Problem 67
Question
(a) Find the values of \(\alpha \in[0,2 \pi)\) that satisfy \(\cos \alpha=-\frac{1}{2} \sqrt{3}\). (b) Find the values of \(\alpha \in[0,2 \pi)\) that satisfy \(\tan \alpha=\frac{1}{\sqrt{3}}\).
Step-by-Step Solution
Verified Answer
(a) None; (b) \( \alpha = \frac{\pi}{6}, \frac{7\pi}{6} \).
1Step 1: Understanding the Range
We are looking for the angles \( \alpha \) such that \( \alpha \) lies within the interval \([0, 2\pi)\). This means the angle should be in radians, starting from 0 and less than \(2\pi\).
2Step 2: Solving \( \cos \alpha = -\frac{1}{2} \sqrt{3} \)
Check the interval and known cosine values. The standard cosine values are \( \cos 30^\circ = \cos(\pi/6) = \sqrt{3}/2 \). Therefore, \(\cos \alpha = -\frac{1}{2} \sqrt{3}\) is incorrect as it doesn't match known cosine values. Double-check the formula to proceed.
3Step 3: Solving \( \tan \alpha = \frac{1}{\sqrt{3}} \)
Recognize that \( \tan(\pi/6) = \frac{1}{\sqrt{3}} \). The tangent function is periodic with period \( \pi \), so possible angles are \( \alpha = \frac{\pi}{6} + n\pi \) for \(n\in\mathbb{Z}\).
4Step 4: Applying Range to \( \tan \alpha = \frac{1}{\sqrt{3}} \)
Since we need \( \alpha \in [0, 2\pi) \), calculate the values:\( \alpha = \frac{\pi}{6} + 0\cdot\pi = \frac{\pi}{6} \) and \( \alpha = \frac{\pi}{6} + 1\cdot\pi = \frac{7\pi}{6} \). Both values lie in the desired range.
Key Concepts
Cosine FunctionTangent FunctionTrigonometric IdentitiesAngle in Radians
Cosine Function
The cosine function is one of the basic trigonometric functions, often denoted as \( \cos \alpha \). It represents the x-coordinate of a point on the unit circle corresponding to an angle \( \alpha \). This function is periodic with a period of \( 2\pi \), meaning it repeats its values every \( 2\pi \) radians.
For an angle \( \alpha \), the cosine function returns a value between -1 and 1. Commonly, cosine values for standard angles are often used, such as \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \) or \( \cos \frac{\pi}{3} = \frac{1}{2} \). In this exercise, we are focused on finding values of \( \alpha \) where the cosine equals specific values.
To discern correct cosine values, always cross-check against known standard angles. If a calculated cosine value such as \(-\frac{1}{2} \sqrt{3}\) doesn't fit the standard values, you may need to reassess your approach to ensure the integrity of your solution.
For an angle \( \alpha \), the cosine function returns a value between -1 and 1. Commonly, cosine values for standard angles are often used, such as \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \) or \( \cos \frac{\pi}{3} = \frac{1}{2} \). In this exercise, we are focused on finding values of \( \alpha \) where the cosine equals specific values.
To discern correct cosine values, always cross-check against known standard angles. If a calculated cosine value such as \(-\frac{1}{2} \sqrt{3}\) doesn't fit the standard values, you may need to reassess your approach to ensure the integrity of your solution.
Tangent Function
The tangent function, denoted by \( \tan \alpha \), is another fundamental trigonometric function. It is defined as the ratio of the sine and cosine of a given angle, so \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \). This function is periodic with a period of \( \pi \), meaning it repeats its values every \( \pi \) radians.
The tangent function has unique points where its value is undefined, specifically whenever the cosine is zero. For example, \( \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \), which is a typical known value used in this exercise. It is essential to use the periodic nature of the tangent function to find all possible angles \( \alpha \) that satisfy the equation in a given range. In our case, \( \alpha = \frac{\pi}{6} \) and \( \alpha = \frac{7\pi}{6} \) both satisfy the condition because \( \tan(\alpha + n\pi) = \tan \alpha \) with \( n \) as an integer.
The tangent function has unique points where its value is undefined, specifically whenever the cosine is zero. For example, \( \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \), which is a typical known value used in this exercise. It is essential to use the periodic nature of the tangent function to find all possible angles \( \alpha \) that satisfy the equation in a given range. In our case, \( \alpha = \frac{\pi}{6} \) and \( \alpha = \frac{7\pi}{6} \) both satisfy the condition because \( \tan(\alpha + n\pi) = \tan \alpha \) with \( n \) as an integer.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are true for all possible angles in their domains. These identities are useful for simplifying trigonometric equations and can help in solving complex trigonometric problems.
Some common trigonometric identities include:
Some common trigonometric identities include:
- Pythagorean Identity: \( \sin^2\alpha + \cos^2\alpha = 1 \)
- Angle Sum and Difference Identities: e.g., \( \cos(a \pm b) = \cos a \cos b \mp \sin a \sin b \)
- Double Angle Identities: e.g., \( \cos 2\alpha = \cos^2\alpha - \sin^2\alpha \)
Angle in Radians
The concept of radians is vital in trigonometry, as it provides a natural measure of angles as the ratio of the arc length to the radius of a circle. Unlike degrees, radians relate directly to arc lengths, making them a preferred unit in mathematical computations.
One complete revolution around a circle is \( 2\pi \) radians, which corresponds to 360 degrees. Therefore, one radian is approximately equal to \( 57.3 \) degrees. This conversion factor helps to understand angles in contexts where the unit used is degrees.
When dealing with trigonometric equations, such as the ones in the exercise, understanding the interval \([0, 2\pi)\) is crucial. This interval defines the range of angles you'll consider for solutions, ensuring that potential solutions are within one full rotation around the circle. Using radians allows for easier manipulation of periodic functions, especially in scenarios involving calculus and higher mathematics.
One complete revolution around a circle is \( 2\pi \) radians, which corresponds to 360 degrees. Therefore, one radian is approximately equal to \( 57.3 \) degrees. This conversion factor helps to understand angles in contexts where the unit used is degrees.
When dealing with trigonometric equations, such as the ones in the exercise, understanding the interval \([0, 2\pi)\) is crucial. This interval defines the range of angles you'll consider for solutions, ensuring that potential solutions are within one full rotation around the circle. Using radians allows for easier manipulation of periodic functions, especially in scenarios involving calculus and higher mathematics.
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