Problem 12 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-22: Find the exact value of the expression whenever it is defined. (a) \(\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)\) (b) \(\cos ^{-1}\left(\cos \frac{4 \pi}{3}\right)\) (c) \(\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)\)

Step-by-Step Solution

Answer

(a) \(-\frac{\pi}{3}\), (b) \(\frac{2\pi}{3}\), (c) \(\frac{\pi}{6}\).

Step 1: Understanding Inverse Sine Function

The expression \( \sin^{-1} \left( \sin \frac{2\pi}{3} \right) \) requires us to find the angle \( x \) such that \( \sin(x) = \sin \frac{2\pi}{3} \) and \( -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \). Since \( \frac{2\pi}{3} \) itself is not in this range, we need the angle in the range that has the same sine value.

Step 2: Finding Equivalent Angle for Inverse Sine

The sine of an angle \( \theta \) in the second quadrant, such as \( \frac{2\pi}{3} \), can be reflected into the first quadrant to find a corresponding angle in the principal range of the inverse sine function. The angle \( \frac{2\pi}{3} \) is equivalent to \( \frac{\pi}{3} \) in terms of sine value but lies within \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). Thus, \( \sin^{-1} \left( \sin \frac{2\pi}{3} \right) = \frac{\pi}{3} - \frac{\pi} = -\frac{\pi}{3} \).

Step 3: Understanding Inverse Cosine Function

For \( \cos^{-1} \left( \cos \frac{4\pi}{3} \right) \), we need to find \( x \) such that \( \cos(x) = \cos \frac{4\pi}{3} \) and \( 0 \leq x \leq \pi \). The angle \( \frac{4\pi}{3} \) is not in this range; however, its cosine is equal to that of an angle within the range.

Step 4: Finding Equivalent Angle for Inverse Cosine

Recognize that \( \frac{4\pi}{3} \) is in the third quadrant where cosine values are negative. The equivalent angle in the range \([0, \pi]\) with the same cosine value is \( \pi - \frac{4\pi}{3} + \pi = \frac{2\pi}{3} \). Thus, \( \cos^{-1} \left( \cos \frac{4\pi}{3} \right) = \frac{2\pi}{3} \).

Step 5: Understanding Inverse Tangent Function

For \( \tan^{-1} \left( \tan \frac{7\pi}{6} \right) \), find \( x \) such that \( \tan(x) = \tan \frac{7\pi}{6} \) and \( -\frac{\pi}{2} < x < \frac{\pi}{2} \). Since \( \tan \) has period \( \pi \), we can find an angle equivalent to \( \frac{7\pi}{6} \) within this range.

Step 6: Finding Equivalent Angle for Inverse Tangent

The angle \( \frac{7\pi}{6} \) corresponds to an angle in the third quadrant. By subtracting \( \pi \), we adjust it to an equivalent angle within \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). Therefore, \( \tan^{-1} \left( \tan \frac{7\pi}{6} \right) = \frac{7\pi}{6} - \pi = \frac{\pi}{6} \).

Key Concepts

Inverse Sine

The inverse sine function, denoted as \( \sin^{-1}(x) \), is an important concept when dealing with trigonometric expressions. This function allows you to determine the angle \( y \) such that \( \sin(y) = x \). The domain for \( x \) in \( \sin^{-1}(x) \) is between \(-1\) and \(1\), and the range is restricted to \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \) to ensure it is a proper function.
The challenge with inverse sine is when the original angle given does not fall into this range, as seen in the expression \( \sin^{-1}\left( \sin \frac{2\pi}{3} \right) \). Here, \( \frac{2\pi}{3} \) is outside the principal range, since \( \frac{\pi}{2} < \frac{2\pi}{3} \). Therefore, we need to find an equivalent angle in the principal range that has the same sine value.
For angles in the second quadrant, like \( \frac{2\pi}{3} \), we use symmetry properties of the sine function to locate a corresponding angle in the first quadrant. This results in \( \sin^{-1}\left( \sin \frac{2\pi}{3} \right) = -\frac{\pi}{3} \), ensuring that it stays within the correct range while maintaining the original sine value.

Inverse Cosine

The inverse cosine function, represented as \( \cos^{-1}(x) \), helps us find the angle \( y \) for which \( \cos(y) = x \). The domain for \( \cos^{-1} \) spans from \(-1\) to \(1\) for \( x \), and the range for \( y \) is confined to \( 0 \leq y \leq \pi \). This range encompasses all valid and unique angles with non-negative cosine values.
In solving \( \cos^{-1}\left( \cos \frac{4\pi}{3} \right) \), the angle \( \frac{4\pi}{3} \) originally appears in the third quadrant, which is beyond the specified range. To resolve this, we seek an equivalent angle within \([0, \pi]\). The cosine function's periodicity and symmetry allow us to identify \( \frac{2\pi}{3} \) as the correct angle in range, such that \( \cos^{-1}\left( \cos \frac{4\pi}{3} \right) = \frac{2\pi}{3} \). This ensures that we correctly identify an angle whose cosine matches while agreeing with the function’s range limitations.

Inverse Tangent

The inverse tangent function, or \( \tan^{-1}(x) \), is utilized to discover the angle \( y \) where \( \tan(y) = x \). This function has a domain including all real numbers for \( x \), with the range limited to \( -\frac{\pi}{2} < y < \frac{\pi}{2} \). This specific range makes \( \tan^{-1} \) particularly useful, as it helps deal with any real number inputs and returns angles in a consistent manner.
When evaluating \( \tan^{-1}\left( \tan \frac{7\pi}{6} \right) \), we seek an equivalent angle within \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). The original angle \( \frac{7\pi}{6} \) lands in the third quadrant, exceeding the allowed limits. Because tangent repeats every \( \pi \), subtracting \( \pi \) adjusts \( \frac{7\pi}{6} \) into the principal range, yielding \( \tan^{-1}\left( \tan \frac{7\pi}{6} \right) = \frac{\pi}{6} \). This process aligns with the need for \( \tan^{-1} \) to yield a result within its designated range while keeping the tangent value consistent.