Problem 36 - 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. $$ \vec{v}=\left\langle-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right\rangle $$

Step-by-Step Solution

Answer

The magnitude of \( \vec{v} \) is 1, and the angle \( \theta \) is \( 225^{\circ} \).

Step 1: Calculate Magnitude

To find the magnitude of the vector \( \vec{v} \), use the formula for the magnitude of a vector: \( \|\vec{v}\| = \sqrt{x^2 + y^2} \), where \( \vec{v} = \left\langle x, y \right\rangle \). Substitute the components of \( \vec{v} = \left\langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle \). The calculation is \[ \|\vec{v}\| = \sqrt{\left(-\frac{\sqrt{2}}{2}\right)^2 + \left(-\frac{\sqrt{2}}{2}\right)^2} = \sqrt{\frac{2}{4} + \frac{2}{4}} = \sqrt{1} = 1. \] Thus, the magnitude is 1.

Step 2: Determine the Reference Angle

To determine the angle \( \theta \), find the reference angle \( \theta_0 \) in the first quadrant using inverse trigonometric functions. Here, you can use either sine or cosine because \( \cos(\theta_0) = \sin(\theta_0) = \frac{\sqrt{2}}{2} \). Therefore, \( \theta_0 = 45^{\circ} \).

Step 3: Determine the Correct Quadrant

Given \( \vec{v} = \left\langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle \), both sine and cosine components are negative, indicating that \( \theta \) is in the third quadrant. Therefore, \( \theta = 180^{\circ} + \theta_0 = 180^{\circ} + 45^{\circ} = 225^{\circ} \).

Step 4: Verify the Vector Representation

Verify that the vector \( \vec{v} \) can be represented as the product of its magnitude and its direction: \[ \vec{v} = \| \vec{v} \| \langle \cos(\theta), \sin(\theta) \rangle = 1 \times \langle \cos(225^{\circ}), \sin(225^{\circ}) \rangle = \langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \rangle. \] This confirms that the calculations for \( \theta \) and \( \|\vec{v}\| \) are correct.

Key Concepts

Angle Determination

Finding the angle of a vector is crucial for understanding its direction. The angle tells us how far the vector is rotated from the reference axis, usually the positive x-axis for 2D vectors.
To determine the angle, we start by finding the "reference angle," which is the angle in the first quadrant. It's calculated using the trigonometric functions sine (\(\sin\)) and cosine (\(\cos\)). With vectors that have components \(\left\langle x, y \right\rangle\), the reference angle is found by:

• Using inverse trigonometric functions like \(\cos^{-1}\) or \(\sin^{-1}\).
• Ensuring that the reference value fits the known \(\sin\) and \(\cos\) relationships, such as using \(\cos(\theta_0) = \frac{\sqrt{2}}{2}\) for angles like \(45^{\circ}\).

However, determining the vector's true angle \(\theta\) requires looking at which quadrant the vector lies in. If both the x and y components are negative, the angle belongs in the third quadrant. That's why, in this exercise, \(\theta = 225^{\circ}\) was found by adding \(180^{\circ}\) to the reference angle \(\theta_0 = 45^{\circ}\). This yields \(\theta = 180^{\circ} + 45^{\circ} = 225^{\circ}\).
Understanding the quadrant-specific adjustments is key for correctly classifying the angle of any given vector.

Trigonometric Functions

Trigonometric functions, namely sine and cosine, are essential when dealing with vectors. They help convert between magnitudes and directional angles. These functions relate the components of a vector to the angle it makes with a reference axis.
The cosine of an angle in a vector usually relates to the x-component, and the sine to the y-component. This relationship helps us break down vectors or reconstruct them from their magnitudes and angles. For example, if \(\vec{v} = \|\vec{v}\|\langle \cos(\theta), \sin(\theta) \rangle\), each vector's direction can be retrieved using trigonometric functions:

• \(\cos(\theta) = x/\|\vec{v}\|\), representing the x-component, is determined by dividing the x-value by the vector's magnitude.
• \(\sin(\theta) = y/\|\vec{v}\|\), representing the y-component, is derived similarly.

For vectors such as \(\left\langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle\), using these functions confirms correct angles like \(225^{\circ}\) when considering the relevant quadrant. These concepts ensure consistent and accurate vector analysis and representation.

Vector Representation

Vectors are represented by components or through their magnitude and direction, providing flexibility in how they are described.
When representing vectors using magnitude and direction, we often utilize the trigonometric format \(\vec{v} = \|\vec{v}\|\langle \cos(\theta), \sin(\theta) \rangle\). This format is helpful to visualize them in geometry or physics. The magnitude \(\|\vec{v}\|\) describes the vector's strength or length, while \(\theta\) specifies its direction.
This dual description allows vectors to be easily manipulated in calculations, especially when changing reference frames or combining multiple vectors. Components can be recalculated anytime if the angle and magnitude are known and vice versa. For example, the vector \(\left\langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle\) can be represented according to its magnitude (which is 1) and direction by combining the angle \(225^{\circ}\) with trigonometric values. This versatile representation is key to solving problems across disciplines, from physics to computer graphics.