Problem 22 - Precalculus: Functions and Graphs

Question

Find the exact value of the expression whenever It is defined. (a) \(\sin \left[\frac{1}{2} \cos ^{-1}\left(-\frac{3}{5}\right)\right]\) (b) \(\cos \left(\frac{1}{2} \sin ^{-1} \frac{12}{13}\right)\) (c) \(\tan \left(\frac{1}{2} \tan ^{-1} \frac{40}{9}\right)\)

Step-by-Step Solution

Answer

(a) \(\frac{2\sqrt{5}}{5}\), (b) \(\frac{3\sqrt{13}}{13}\), (c) \(\frac{4}{5}\)."

Step 1: Understand the Problem

We are asked to find the exact values of expressions involving inverse trigonometric functions with angles halved. We need to leverage trigonometric identities such as half-angle formulas to evaluate these expressions.

Step 2: Solve (a) using Half-Angle Formula

The expression to evaluate is \( \sin \left[ \frac{1}{2} \cos ^{-1}\left(-\frac{3}{5}\right) \right] \). The half-angle formula for sine is \( \sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos\theta}{2}} \). First, determine \( \cos\theta \). Since \( \theta = \cos^{-1}\left(-\frac{3}{5}\right) \), \( \cos\theta = -\frac{3}{5} \). Using a right triangle, the opposite side is 4 (since \( 4^2 + (-3)^2 = 5^2 \)). Thus, \( \sin\theta = \frac{4}{5} \). Now apply the half-angle formula: \[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - (-\frac{3}{5})}{2}} = \sqrt{\frac{1 + \frac{3}{5}}{2}} = \sqrt{\frac{8}{10}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}. \]

Step 3: Solve (b) using Half-Angle Formula

The expression to evaluate is \( \cos \left(\frac{1}{2} \sin ^{-1} \frac{12}{13}\right) \). The half-angle formula for cosine is \( \cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos\theta}{2}} \). We have \( \theta = \sin^{-1}\left(\frac{12}{13}\right) \), therefore \( \sin\theta = \frac{12}{13} \). From a right triangle, the adjacent side is 5 (since \( 5^2 + 12^2 = 13^2 \)), so \( \cos\theta = \frac{5}{13} \). Apply the formula: \[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \frac{5}{13}}{2}} = \sqrt{\frac{18}{26}} = \sqrt{\frac{9}{13}} = \frac{3}{\sqrt{13}} = \frac{3\sqrt{13}}{13}. \]

Step 4: Solve (c) using Half-Angle Formula

The expression to evaluate is \( \tan \left(\frac{1}{2} \tan ^{-1} \frac{40}{9}\right) \). The half-angle formula for tangent is \( \tan\left(\frac{\theta}{2}\right) = \frac{\sin\theta}{1 + \cos\theta} \). First, \( \theta = \tan^{-1}\left(\frac{40}{9}\right) \) implies \( \tan\theta = \frac{40}{9} \). Using a right triangle, the hypotenuse is \( \sqrt{40^2 + 9^2} = 41 \), so \( \sin\theta = \frac{40}{41} \) and \( \cos\theta = \frac{9}{41} \). Apply the formula: \[ \tan\left(\frac{\theta}{2}\right) = \frac{\frac{40}{41}}{1 + \frac{9}{41}} = \frac{\frac{40}{41}}{\frac{50}{41}} = \frac{40}{50} = \frac{4}{5}. \]

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \( \cos^{-1}(x) \), \( \sin^{-1}(x) \), and \( \tan^{-1}(x) \), allow us to find angles when given a certain trigonometric ratio. These functions are very useful when we need to reverse trigonometric functions to determine angles.
For example, \( \cos^{-1}(-3/5) \) means "what angle \( \theta \) has a cosine of \(-3/5\)." This concept is crucial when working with triangles or waves because it helps us manipulate and find angles using known ratios.
It's important to remember that these functions only produce angles within certain ranges:

• \( \cos^{-1}(x) \) gives angles between \(0\) and \(\pi\)
• \( \sin^{-1}(x) \) gives angles between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\)
• \( \tan^{-1}(x) \) results in angles between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\)

This restriction is needed to maintain each function as a proper inverse that gives a unique value.

Trigonometric Identities

Trigonometric identities are equations that are true for all values of the involved variables where the functions are defined. These identities include the half-angle formulas, which are particularly useful in reducing complex expressions involving halved angles.
The half-angle formulas are derived from other trigonometric identities and allow us to find the sine, cosine, or tangent of half an angle when the trigonometric value of the full angle is known. For sine and cosine, the half-angle identities are:

• \( \sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos\theta}{2}} \)
• \( \cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos\theta}{2}} \)

Meanwhile, the half-angle formula for tangent is:

• \( \tan\left(\frac{\theta}{2}\right) = \frac{\sin\theta}{1 + \cos\theta} \)

These identities are indispensable when solving expressions like \( \sin\left(\frac{1}{2}\cos^{-1}(-3/5)\right)\) because they allow computing exact values without a calculator.

Right Triangle Trigonometry

Right triangle trigonometry revolves around the relationships between the angles and sides of a right triangle. For any right triangle, the fundamental trigonometric ratios are sine, cosine, and tangent, which are defined as follows:

• \( \sin\theta = \frac{\text{opposite side}}{\text{hypotenuse}} \)
• \( \cos\theta = \frac{\text{adjacent side}}{\text{hypotenuse}} \)
• \( \tan\theta = \frac{\text{opposite side}}{\text{adjacent side}} \)

These basic definitions are crucial in solving problems involving right triangles.
When given an inverse trigonometric function like \( \sin^{-1}(12/13) \), right triangle trigonometry can help visualize this as an angle \( \theta \) for which the sine ratio is \( \frac{12}{13} \). This means in a triangle, the opposite side could be 12 units and the hypotenuse 13 units. From there, using the Pythagorean theorem, the adjacent side can be calculated, enabling computation of \( \cos\theta\) or \( \tan\theta\).

Exact Values of Trigonometric Expressions

Exact values of trigonometric expressions are values that are found without any approximation, often involving simplifying expressions with radicals. In problems where these exact values are needed, knowledge of both inverse trigonometric functions and trigonometric identities is essential.
For example, the exact value for \( \sin\left( \frac{1}{2} \cos^{-1}(-3/5) \right) \) involves using the half-angle sine identity, and calculating manually without a calculator ensures precision. In terms of right triangle trigonometry, it involves knowing the full angle's trigonometric values and correctly applying identities.
Using these calculations and techniques, we can determine values like \( \sin\left( x/2 \right) = \frac{2\sqrt{5}}{5} \) or \( \tan\left( \frac{\theta}{2} \right) = \frac{4}{5} \). These exact values allow deeper insights and are part of the foundation for more advanced problems in trigonometry and calculus.