Problem 49 - Precalculus Mathematics for Calculus

Question

Find the exact value of the expression. $$\tan \left(\sin ^{-1} \frac{3}{4}-\cos ^{-1} \frac{1}{3}\right)$$

Step-by-Step Solution

Answer

The exact value is \( -\frac{3\sqrt{7} + 20\sqrt{14}}{67} \).

Step 1: Understand the Problem

We need to find the exact value of \( \tan \left( \sin^{-1} \frac{3}{4} - \cos^{-1} \frac{1}{3} \right) \). This involves inverse trigonometric functions and their differences.

Step 2: Convert Inverse Sine to Angle

Let \( \theta = \sin^{-1} \frac{3}{4} \). Then \( \sin \theta = \frac{3}{4} \). By the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \), compute \( \cos \theta = \sqrt{1 - \left(\frac{3}{4}\right)^2} = \sqrt{\frac{16}{16} - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \).

Step 3: Convert Inverse Cosine to Angle

Let \( \phi = \cos^{-1} \frac{1}{3} \). Then \( \cos \phi = \frac{1}{3} \). Using the Pythagorean identity \( \cos^2 \phi + \sin^2 \phi = 1 \), compute \( \sin \phi = \sqrt{1 - \left(\frac{1}{3}\right)^2} = \sqrt{\frac{9}{9} - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \).

Step 4: Use the Tangent of a Difference Formula

We use \( \tan(a-b) = \frac{\tan a - \tan b}{1 + \tan a \cdot \tan b} \) with \( a = \theta \) and \( b = \phi \). We already found \( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/4}{\sqrt{7}/4} = \frac{3}{\sqrt{7}} \) and \( \tan \phi = \frac{\sin \phi}{\cos \phi} = \frac{2\sqrt{2}/3}{1/3} = 2\sqrt{2} \).

Step 5: Calculate the Tangent of the Difference

Substitute the values into the formula: \( \tan \left( \sin^{-1} \frac{3}{4} - \cos^{-1} \frac{1}{3} \right) = \frac{\frac{3}{\sqrt{7}} - 2\sqrt{2}}{1 + \frac{3}{\sqrt{7}} \cdot 2\sqrt{2}} \). Simplify this expression by rationalizing and calculating the denominator.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are used to find the angle when a specific trigonometric ratio is known. For instance, with \(\sin^{-1}\), we determine the angle whose sine is a given value. When you see \(\sin^{-1} \frac{3}{4}\), it asks which angle within the range of \([-\frac{\pi}{2}, \frac{\pi}{2}]\) has a sine of \(\frac{3}{4}\).
\(\cos^{-1}\), on the other hand, deals with cosine values, finding angles within \([0, \pi]\). If \(\cos^{-1} \frac{1}{3}\) is given, we're asking which angle has cosine \(\frac{1}{3}\).
Inverse trigonometric functions are commonly used in conjunction with identities to solve problems, helping to shift between different types of trigonometric expressions smoothly.

Tangent of a Difference Formula

The tangent of a difference formula comes in handy when you encounter terms like \(\tan(a-b)\). This formula states: \[\tan(a-b) = \frac{\tan a - \tan b}{1 + \tan a \cdot \tan b}\] This formula simplifies the calculation of the tangent for expressions involving the difference between two angles. For example, in our exercise, \(a = \sin^{-1} \frac{3}{4}\) and \(b = \cos^{-1} \frac{1}{3}\).
First, find \(\tan a\) and \(\tan b\) by calculating \(\tan a = \frac{\sin a}{\cos a}\) and \(\tan b = \frac{\sin b}{\cos b}\). Then, plug these values into the formula.
This method avoids directly calculating the angles and is particularly useful in exact trigonometric value problems.

Pythagorean Identity

The Pythagorean identity is a key tool in trigonometry, which relates the square of sine and cosine functions: \[\cos^2 \theta + \sin^2 \theta = 1\] This identity is useful when you have either \(\sin \theta\) or \(\cos \theta\) and need to find the other. In our solution, knowing \(\sin \theta = \frac{3}{4}\), we used this identity to determine \(\cos \theta\) by rearranging it: \[\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{3}{4}\right)^2} = \frac{\sqrt{7}}{4}\] Likewise, knowing \(\cos \phi = \frac{1}{3}\), we found \(\sin \phi\) as: \[\sin \phi = \sqrt{1 - \cos^2 \phi} = \frac{2\sqrt{2}}{3}\] Use this identity whenever trigonometric expressions involving sine and cosine need simplifying or solving.