Problem 19 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-22: Find the exact value of the expression whenever it is defined. (a) \(\sin \left[2 \arccos \left(-\frac{3}{5}\right)\right]\) (b) \(\cos \left(2 \sin ^{-1} \frac{15}{17}\right)\) (c) \(\tan \left(2 \tan ^{-1} \frac{3}{4}\right)\)

Step-by-Step Solution

Answer

(a) \(-\frac{24}{25}\), (b) \(-\frac{7}{289}\), (c) \(\frac{24}{7}\)."

Step 1: Recognize the Expression

The expression given for part (a) is \(\sin \left[2 \arccos \left(-\frac{3}{5}\right)\right]\). This involves using the double angle identity for \(\sin(2\theta)\), where \(\theta = \arccos\left(-\frac{3}{5}\right)\).

Step 2: Apply the Double Angle Identity

The double angle identity for sine is \(\sin(2\theta) = 2 \sin(\theta) \cos(\theta)\). We need to find \(\sin(\theta)\) and \(\cos(\theta)\) given \(\theta = \arccos\left(-\frac{3}{5}\right)\).

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

Since \(\cos(\theta) = -\frac{3}{5}\), we use the Pythagorean identity \(\sin^2(\theta) + \cos^2(\theta) = 1\) to find \(\sin(\theta)\). This gives us \(\sin(\theta) = \sqrt{1 - \left(-\frac{3}{5}\right)^2} = \frac{4}{5}\).

Step 4: Evaluate the Expression for Part (a)

Substitute \(\sin(\theta)\) and \(\cos(\theta)\) into the double angle formula: \(\sin\left(2 \theta\right) = 2 \left(\frac{4}{5}\right) \left(-\frac{3}{5}\right) = -\frac{24}{25}\).

Step 5: Recognize the Expression for Part (b)

The expression given is \(\cos \left(2 \sin^{-1} \frac{15}{17}\right)\). Use the double angle identity for \(\cos(2\theta)\), where \(\theta = \sin^{-1} \frac{15}{17}\).

Step 6: Apply the Double Angle Identity for Cosine

The double angle identity for cosine is \(\cos(2\theta) = 1 - 2 \sin^2(\theta)\). First, find \(\sin(\theta)\) as given, and use it to find \(\cos(\theta)\).

Step 7: Calculate \(\cos(\theta)\) Given \(\sin(\theta)\)

Given \(\sin(\theta) = \frac{15}{17}\), we use \(\cos^2(\theta) = 1 - \sin^2(\theta)\), leading to \(\cos(\theta) = \frac{8}{17}\).

Step 8: Evaluate the Expression for Part (b)

Using the identity, \(\cos(2\theta) = 1 - 2 \left(\frac{15}{17}\right)^2 = -\frac{7}{289}\).

Step 9: Recognize the Expression for Part (c)

The expression is \(\tan\left(2 \tan^{-1} \frac{3}{4}\right)\). Use the double angle identity for \(\tan(2\theta)\), where \(\theta = \tan^{-1} \frac{3}{4}\).

Step 10: Apply the Double Angle Identity for Tangent

The double angle identity for tangent is \(\tan(2\theta) = \frac{2 \tan(\theta)}{1 - \tan^2(\theta)}\). Here, \(\tan(\theta) = \frac{3}{4}\).

Step 11: Evaluate the Expression for Part (c)

Substitute \(\tan(\theta)\) into the formula: \(\tan(2\theta) = \frac{2 \cdot \frac{3}{4}}{1 - \left(\frac{3}{4}\right)^2} = \frac{24}{7}\).

Key Concepts

double angle identities

Double angle identities are useful formulas in trigonometry that help you find the sine, cosine, or tangent of double angles. This means you can simplify complex expressions by using these identities.
For the sine double angle, the formula is:

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

This identity shows how to express \(2\theta\) in terms of \(\sin(\theta)\) and \(\cos(\theta)\). It is used when you know these two trigonometric values or can derive them using other identities.
When looking at cosine, the double angle identity is:

• \( \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) \)
• Alternatively, \( \cos(2\theta) = 2\cos^2(\theta) - 1 \)
• Or, \( \cos(2\theta) = 1 - 2\sin^2(\theta) \)

These variations allow flexibility depending on which trigonometric function values you know.
For tangent, the identity is:

• \( \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \)

It helps to simplify expressions involving tangents of double angles and is especially useful when solving more complex trigonometric equations.
Understanding and applying these formulas can make solving trigonometry problems involving double angles more straightforward.

inverse trigonometric functions

Inverse trigonometric functions play a crucial role in finding angles given a trigonometric value.
These functions include \(\sin^{-1}(x)\), \(\cos^{-1}(x)\), and \(\tan^{-1}(x)\). They reverse the process of standard trigonometric functions, turning ratios back into angles.
For instance:

• \( \sin^{-1}(x) \) gives an angle, \(\theta\), whose sine value is \(x\).
• \( \cos^{-1}(x) \) provides an angle or \(\theta\), where \(\cos(\theta) = x\).
• \( \tan^{-1}(x) \) results in an angle where \(\tan(\theta) = x\).

These functions are essential for solving trigonometric equations where the angle itself is the unknown.
Inverse trigonometric functions are bounded within specific ranges to ensure they are well-defined single values. For example, \( \sin^{-1}(x) \) returns values from \(-\frac{\pi}{2} \) to \( \frac{\pi}{2} \).
Recognizing these ranges allows you to accurately solve problems and interpret solutions in the correct context.

Pythagorean identity

The Pythagorean identity is a fundamental trigonometric formula:

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

This identity reveals the intrinsic relationship between sine and cosine, rooted in the Pythagorean theorem for right triangles.
It's invaluable for finding either \(\sin(\theta)\) or \(\cos(\theta)\) when you know the other, aiding in a variety of trigonometric simplifications.
Derived identities include:

• \( \tan^2(\theta) + 1 = \sec^2(\theta) \)
• \( 1 + \cot^2(\theta) = \csc^2(\theta) \)

These variations provide further insights into relationships between other trigonometric functions.
Using the Pythagorean identity can simplify trigonometric expressions and solve equations, especially when dealing with squared terms. It acts as a bridge for transformations and aids in converting one form to another, enhancing your problem-solving toolkit in trigonometry.