Problem 215 - Calculus Volume 1

Question

For the following exercises, evaluate the functions. Give the exact value. $$ \tan ^{-1}\left(\tan \left(-\frac{\pi}{6}\right)\right) $$

Step-by-Step Solution

Answer

The exact value is \( -\frac{\pi}{6} \).

Step 1: Understanding the problem

We need to evaluate the expression \( \tan^{-1}\left(\tan\left(-\frac{\pi}{6}\right)\right) \). This involves finding the tangent of \( -\frac{\pi}{6} \) and then applying the inverse tangent function to the result.

Step 2: Find \( \tan\left(-\frac{\pi}{6}\right) \)

The tangent of an angle is defined as the ratio of the sine to the cosine of that angle. Knowing the unit circle values, \( \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \) and \( \cos\left(-\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \). Thus, \( \tan\left(-\frac{\pi}{6}\right) = \frac{\sin\left(-\frac{\pi}{6}\right)}{\cos\left(-\frac{\pi}{6}\right)} = \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} \).

Step 3: Apply the inverse tangent function

Now, we need to find \( \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) \). The inverse tangent function \( \tan^{-1} \) or \( \arctan \) will give us an angle \( \theta \) such that \( \tan(\theta) = -\frac{1}{\sqrt{3}} \). In the range of \( \tan^{-1} \), which is \((-\frac{\pi}{2}, \frac{\pi}{2})\), this angle is \( -\frac{\pi}{6} \).

Step 4: Conclusion

Therefore, \( \tan^{-1}\left(\tan\left(-\frac{\pi}{6}\right)\right) = -\frac{\pi}{6} \). This is because the range of the inverse tangent function ensures the output will be within \((-\frac{\pi}{2}, \frac{\pi}{2})\), and \( -\frac{\pi}{6} \) is within this range.

Key Concepts

Unit Circle

The unit circle is a crucial concept in understanding trigonometric functions. It is a circle with a radius of 1, centered at the origin of a coordinate plane. The unit circle allows us to define the sine, cosine, and tangent of an angle using the coordinates of points on the circle.

- The angle is measured from the positive x-axis, moving counter-clockwise.- The sine of an angle is the y-coordinate of its corresponding point.- The cosine of an angle is the x-coordinate of its corresponding point.- For tangent, we take the ratio of the sine to the cosine.In this exercise, the unit circle helps us determine that:- The sine of \(-\frac{\pi}{6}\) is \(-\frac{1}{2}\).- The cosine of \(-\frac{\pi}{6}\) is \(\frac{\sqrt{3}}{2}\).Using these values, we can easily find the tangent by calculating \( \frac{\sin(-\frac{\pi}{6})}{\cos(-\frac{\pi}{6})} \). These calculations are foundational to solving problems involving inverse trigonometric functions.

Tangent Function

The tangent function is a trigonometric function represented as \( \tan(\theta) \). It is defined as the ratio of the sine to the cosine of an angle \( \theta \). Mathematically, it is expressed as:\[\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\]The tangent function is periodic, meaning it repeats its values in intervals of \( \pi \). This periodic nature is important when evaluating expressions involving tangent functions.A unique feature of the tangent is its undefined points at angles where the cosine is zero, such as \( \frac{\pi}{2} \) and \( -\frac{\pi}{2} \). These points are asymptotes on the graph of the tangent function. In this exercise, we used the tangent function to calculate \( \tan\left(-\frac{\pi}{6}\right) \), which led to understanding its role in inverse trigonometric calculations.

Arctan

The inverse tangent function, often called "arctan," is represented as \( \tan^{-1}(x) \) or \( \arctan(x) \). It serves as the inverse operation of the tangent function. Given a value of \( x \), the arctan finds the angle \( \theta \) such that \( \tan(\theta) = x \).The range of the inverse tangent function is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), which ensures unique solutions for all real numbers. Within this range, for a given \( x \), \( \arctan(x) \) returns an angle \( \theta \).

- When \( x \) is positive, \( \theta \) will be positive.- When \( x \) is negative, \( \theta \) will be negative.In the exercise we tackled, applying \( \tan^{-1} \) to \( -\frac{1}{\sqrt{3}} \) gave us the precise angle \( -\frac{\pi}{6} \), perfectly illustrating how the inverse tangent function works within its specified range.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values of the variables involved. They are vital tools in simplifying and solving trigonometric problems.

Some fundamental identities include:

• Reciprocal identities, such as \( \tan(\theta) = \frac{1}{\cot(\theta)} \)
• Pythagorean identities, like \( \sin^2(\theta) + \cos^2(\theta) = 1 \)
• Angle sum and difference identities

In our exercise, we relied on the identity \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \) to find the tangent of \( -\frac{\pi}{6} \). This identity demonstrated the dependency of tangent on both sine and cosine, helping us navigate through the inverse operation with \( \arctan \). By understanding these identities, simplifying complex expressions or evaluating trigonometric functions becomes more manageable and intuitive.