Inverse Trigonometric Functions: Step-by-Step Solved Problems & Examples

Let \( y = \arcsin\left(-\dfrac{\sqrt{3}}{2}\right) \). By Theorem 1:

\[ \sin y = -\dfrac{\sqrt{3}}{2}, \quad -\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2} \]

From special angles, we know \( \sin\left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{2} \).

Using the odd function property \( \sin(-x) = -\sin x \):

\[ \sin\left(-\dfrac{\pi}{3}\right) = -\dfrac{\sqrt{3}}{2} \]

Comparing with \( \sin y = -\dfrac{\sqrt{3}}{2} \), we conclude:

\[ y = -\dfrac{\pi}{3} \]

Let \( y = \arctan\left(-1\right) \). By Theorem 3:

\[ \tan y = -1, \quad -\dfrac{\pi}{2} < y < \dfrac{\pi}{2} \]

From special angles, \( \tan\left(\dfrac{\pi}{4}\right) = 1 \).

Using the odd function property \( \tan(-x) = -\tan x \):

\[ \tan\left(-\dfrac{\pi}{4}\right) = -1 \]

Comparing with \( \tan y = -1 \), we obtain:

\[ y = -\dfrac{\pi}{4} \]

Let \( y = \arccos\left(-\dfrac{1}{2}\right) \). By Theorem 2:

\[ \cos y = -\dfrac{1}{2}, \quad 0 \le y \le \pi \]

From special angles, \( \cos\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2} \).

Using the supplement identity \( \cos(\pi - x) = -\cos x \):

\[ \cos\left(\pi - \dfrac{\pi}{3}\right) = -\dfrac{1}{2} \]

Comparing with \( \cos y = -\dfrac{1}{2} \), we obtain:

\[ y = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3} \]

Let \( z = \cos(\arcsin x) \) and \( y = \arcsin x \), so \( z = \cos y \).

By Theorem 1, \( y = \arcsin x \) is equivalent to:

\[ \sin y = x, \quad -\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2} \]

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 \( -\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2} \), \( \cos y \) is non-negative:

\[ z = \cos(\arcsin x) = \sqrt{1 - x^2} \]

Let \( z = \csc(\arctan x) \) and \( y = \arctan x \), so \( z = \csc y = \dfrac{1}{\sin y} \).

By Theorem 3, \( y = \arctan x \) is equivalent to:

\[ \tan y = x, \quad -\dfrac{\pi}{2} < y < \dfrac{\pi}{2} \]

Using trigonometric relations:

\[ \tan^2 y = \dfrac{\sin^2 y}{\cos^2 y} = \dfrac{\sin^2 y}{1 - \sin^2 y} \]

Solve for \( \sin y \):

\[ \sin y = \pm \sqrt{ \dfrac{\tan^2 y}{1 + \tan^2 y} } \]

For \( -\dfrac{\pi}{2} < y < \dfrac{\pi}{2} \), \( \sin y \) and \( \tan y \) share the same sign. Therefore:

\[ \sin y = \dfrac{\tan y}{\sqrt{1 + \tan^2 y}} \]

Substitute \( \tan y = x \):

\[ \sin y = \dfrac{x}{\sqrt{1 + x^2}} \]

Thus:

\[ z = \csc(\arctan x) = \dfrac{1}{\sin y} = \dfrac{\sqrt{1 + x^2}}{x} \]

1. \( \arcsin\left( \sin \left( \dfrac{7\pi}{4} \right) \right) \)

Recall that \( \arcsin(\sin y) = y \) only for \( -\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2} \).

Since \( \sin\left(\dfrac{7\pi}{4}\right) = \sin\left(-\dfrac{\pi}{4}\right) \), and \( -\dfrac{\pi}{4} \) falls within the required range:

\[ \arcsin\left( \sin \left( \dfrac{7\pi}{4} \right) \right) = \arcsin\left( \sin \left( -\dfrac{\pi}{4} \right) \right) = -\dfrac{\pi}{4} \]

2. \( \arccos\left( \cos \left( \dfrac{4\pi}{3} \right) \right) \)

Recall that \( \arccos(\cos y) = y \) only for \( 0 \le y \le \pi \).

Since \( \cos\left(\dfrac{4\pi}{3}\right) = \cos\left(\dfrac{2\pi}{3}\right) \), and \( \dfrac{2\pi}{3} \) is in the required interval:

\[ \arccos\left( \cos \left( \dfrac{4\pi}{3} \right) \right) = \arccos\left( \cos \left( \dfrac{2\pi}{3} \right) \right) = \dfrac{2\pi}{3} \]