This tutorial provides detailed solutions to inverse trigonometric functions problems. Exercises with answers are included. We first review key theorems and properties of inverse trigonometric functions.
\( y = \arcsin x \ \Leftrightarrow\ \sin y = x \)
with \(\ -1 \le x \le 1 \ \) and \(\ -\frac{\pi}{2} \le y \le \frac{\pi}{2} \).
\( y = \arccos x \ \Leftrightarrow\ \cos y = x \)
with \(\ -1 \le x \le 1 \ \) and \(\ 0 \le y \le \pi \).
\( y = \arctan x \ \Leftrightarrow\ \tan y = x \)
with \(\ -\frac{\pi}{2} < y < \frac{\pi}{2} \).
Evaluate the following:
1. \( \arcsin\left(-\frac{\sqrt{3}}{2}\right) \)
Let \( y = \arcsin\left(-\frac{\sqrt{3}}{2}\right) \). By Theorem 1:
\[ \sin y = -\frac{\sqrt{3}}{2}, \quad -\frac{\pi}{2} \le y \le \frac{\pi}{2} \]From special angles, \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \).
Using \( \sin(-x) = -\sin x \):
\[ \sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \]Comparing with \( \sin y = -\frac{\sqrt{3}}{2} \), we conclude:
\[ y = -\frac{\pi}{3} \]2. \( \arctan\left(-1\right) \)
Let \( y = \arctan\left(-1\right) \). By Theorem 3:
\[ \tan y = -1, \quad -\frac{\pi}{2} < y < \frac{\pi}{2} \]From special angles, \( \tan\left(\frac{\pi}{4}\right) = 1 \).
Using \( \tan(-x) = -\tan x \):
\[ \tan\left(-\frac{\pi}{4}\right) = -1 \]Comparing with \( \tan y = -1 \), we obtain:
\[ y = -\frac{\pi}{4} \]3. \( \arccos\left(-\frac{1}{2}\right) \)
Let \( y = \arccos\left(-\frac{1}{2}\right) \). By Theorem 2:
\[ \cos y = -\frac{1}{2}, \quad 0 \le y \le \pi \]From special angles, \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \).
Using \( \cos(\pi - x) = -\cos x \):
\[ \cos\left(\pi - \frac{\pi}{3}\right) = -\frac{1}{2} \]Comparing with \( \cos y = -\frac{1}{2} \), we obtain:
\[ y = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]Let \( z = \cos(\arcsin x) \) and \( y = \arcsin x \), so \( z = \cos y \).
By Theorem 1, \( y = \arcsin x \) is equivalent to:
\[ \sin y = x, \quad -\frac{\pi}{2} \le y \le \frac{\pi}{2} \]Using the Pythagorean identity:
\[ \sin^2 y + \cos^2 y = 1 \]Substitute \( \sin y = x \) and solve for \( \cos y \):
\[ \cos y = \pm \sqrt{1 - x^2} \]Since \( -\frac{\pi}{2} \le y \le \frac{\pi}{2} \), \( \cos y \) is non-negative.
\[ z = \cos(\arcsin x) = \sqrt{1 - x^2} \]Let \( z = \csc(\arctan x) \) and \( y = \arctan x \), so \( z = \csc y = \frac{1}{\sin y} \).
By Theorem 3, \( y = \arctan x \) is equivalent to:
\[ \tan y = x, \quad -\frac{\pi}{2} < y < \frac{\pi}{2} \]We know:
\[ \tan^2 y = \frac{\sin^2 y}{\cos^2 y} = \frac{\sin^2 y}{1 - \sin^2 y} \]Solve for \( \sin y \):
\[ \sin y = \pm \sqrt{ \frac{\tan^2 y}{1 + \tan^2 y} } \]For \( -\frac{\pi}{2} < y < \frac{\pi}{2} \), \( \sin y \) and \( \tan y \) have the same sign. Therefore:
\[ \sin y = \frac{\tan y}{\sqrt{1 + \tan^2 y}} \]Substitute \( \tan y = x \):
\[ \sin y = \frac{x}{\sqrt{1 + x^2}} \]Thus:
\[ z = \csc(\arctan x) = \frac{1}{\sin y} = \frac{\sqrt{1 + x^2}}{x} \]Evaluate the following:
1. \( \arcsin\left( \sin \left( \frac{7\pi}{4} \right) \right) \)
Recall: \( \arcsin(\sin y) = y \) only for \( -\frac{\pi}{2} \le y \le \frac{\pi}{2} \).
Since \( \sin\left(\frac{7\pi}{4}\right) = \sin\left(-\frac{\pi}{4}\right) \), and \( -\frac{\pi}{4} \) is in the required range:
\[ \arcsin\left( \sin \left( \frac{7\pi}{4} \right) \right) = \arcsin\left( \sin \left( -\frac{\pi}{4} \right) \right) = -\frac{\pi}{4} \]2. \( \arccos\left( \cos \left( \frac{4\pi}{3} \right) \right) \)
Recall: \( \arccos(\cos y) = y \) only for \( 0 \le y \le \pi \).
Since \( \cos\left(\frac{4\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right) \), and \( \frac{2\pi}{3} \) is in the required range:
\[ \arccos\left( \cos \left( \frac{4\pi}{3} \right) \right) = \arccos\left( \cos \left( \frac{2\pi}{3} \right) \right) = \frac{2\pi}{3} \]