Problem 44
Question
Is there only one way to evaluate \(\cos \left(\frac{5 \pi}{4}\right)\) ? Explain how to set up the solution in two different ways, and then compute to make sure they give the same answer.
Step-by-Step Solution
Verified Answer
Two ways: using reference angles and trigonometric identities, both result in \(-\frac{\sqrt{2}}{2}\).
1Step 1: Understanding the Angle
First, we need to understand that \( \frac{5\pi}{4} \) is an angle in the unit circle. Specifically, it is located in the third quadrant where both sine and cosine are negative. This angle is equivalent to \( 225^\circ \). It can be represented as \( \pi + \frac{\pi}{4} \).
2Step 2: Using Reference Angles
We approach the calculation using reference angles. The reference angle for \( \frac{5\pi}{4} \) is \( \frac{\pi}{4} \), or \( 45^\circ \). We know \( \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \). Since \( \frac{5\pi}{4} \) is in the third quadrant, \( \cos \left( \frac{5\pi}{4} \right) \) is negative: \( \cos \left( \frac{5\pi}{4} \right) = -\frac{\sqrt{2}}{2} \).
3Step 3: Using Trigonometric Identities
Another method to find \( \cos \left( \frac{5\pi}{4} \right) \) involves using the trigonometric identity for cosine: \( \cos(\pi + \theta) = -\cos(\theta) \). Here, \( \theta = \frac{\pi}{4} \), so \( \cos \left( \frac{5\pi}{4} \right) = -\cos \left( \frac{\pi}{4} \right) = -\frac{\sqrt{2}}{2} \).
4Step 4: Verification
Both methods used the properties of the unit circle and trigonometric identities. In both situations, we calculated that \( \cos \left( \frac{5\pi}{4} \right) = -\frac{\sqrt{2}}{2} \). Thus, both approaches give the same result, verifying the consistent evaluation of this cosine value.
Key Concepts
Reference AnglesCosineUnit CircleTrigonometric Identities
Reference Angles
Reference angles are crucial for simplifying trigonometric evaluations. They help us determine the actual angle's trigonometric values by referring to an equivalent angle in the first quadrant.
To find the reference angle, we usually subtract the actual angle from the closest x-axis boundary (either 0, \(\pi\), or \(2\pi\)). These angles, ranging from 0 to \(\pi/2\), help in identifying the sine, cosine, and other trigonometric values based on known first-quadrant values.
For example, the reference angle for \(\frac{5\pi}{4}\) is \(\frac{\pi}{4}\). Though the angle is in the third quadrant, its reference angle \(\frac{\pi}{4}\) is what guides us to evaluate its trigonometric functions.
To find the reference angle, we usually subtract the actual angle from the closest x-axis boundary (either 0, \(\pi\), or \(2\pi\)). These angles, ranging from 0 to \(\pi/2\), help in identifying the sine, cosine, and other trigonometric values based on known first-quadrant values.
For example, the reference angle for \(\frac{5\pi}{4}\) is \(\frac{\pi}{4}\). Though the angle is in the third quadrant, its reference angle \(\frac{\pi}{4}\) is what guides us to evaluate its trigonometric functions.
Cosine
Cosine is one of the fundamental trigonometric functions. It represents the x-coordinate of points on the unit circle.
Its value can be derived using the angle and its reference angle, especially when dealing with the unit circle. This function varies between -1 and 1 as it revolves through the quadrants.
In our given angle, \(\frac{5\pi}{4}\), the cosine is negative because it lies in the third quadrant. Using the reference angle, we find that \(\cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}\).
Its value can be derived using the angle and its reference angle, especially when dealing with the unit circle. This function varies between -1 and 1 as it revolves through the quadrants.
In our given angle, \(\frac{5\pi}{4}\), the cosine is negative because it lies in the third quadrant. Using the reference angle, we find that \(\cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}\).
Unit Circle
The unit circle is a powerful tool for understanding and visualizing trigonometric functions. It's a circle with a radius of 1, centered at the origin of a coordinate plane.
Each angle drawn from the origin to the circle's perimeter provides corresponding sine and cosine values, determined by the coordinates of the adjacent point on the circle.
This circle is particularly useful for analyzing the periodic nature of trigonometric functions.
- In the third quadrant, angles result in negative sine and cosine values.
- The unit circle helps us see that \(\frac{5\pi}{4}\) corresponds to the coordinates \(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\). Thus, confirming \(\cos(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2}\).
Each angle drawn from the origin to the circle's perimeter provides corresponding sine and cosine values, determined by the coordinates of the adjacent point on the circle.
This circle is particularly useful for analyzing the periodic nature of trigonometric functions.
- In the third quadrant, angles result in negative sine and cosine values.
- The unit circle helps us see that \(\frac{5\pi}{4}\) corresponds to the coordinates \(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\). Thus, confirming \(\cos(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2}\).
Trigonometric Identities
Trigonometric identities are equations that hold true for angles. They simplify and relate different trigonometric functions.
These identities are essential for calculating values across different quadrants and solving complex problems.
- One example is \(\cos(\pi + \theta) = -\cos(\theta)\). This reflects on symmetry and transformation about the origin.
Applied to our problem, with \(\theta = \frac{\pi}{4}\), it confirms that \(\cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}\). Emphasizing how transformations keep trigonometric evaluations consistent.
These identities are essential for calculating values across different quadrants and solving complex problems.
- One example is \(\cos(\pi + \theta) = -\cos(\theta)\). This reflects on symmetry and transformation about the origin.
Applied to our problem, with \(\theta = \frac{\pi}{4}\), it confirms that \(\cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}\). Emphasizing how transformations keep trigonometric evaluations consistent.
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