Problem 85 - A Graphical Approach to Precalculus with Limits

Question

Give the exact real number value of each expression. Do not use a calculator. $$\cos \left(2 \arctan \frac{4}{3}\right)$$

Step-by-Step Solution

Answer

The exact value is \( \frac{-7}{25} \).

Step 1: Identify the inverse trigonometric function

The expression involves \( \arctan \frac{4}{3} \). This means we are considering an angle, let’s call it \( \theta \), such that \( \tan \theta = \frac{4}{3} \).

Step 2: Represent \( \theta \) in a right triangle

Visualize a right triangle where the opposite side to angle \( \theta \) is 4 and the adjacent side is 3. Using the Pythagorean theorem, find the hypotenuse: \((\text{hypotenuse})^2 = 3^2 + 4^2 = 9 + 16 = 25 \). So, \( \text{hypotenuse} = 5 \).

Step 3: Express \( \sin \theta \) and \( \cos \theta \)

Using the triangle, \( \sin \theta = \frac{4}{5} \) and \( \cos \theta = \frac{3}{5} \).

Step 4: Utilize the double angle identity for cosine

We use the identity \( \cos(2\theta) = \cos^2 \theta - \sin^2 \theta \). Substitute the values from Step 3: \( \cos(2\theta) = \left(\frac{3}{5}\right)^2 - \left(\frac{4}{5}\right)^2 = \frac{9}{25} - \frac{16}{25} \).

Step 5: Simplify the expression

Calculate the expression \( \frac{9}{25} - \frac{16}{25} = \frac{9 - 16}{25} = \frac{-7}{25} \). This gives us \( \cos(2\theta) = \frac{-7}{25} \).

Key Concepts

Double Angle Identities

Understanding double angle identities in trigonometry is crucial for solving problems like the one you're facing. Double angle identities provide formulas to calculate the trigonometric functions of double angles (i.e., twice a given angle). One important identity is the double angle formula for cosine:

• \( \cos(2\theta) = \cos^2\theta - \sin^2\theta \)

This formula allows us to express the cosine of double angles in terms of the cosine and sine of the angle itself. In the given exercise, we were tasked to find \( \cos(2\theta) \) where \( \theta \) is determined using the inverse tangent function. By breaking \( \cos(2\theta) \) into \( \cos^2\theta - \sin^2\theta \), we can substitute the known values of \( \cos\theta \) and \( \sin\theta \) to find the solution.This identity is useful because it translates a more complex double angle problem into a simpler calculation involving only the trigonometric functions of a single angle.

Inverse Trigonometric Functions

Inverse trigonometric functions are powerful tools in trigonometry that enable us to find angles when given certain trigonometric ratio values. For instance, the inverse tangent of a ratio \( \tan(\theta) = \frac{4}{3} \), yields the angle \( \theta \) in the context of our problem.

• \( \arctan \left(\frac{4}{3}\right) = \theta \)

This means we are looking for an angle where the ratio of the opposite side to the adjacent side is \( \frac{4}{3} \). In the exercise, knowing this angle helps us establish relationships within a right triangle. Inverse trigonometric functions effectively allow us to reverse the typical process: from knowing an angle to knowing a side length, and vice versa. They are essential for cases where determining the angle directly from a trigonometric equation isn't feasible without these functions.

Pythagorean Theorem

The Pythagorean Theorem is an essential part of trigonometry that connects the sides of a right triangle. It states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides:

• \( a^2 + b^2 = c^2 \)

In the exercise, once we assumed \( \tan \theta = \frac{4}{3} \), it was useful to visualize \( \theta \) in a right triangle with an opposite side of 4 and an adjacent side of 3. We applied the Pythagorean Theorem to find the hypotenuse:\[(3^2) + (4^2) = 9 + 16 = 25\]Hence, the hypotenuse is 5. Knowing all sides of the triangle lets us easily find \( \sin \theta = \frac{4}{5} \) and \( \cos \theta = \frac{3}{5} \). Applying the Pythagorean Theorem helps break down geometry problems in trigonometry by providing a solvable framework to calculate unknown lengths, ultimately enabling other trigonometric calculations.