Problem 50 - Precalculus Mathematics for Calculus

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 \( \frac{10 - 2\sqrt{5}}{15} \).

Step 1: Understand the given expression

The problem asks us to find the exact value of the expression \( \sin \left( \cos^{-1} \frac{2}{3} - \tan^{-1} \frac{1}{2} \right) \). This expression involves subtracting two inverse trigonometric functions inside a sine function.

Step 2: Simplify using identities

To simplify the expression, we start by using the sine difference identity: \[\sin(a - b) = \sin a \cos b - \cos a \sin b\] Here, let \( a = \cos^{-1} \frac{2}{3} \) and \( b = \tan^{-1} \frac{1}{2} \).

Step 3: Evaluate \( \sin(\cos^{-1}(\frac{2}{3})) \)

From \( a = \cos^{-1} \frac{2}{3} \), we know \( \cos a = \frac{2}{3} \). Use the identity \( \sin^2 a + \cos^2 a = 1 \), which gives: \[\sin^2 a = 1 - (\frac{2}{3})^2 = 1 - \frac{4}{9} = \frac{5}{9}\]Thus, \( \sin a = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \).

Step 4: Evaluate \( \cos(\tan^{-1}(\frac{1}{2})) \)

From \( b = \tan^{-1} \frac{1}{2} \), we know \( \tan b = \frac{1}{2} \). Think of this as a right triangle where opposite side is 1 and adjacent side is 2: \[\text{Hypotenuse} = \sqrt{1^2 + 2^2} = \sqrt{5}\]Thus, \( \cos b = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \).

Step 5: Evaluate \( \sin(\tan^{-1}(\frac{1}{2})) \)

Using the same triangle from Step 4, \( \sin b = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} \).

Step 6: Compute \( \sin(a - b) \) using the angle difference identity

Substituting the values found:\[\sin(a - b) = \sin a \cos b - \cos a \sin b\]Plug in the values:\[\sin(a - b) = \frac{\sqrt{5}}{3} \cdot \frac{2\sqrt{5}}{5} - \frac{2}{3} \cdot \frac{\sqrt{5}}{5}\]Simplify:\[= \frac{2 \cdot 5}{15} - \frac{2\sqrt{5}}{15}\]\[= \frac{10 - 2\sqrt{5}}{15}\]

Step 7: Final Result Verification

Verify if simplifications were done correctly:For both parts of the calculation (before subtraction), check each value by ensuring \( \sin^2x + \cos^2x = 1\) holds true based on sides. Done.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find angles when the value of the trigonometric ratio is known.
These functions are the inverses of the standard trigonometric functions: sine, cosine, and tangent.
In this exercise, we see two inverse trigonometric functions:

• \( \cos^{-1} \left( \frac{2}{3} \right) \), which gives us the angle whose cosine is \( \frac{2}{3} \).
• \( \tan^{-1} \left( \frac{1}{2} \right) \), which gives us the angle whose tangent is \( \frac{1}{2} \).

The angle found with \( \cos^{-1} \) is used to calculate \( \sin a \), with \( a = \cos^{-1} \left( \frac{2}{3} \right) \).
Similarly, the angle found with \( \tan^{-1} \) lets us calculate \( \sin b \) and \( \cos b \), where \( b = \tan^{-1} \left( \frac{1}{2} \right) \).
Understanding how to work with these functions is key to solving the problem accurately.

Sine Difference Identity

The sine difference identity helps break down complex expressions involving sine of angle differences.
This identity is expressed as:

• \( \sin(a - b) = \sin a \cos b - \cos a \sin b \)

In the given problem, this identity simplifies the calculation of \( \sin \left( \cos^{-1} \frac{2}{3} - \tan^{-1} \frac{1}{2} \right) \) by expressing it as a combination of \( \sin a \), \( \cos a \), \( \sin b \), and \( \cos b \).
We find each of these values using the angles provided by the inverse trigonometric functions.
By systematically using the identity, the expression is simplified to reach the desired exact value.

Right Triangle Properties

Right triangle properties are essential for calculating sine and cosine when dealing with inverse trigonometric problems.
In Step 4 of the solution, we consider a right triangle to determine \( \cos b \) and \( \sin b \).
When \( \tan b = \frac{1}{2} \), the triangle has:

• An opposite side of 1 and an adjacent side of 2.
• The hypotenuse is found using the Pythagorean theorem: \( \sqrt{1^2 + 2^2} = \sqrt{5} \).

These side lengths allow us to find other trigonometric functions:

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

Understanding how to use right triangle relationships helps in finding exact values when working with angles obtained from inverse trigonometric functions.