Problem 32 - 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. $$ \vec{v}=\langle 1, \sqrt{3}\rangle $$

Step-by-Step Solution

Answer

The magnitude is 2 and the angle \(\theta\) is \(60^{\circ}\).

Step 1: Find the Magnitude of the Vector

The magnitude of a vector \( \vec{v} = \langle x, y \rangle \) is given by the formula: \[ \|\vec{v}\| = \sqrt{x^2 + y^2} \]. Substitute \( x = 1 \) and \( y = \sqrt{3} \) into this formula: \[ \|\vec{v}\| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]. Therefore, the magnitude \( \|\vec{v}\| = 2 \).

Step 2: Express Vector in Terms of Magnitude and Unit Vector

Given that \( \vec{v} = \|\vec{v}\| \langle \cos(\theta), \sin(\theta) \rangle \), we have \( \vec{v} = 2 \langle \cos(\theta), \sin(\theta) \rangle \). This implies \( 2\cos(\theta) = 1 \) and \( 2\sin(\theta) = \sqrt{3} \). Convert this into equations for \( \cos(\theta) \) and \( \sin(\theta) \): \( \cos(\theta) = \frac{1}{2} \) and \( \sin(\theta) = \frac{\sqrt{3}}{2} \).

Step 3: Determine the Angle \(\theta\)

The values \( \cos(\theta) = \frac{1}{2} \) and \( \sin(\theta) = \frac{\sqrt{3}}{2} \) are standard trigonometric ratios for the angle \( 60^{\circ} \) (or \( \frac{\pi}{3} \) radians). We check this angle measure is within the desired range \( 0 \leq \theta < 360^{\circ} \). Thus, \( \theta = 60^{\circ} \).

Key Concepts

Trigonometric Functions

Trigonometric functions are essential tools in mathematics, especially when dealing with vectors and angles. They help us relate the angles of a triangle to its side lengths.
In the context of vectors, trigonometric functions such as sine and cosine are used to express the vector components in terms of an angle and a magnitude. For a vector \( \vec{v} = \langle x, y \rangle \), the components can be represented using the angle \( \theta \) with formulas:

• \( x = \|\vec{v}\| \cos(\theta) \)
• \( y = \|\vec{v}\| \sin(\theta) \)

These relationships are based on the unit circle and the right triangle definitions of sine and cosine.
Utilizing trigonometric functions simplifies the process of finding unknowns such as angles, by comparing known values with standard trigonometric ratios.

Unit Vector

A unit vector is a vector with a magnitude of 1, and it points in the direction of the vector it represents. Unit vectors are essential because they allow us to express vectors in terms of direction alone, separated from magnitude.
For any vector \( \vec{v} \), the unit vector \( \hat{v} \) pointing in the same direction can be found by dividing each component of \( \vec{v} \) by its magnitude \( \|\vec{v}\| \). Mathematically, this is expressed as:

• \( \hat{v} = \frac{1}{\|\vec{v}\|} \cdot \vec{v} \)

Using unit vectors, we can express any vector \( \vec{v} \) in terms of its magnitude and direction: \( \vec{v} = \|\vec{v}\| \cdot \hat{v} \).
This concept is key when solving problems like the one from the exercise, where we express a vector using its direction determined by trigonometric functions.

Angle Determination

Determining the angle \( \theta \) of a vector involves using trigonometric ratios to identify the vector's direction with respect to the origin. Given a vector \( \vec{v} = \langle x, y \rangle \), we use the formulas:

• \( \cos(\theta) = \frac{x}{\|\vec{v}\|} \)
• \( \sin(\theta) = \frac{y}{\|\vec{v}\|} \)

To find \( \theta \), recognize or calculate these ratios and match them to known trigonometric angles.
In our original problem, \( \cos(\theta) = \frac{1}{2} \) and \( \sin(\theta) = \frac{\sqrt{3}}{2} \), correspond to the angle \( 60^{\circ} \), a standard angle in trigonometry. Thus, determining angles often involves recognizing these common ratios for key angles, making memorization of standard trigonometric values very useful.