Problem 41 - Algebra and Trigonometry

Question

Two direction angles of a vector are given. Find the third direction angle, given that it is either obtuse or acute as indicated. (In Exercises 43 and 44, round your answers to the nearest degree.) $$ \alpha=\frac{\pi}{3}, \quad \gamma=\frac{2 \pi}{3} ; \quad \beta \text { is acute } $$

Step-by-Step Solution

Answer

\( \beta = 45 \) degrees.

Step 1: Understanding the Concept of Direction Angles

Direction angles of a vector \( \alpha, \beta, \gamma \) are related to the direction cosines \( \cos \alpha, \cos \beta, \cos \gamma \). By the identity for direction cosines, we have: \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \).

Step 2: Calculate Cosines of Given Angles

Given \( \alpha = \frac{\pi}{3} \) and \( \gamma = \frac{2\pi}{3} \), convert these angles to their cosines. Therefore, \( \cos \alpha = \cos \frac{\pi}{3} = \frac{1}{2} \) and \( \cos \gamma = \cos \frac{2\pi}{3} = -\frac{1}{2} \).

Step 3: Solve for \( \cos \beta \)

Substitute the known values into the identity: \( \cos^2 \frac{\pi}{3} + \cos^2 \beta + \cos^2 \frac{2\pi}{3} = 1 \). Since \( \cos \frac{\pi}{3} = \frac{1}{2} \) and \( \cos \frac{2\pi}{3} = -\frac{1}{2} \), their squares are both \( \frac{1}{4} \). It follows that: \( \frac{1}{4} + \cos^2 \beta + \frac{1}{4} = 1 \), simplifying to \( \cos^2 \beta = \frac{1}{2} \).

Step 4: Determine the Correct Value of \( \cos \beta \)

Since \( \beta \) is acute, \( \cos \beta \) must be positive. Therefore, \( \cos \beta = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2} \).

Step 5: Calculate \( \beta \)

To find \( \beta \), take the inverse cosine: \( \beta = \cos^{-1}\left(\frac{\sqrt{2}}{2}\right) \), which results in \( \beta = \frac{\pi}{4} \) or 45 degrees.

Key Concepts

Direction Cosines

The direction cosines are crucial when discussing direction angles of a vector. If you picture a vector in a three-dimensional space, it forms certain angles with the axes of this space. These angles are called direction angles and are denoted as \(\alpha\), \(\beta\), and \(\gamma\) for the x, y, and z axes respectively.

• Each direction angle has a corresponding direction cosine, which is simply the cosine of the direction angle. This is represented by \(\cos\alpha\), \(\cos\beta\), and \(\cos\gamma\).
• An important identity that these cosines satisfy is: \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\).

This equation is essential because it helps ensure that the vector remains unit. It's like ensuring the total 'energy' along all three axes doesn't exceed what we have. For the problem at hand, directional cosines provide a framework to find the known and unknown angles corresponding to given cosines.

Inverse Trigonometric Functions

Inverse trigonometric functions help us retrieve angles from ratios of sides, which is essential when dealing with direction cosines.

• The function \(\cos^{-1}(x)\), also known as the arccosine, gives us an angle whose cosine is \(x\).
• Because cosine is a periodic function, when taking its inverse to find angles, you may get multiple possibilities. Usually, the principal value (a specific angle) is chosen.

In the context of direction angles, using the inverse cosine function allows us to take the known direction cosine and find the specific angle that it corresponds to. For example, if \(\cos \beta = \frac{\sqrt{2}}{2}\), then using \(\cos^{-1}\), we find that \(\beta = \frac{\pi}{4}\) or 45 degrees. This works because \(\cos\) of an angle in radians matches what trigonometric tables or calculators provide for such cases.

Acute Angles

An acute angle is an angle that is less than 90 degrees, or equivalently, less than \(\frac{\pi}{2}\) radians.

• When solving vector problems or any that involve directional angles, knowing the nature (acute, obtuse, right) of an angle can significantly influence which solutions are valid.
• In the given problem, once the cosine of \(\beta\) was found, it was crucial to notice that \(\beta\) needed to be acute, which eliminated potential angles greater than 90 degrees.

Thus, recognizing the constraint that \(\beta\) is acute guided us towards the correct solution. So even though \(\cos \beta = \frac{\sqrt{2}}{2}\) could refer to different angles (like \(\frac{3\pi}{4}\)), the fact that it was stated as acute narrowed it down to \(\frac{\pi}{4}\). This knowledge is pivotal in ensuring a precise understanding of the problem and its constraints.