Problem 47 - Precalculus Mathematics for Calculus

Question

Find the exact value of the given expression. $$\sin \left(2 \cos ^{-1} \frac{7}{25}\right)$$

Step-by-Step Solution

Answer

\(\frac{336}{625}\)

Step 1: Understand the Problem

We need to find the exact value of \( \sin \left(2 \cos^{-1} \frac{7}{25}\right) \). This involves using a trigonometric identity for double angles and understanding how to handle inverse trigonometric functions.

Step 2: Use the Inverse Cosine Function

We start by letting \( \theta = \cos^{-1} \frac{7}{25} \). This implies that \( \cos \theta = \frac{7}{25} \). We are looking for \( \sin(2\theta) \).

Step 3: Apply the Double Angle Identity for Sine

The double angle identity for sine is \( \sin(2\theta) = 2\sin \theta \cos \theta \). We already found \( \cos \theta = \frac{7}{25} \). Now, we need to find \( \sin \theta \).

Step 4: Find Sine Using Pythagorean Identity

Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we solve for \( \sin \theta \): \[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{7}{25}\right)^2 = 1 - \frac{49}{625} = \frac{576}{625} \]. Thus, \( \sin \theta = \sqrt{\frac{576}{625}} = \frac{24}{25} \).

Step 5: Substitute Values into the Double Angle Identity

Substituting the values we found into \( \sin(2\theta) = 2 \sin \theta \cos \theta \), we get \( \sin(2\theta) = 2 \times \frac{24}{25} \times \frac{7}{25} \).

Step 6: Simplify the Expression

The expression becomes \( \frac{336}{625} \). Therefore, \( \sin(2 \cos^{-1} \frac{7}{25}) = \frac{336}{625} \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are a really interesting part of trigonometry. They allow us to find angles when we know the sides of a triangle. These functions include \( \cos^{-1}, \sin^{-1}, \tan^{-1} \) and others.
\( \cos^{-1} \frac{7}{25} \) means we're looking for an angle \( \theta \) whose cosine is \( \frac{7}{25} \). This angle's cosine value tells us how the adjacent side compares to the hypotenuse in a right triangle.
The use of inverse trigonometric functions is important because it helps us shift from the length-focused study of triangles to an angle-focused study. This is essential for solving many types of geometry problems, including those that involve circles and waves.
This knowledge allows us to calculate values in expressions like \( \sin \left(2 \cos^{-1} \frac{7}{25}\right) \). It's the foundation for finding angles, which we then use with other trigonometric identities.

Double Angle Formula

The double angle formulas are very handy. They help us find trigonometric values of double angles using single angle values.
The double angle formula for sine is \( \sin(2\theta) = 2\sin \theta \cos \theta \). This means if you know \( \sin \theta \) and \( \cos \theta \), you can find \( \sin(2\theta) \) easily.

• Multiply \( \sin \theta \) by \( \cos \theta \).
• Then, double that result for \( \sin(2\theta) \).

In the given problem, once we knew \( \cos \theta = \frac{7}{25} \) and found \( \sin \theta = \frac{24}{25} \), we simply plugged these into the double angle formula.
This gave us the result \( \sin(2\theta) = \frac{336}{625} \). Double angle formulas like this one simplify the process of handling expressions involving trigonometric functions.

Pythagorean Identity

The Pythagorean identity is a core concept in trigonometry. It states that \( \sin^2 \theta + \cos^2 \theta = 1 \). This is extremely useful for finding unknown side lengths or angles in triangles.
This identity essentially derives from the Pythagorean theorem applied to a unit circle. It helps us find \( \sin \theta \) or \( \cos \theta \) when we know one of them.

• For example, if you know \( \cos \theta \), you can find \( \sin \theta \) using the identity.
• Just rearrange the formula: \( \sin^2 \theta = 1 - \cos^2 \theta \).

In our exercise, knowing \( \cos \theta = \frac{7}{25} \), we used this identity to find that \( \sin \theta = \frac{24}{25} \).
These identities are fundamental because they connect the different trigonometric functions together, and allow you to work out otherwise unknown values.