Problem 20 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-22: Find the exact value of the expression whenever it is defined. (a) \(\sin \left(2 \tan ^{-1} \frac{5}{12}\right)\) (b) \(\cos \left(2 \arccos \frac{9}{41}\right)\) (c) \(\tan \left[2 \arcsin \left(-\frac{8}{17}\right)\right]\)

Step-by-Step Solution

Answer

(a) \(\frac{120}{169}\), (b) \(\frac{-1357}{1681}\), (c) \(\frac{-240}{161}\).

Step 1: Identify the Trigonometric Identity for Part (a)

The expression requires using the double angle identity for sine. Recall that \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \). Here, \( \theta = \tan^{-1} \frac{5}{12} \).

Step 2: Calculate \( \sin(\theta) \) and \( \cos(\theta) \) for Part (a)

Given \( \theta = \tan^{-1} \frac{5}{12} \), form a right triangle where the opposite side is \(5\) and the adjacent side is \(12\). Use the Pythagorean theorem to find the hypotenuse: \( \sqrt{5^2 + 12^2} = 13 \). Thus, \( \sin(\theta) = \frac{5}{13} \) and \( \cos(\theta) = \frac{12}{13} \).

Step 3: Calculate \( \sin(2\theta) \) for Part (a)

Apply the identity: \[ \sin(2\theta) = 2 \times \frac{5}{13} \times \frac{12}{13} \]This simplifies to: \[ 2 \times \frac{60}{169} = \frac{120}{169} \]

Step 4: Identify the Trigonometric Identity for Part (b)

The task requires using the double angle identity for cosine. Recall that \( \cos(2\theta) = 2 \cos^2(\theta) - 1 \). Here, \( \theta = \arccos \frac{9}{41} \).

Step 5: Calculate \( \cos(2\theta) \) for Part (b)

Given that \( \cos(\theta) = \frac{9}{41} \), substitute into the double angle formula:\[ \cos(2\theta) = 2 \left(\frac{9}{41}\right)^2 - 1 \]Calculate:\[ 2 \times \frac{81}{1681} - 1 = \frac{162}{1681} - \frac{1681}{1681} = \frac{-1357}{1681} \]

Step 6: Identify the Trigonometric Identity for Part (c)

The expression requires using the double angle identity for tangent. Recall that \( \tan(2\theta) = \frac{2 \tan(\theta)}{1 - \tan^2(\theta)} \). Here, \( \theta = \arcsin \left(-\frac{8}{17}\right) \).

Step 7: Calculate \( \tan(\theta) \) for Part (c)

Given \( \arcsin \theta = -\frac{8}{17} \), form a right triangle with opposite side \(-8\) and hypotenuse \(17\). Use Pythagorean theorem for adjacent side: \( \sqrt{17^2 - (-8)^2} = \sqrt{289 - 64} = 15 \). Thus, \( \tan(\theta) = \frac{-8}{15} \).

Step 8: Calculate \( \tan(2\theta) \) for Part (c)

Substitute \( \tan(\theta) = \frac{-8}{15} \) into the double angle formula: \[ \tan(2\theta) = \frac{2 \times \frac{-8}{15}}{1 - \left(\frac{-8}{15}\right)^2} \]Calculate:\[ \frac{-16}{15} \div \left(1 - \frac{64}{225}\right) = \frac{-16}{15} \div \frac{161}{225} = \frac{-16}{15} \times \frac{225}{161} = \frac{-240}{161} \]

Key Concepts

Double Angle Formulas

Double angle formulas are a set of trigonometric identities that express trigonometric functions of double angles, like \(2\theta\), in terms of single angles. They are especially useful for simplifying complex trigonometric expressions. Here are the key double angle formulas:

• For sine: \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \)
• For cosine: \( \cos(2\theta) = 2 \cos^2(\theta) - 1 = 1 - 2 \sin^2(\theta) \)
• For tangent: \( \tan(2\theta) = \frac{2 \tan(\theta)}{1 - \tan^2(\theta)} \)

Understanding these formulas helps in converting expressions and in solving problems where a trigonometric function is used with a doubled angle.When approaching problems with these formulas, identify your \(\theta\) first. Rewrite the expression using the appropriate double angle formula. Then, simplify the expression using algebraic techniques.

Inverse Trigonometric Functions

Inverse trigonometric functions, like \(\tan^{-1}(x)\), \(\cos^{-1}(x)\), and \(\sin^{-1}(x)\) are functions that reverse the operation of the basic trigonometric ratios. They are used to find the angle when the sine, cosine, or tangent values are known. These functions are critical for solving problems involving angles and right triangles.

• \( \tan^{-1}(x) \) gives you an angle whose tangent is \(x\).
• \( \cos^{-1}(x) \) provides an angle whose cosine is \(x\).
• \( \sin^{-1}(x) \) results in an angle whose sine is \(x\).

These functions have defined ranges, for example, \( \tan^{-1}(x) \) has a range of \((-\frac{\pi}{2}, \frac{\pi}{2})\), which is essential for ensuring that inverse values are correctly interpreted as angles. In the exercises, they are used to determine an initial angle \(\theta\), which can then be manipulated further using double angle identities.

Right Triangle Trigonometry

Right triangle trigonometry is foundational for understanding and using trigonometric identities and functions. It all begins with the right triangle, where the basic trigonometric ratios are defined based on a triangle’s sides.The primary ratios are:

• \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \)
• \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \)
• \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \)

In the exercise, these ratios are crucial for calculating known values of sine, cosine, and tangent for a given angle \(\theta\). By constructing a right triangle with given ratios, one calculates missing sides using the Pythagorean theorem when two sides are known. From here, these sides help calculate the required trigonometric functions for solving the problem.