Problem 50 - Algebra and Trigonometry

Question

Find the exact value of the expression. $$ \sin \left(\cos ^{-1} \frac{2}{3}-\tan ^{-1} \frac{1}{2}\right) $$

Step-by-Step Solution

Answer

The exact value of the expression is \( \frac{2(\sqrt{5} - 1)}{3\sqrt{5}} \).

Step 1: Identify the Trigonometric Values

First, let us label the angles: let \( \theta = \cos^{-1} \frac{2}{3} \) and \( \phi = \tan^{-1} \frac{1}{2} \). We need to find \( \sin(\theta - \phi) \). We will use the identities of trigonometric functions to find \( \sin \theta \), \( \cos \theta \), \( \sin \phi \), and \( \cos \phi \).

Step 2: Find \( \sin \theta \) and \( \cos \theta \)

Since \( \theta = \cos^{-1} \frac{2}{3} \), we have \( \cos \theta = \frac{2}{3} \). Use \( \sin^2 \theta + \cos^2 \theta = 1 \) to find \( \sin \theta \):\[ \sin \theta = \sqrt{1 - \left(\frac{2}{3}\right)^2} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

Step 3: Find \( \sin \phi \) and \( \cos \phi \)

Since \( \phi = \tan^{-1} \frac{1}{2} \), we have \( \tan \phi = \frac{1}{2} \). Use the right triangle concept where opposite = 1, adjacent = 2:\[ \sin \phi = \frac{1}{\sqrt{1^2 + 2^2}} = \frac{1}{\sqrt{5}} \]\[ \cos \phi = \frac{2}{\sqrt{1^2 + 2^2}} = \frac{2}{\sqrt{5}} \]

Step 4: Use Trigonometric Identity for Difference of Angles

Now that we have \( \sin \theta, \cos \theta, \sin \phi, \cos \phi \), apply the identity:\[ \sin(\theta - \phi) = \sin \theta \cos \phi - \cos \theta \sin \phi \]Substitute the values:\[ = \left(\frac{\sqrt{5}}{3}\right) \left(\frac{2}{\sqrt{5}}\right) - \left(\frac{2}{3}\right) \left(\frac{1}{\sqrt{5}}\right) \]

Step 5: Simplify the Expression

Simplify the expression:\[ \sin(\theta - \phi) = \frac{2\sqrt{5}}{3\sqrt{5}} - \frac{2}{3\sqrt{5}} \]Because both terms are over the same denominator \(3\sqrt{5}\), we can subtract the numerators:\[ = \frac{2\sqrt{5} - 2}{3\sqrt{5}} \]

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are special functions that help us find angles when given the value of a trigonometric ratio. They are the inverses of the six primary trigonometric functions, namely sine, cosine, and tangent, among others. These functions are often used to solve equations where the variable is an angle. For example, in our exercise, we are given

• \(\cos^{-1} \frac{2}{3}\), which is the inverse of the cosine function, helping to find angle \(\theta\) such that \(\cos \theta = \frac{2}{3}\).
• \(\tan^{-1} \frac{1}{2}\), which is the inverse of the tangent function, helping to find angle \(\phi\) such that \(\tan \phi = \frac{1}{2}\).

inverse trigonometric functions are crucial in problems involving trigonometric identities and values because they provide the angle needed to apply these identities or calculate specific trigonometric values. These functions help bridge the gap between the angle and its trigonometric value.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the involved variables. They are immensely valuable tools in simplifying expressions and solving trigonometric equations. One common and useful identity used in this exercise is the difference of angles identity:

• \(\sin(\theta - \phi) = \sin \theta \cos \phi - \cos \theta \sin \phi\).

This identity helps express trigonometric functions of combined angles in terms of the individual angles' functions, greatly simplifying complex trigonometric expressions.
In our exercise, this identity allows us to calculate \(\sin(\theta - \phi)\) using known values for \(\sin \theta, \cos \theta, \sin \phi,\) and \(\cos \phi\). Learning and practicing the use of these identities makes trigonometry much more manageable and increases problem-solving efficiency.

Trigonometric Values

Trigonometric values relate to the specific sine, cosine, and tangent values of angles, often found within right triangles or calculated using known identities. Knowing these values is essential in evaluating trigonometric expressions.
For example, given \(\cos^{-1} \frac{2}{3}\), we used the Pythagorean identity:

• \(\sin^2 \theta + \cos^2 \theta = 1\)

to find \(\sin \theta = \frac{\sqrt{5}}{3}\). Similarly, for \(\tan^{-1} \frac{1}{2}\), we imagined a right triangle where our known tangent ratio was opposite over adjacent, i.e., \(1/2\), and calculated

• \(\sin \phi = \frac{1}{\sqrt{5}}\)
• \(\cos \phi = \frac{2}{\sqrt{5}}\)

Understanding how to derive these values is critical to solving problems involving trigonometric functions, as they serve as the foundational building blocks for more complex calculations in trigonometry.