exercise, the vector \( \vec{v} = \langle -\frac{1}{2}, -\frac{\sqrt{3}}{2} \rangle \) was expressed in terms of these functions. With \( \cos(\theta) = -\frac{1}{2} \) and \( \sin(\theta) = -\frac{\sqrt{3}}{2} \), it corresponds to angles where the trigonometric functions produce negative results. This specifically indicates that the angle is located in the third quadrant of the unit circle, precisely at \( 240^{\circ} \).
Using trigonometric functions makes connecting vector components with their directional angles intuitive and systematic.

Magnitude of a Vector

The magnitude of a vector provides a measure of its length or size, without regard to its direction. For a vector \( \vec{v} = \langle a, b \rangle \), the magnitude \( \|\vec{v}\| \) is calculated using the Pythagorean theorem:
\[\|\vec{v}\| = \sqrt{a^2 + b^2}\]
In the exercise, the vector \( \vec{v} = \langle -\frac{1}{2}, -\frac{\sqrt{3}}{2} \rangle \) has a magnitude calculated as:\[\|\vec{v}\| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1\]
Understanding the magnitude allows us to know how far a vector reaches in space, and when normalized (as is in this example), it simplifies directly to the direction represented by angles and trigonometric values.

Unit Circle

The unit circle is a circle centered at the origin \((0,0)\) in the Cartesian coordinate plane with a radius of 1. It's significant in analyzing trigonometric functions as every point \((x, y)\) on the unit circle satisfies the equation \(x^2 + y^2 = 1\).

• It's used to define the trigonometric functions sine and cosine, where the angle \( \theta \) represents a rotation.
• The coordinates \( (\cos(\theta), \sin(\theta)) \) are associated with any angle \( \theta \) as a point on this circle.

In the context of the exercise, by translating \( \vec{v} = \langle -\frac{1}{2}, -\frac{\sqrt{3}}{2} \rangle \) to the unit circle, we identify the angle \(\theta = 240^{\circ}\). This angle ensures that \( \, \cos(\theta) = -\frac{1}{2} \, \), and \( \, \sin(\theta) = -\frac{\sqrt{3}}{2} \, \) are both accurately reflected as coordinates on the unit circle.
The unit circle simplifies understanding the relationship between angular measures and the position of vectors, serving as a bridge between algebraic and geometric interpretations of trigonometric functions.