Problem 7 - Precalculus Mathematics for Calculus

Question

Find the exact value of each expression, if it is defined. (a) \(\tan ^{-1}(-1)\) (b) \(\tan ^{-1} \sqrt{3}\) (c) \(\tan ^{-1} \frac{\sqrt{3}}{3}\)

Step-by-Step Solution

Answer

(a) \(-\frac{\pi}{4}\), (b) \(\frac{\pi}{3}\), (c) \(\frac{\pi}{6}\).

Step 1: Understanding Inverse Tangents

The inverse tangent function, denoted as \(\tan^{-1}(x)\), finds the angle whose tangent is \(x\). The range of \(\tan^{-1}\) is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

Step 2: Calculate \(\tan^{-1}(-1)\)

To find \(\tan^{-1}(-1)\), we need an angle in the range \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) whose tangent is -1. The angle is \(-\frac{\pi}{4}\) since \(\tan(-\frac{\pi}{4}) = -1\).

Step 3: Calculate \(\tan^{-1} \sqrt{3}\)

To find \(\tan^{-1} \sqrt{3}\), we look for an angle in the range \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) whose tangent is \(\sqrt{3}\). The angle is \(\frac{\pi}{3}\) because \(\tan(\frac{\pi}{3}) = \sqrt{3}\).

Step 4: Calculate \(\tan^{-1} \frac{\sqrt{3}}{3}\)

For \(\tan^{-1} \frac{\sqrt{3}}{3}\), we need an angle in the range \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) with a tangent of \(\frac{\sqrt{3}}{3}\). This angle is \(\frac{\pi}{6}\), as \(\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\).

Key Concepts

tangent function

The tangent function, often abbreviated as 'tan', is a fundamental concept in trigonometry. It relates an angle within a right triangle to the ratio of the lengths of the opposite side to the adjacent side. This relationship is usually defined on the unit circle for angles ranging from 0 to \(2\pi\). The formula can be represented as \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \\).

• The tangent function repeats every \(\pi\) radians, making it a periodic function.
• It differs from sine and cosine as it can take any real number as a result, extending beyond the typical range of -1 to 1.
• It is undefined at certain points, specifically where the cosine of the angle is zero, as it involves division by zero.

Understanding these properties helps bridge into inverse functions, especially when seeking to find angles given a particular tangent value.

inverse tangent

Inverse trigonometric functions are essential for calculating angles when the value of the trigonometric function (such as tangent) is known. The inverse tangent function, denoted as \(\tan^{-1}(x)\), finds an angle whose tangent is \(x\). It helps us reverse the tangent function under specific conditions.

• The range of the inverse tangent function is constrained to \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), which means it returns angles only within this interval.
• It ensures every outputted angle's tangent results back in the original \(x\) value provided to the inverse function.
• This makes it particularly useful in trigonometry, where solving for unknown angles is necessary.

For example, calculating \( \tan^{-1}(-1) \) means finding an angle within this limited range that results in \( -1 \) when substituted into a tangent function. This is typically \(-\frac{\pi}{4}\), since \( \tan(-\frac{\pi}{4}) = -1\).

angle calculation

Angle calculation with inverse functions like \(\tan^{-1}\) is crucial for applications ranging from geometry to real-world scenarios such as navigation and engineering. Calculating angles involves not just knowing the function but also understanding the typical angle measures within specific quadrants of the unit circle.

• When calculating \( \tan^{-1}(\sqrt{3}) \), you identify an angle \( \theta \) where \( \tan(\theta) = \sqrt{3} \). For \( \tan \theta = \sqrt{3}, \) the angle \( \theta = \frac{\pi}{3} \) is correct because \( \tan(\frac{\pi}{3}) = \sqrt{3} \).
• Similarly, for \( \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) \,\) solve for an angle where \( \tan(\theta) = \frac{\sqrt{3}}{3} \.\) This gives \( \theta = \frac{\pi}{6} \,\) as \( \tan(\frac{\pi}{6}) = \frac{\sqrt{3}}{3} \).
• Using these skills allows you to solve various problems involving angles and their trigonometric functions efficiently.

Mastering angle calculations enriches your understanding of periodicity and the cyclic nature of trigonometric functions, leading to practical math solutions.