Problem 20 - Precalculus: Functions and Graphs

Question

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: Understand Inverse Trigonometric Functions

The expressions given in the problem involve inverse trigonometric functions: \(\tan^{-1}\), \(\arccos\), and \(\arcsin\). Recognize that these functions are used to determine angles whose trigonometric function values are known and will be used in identities for calculating double angles.

Step 2: Use Double Angle Formulas

Recall the double angle identities: 1. \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \)2. \( \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) \) or equivalently \( 2\cos^2(\theta) - 1 \) or \( 1 - 2\sin^2(\theta) \)3. \( \tan(2\theta) = \frac{2\tan(\theta)}{1-\tan^2(\theta)} \)

Step 3: Solve Part (a): \(\sin\left(2 \tan^{-1} \frac{5}{12}\right)\)

Let \( \theta = \tan^{-1} \frac{5}{12} \). This implies \( \tan(\theta) = \frac{5}{12} \). Consider a right triangle where the opposite side is 5 and the adjacent side is 12, making the hypotenuse \( \sqrt{5^2 + 12^2} = 13 \). Thus, \( \sin(\theta) = \frac{5}{13} \) and \( \cos(\theta) = \frac{12}{13} \).Use the identity \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \):\[ \sin\left(2 \tan^{-1} \frac{5}{12}\right) = 2 \left(\frac{5}{13}\right)\left(\frac{12}{13}\right) = \frac{120}{169} \]

Step 4: Solve Part (b): \(\cos\left(2 \arccos \frac{9}{41}\right)\)

Let \( \theta = \arccos \frac{9}{41} \). This implies \( \cos(\theta) = \frac{9}{41} \). Consider a right triangle with adjacent side 9 and hypotenuse 41, making the opposite side \( \sqrt{41^2 - 9^2} = 40 \). Thus, \( \sin(\theta) = \frac{40}{41} \).Use the identity \( \cos(2\theta) = 2\cos^2(\theta) - 1 \):\[ \cos\left(2 \arccos \frac{9}{41}\right) = 2\left(\frac{9}{41}\right)^2 - 1 = 2\left(\frac{81}{1681}\right) - 1 = \frac{162}{1681} - \frac{1681}{1681} = \frac{-1357}{1681} \]

Step 5: Solve Part (c): \(\tan\left[2 \arcsin \left(-\frac{8}{17}\right)\right]\)

Let \( \theta = \arcsin \left(-\frac{8}{17}\right) \). This implies \( \sin(\theta) = -\frac{8}{17} \). Consider a right triangle with opposite side -8 and hypotenuse 17, making the adjacent side \( \sqrt{17^2 - 8^2} = 15 \). Thus, \( \cos(\theta) = \frac{15}{17} \).Use the identity \( \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \), where \( \tan(\theta) = \frac{-8}{15} \):\[ \tan[2 \arcsin(-\frac{8}{17})] = \frac{2(-\frac{8}{15})}{1 - (-\frac{8}{15})^2} = \frac{-\frac{16}{15}}{1 - \frac{64}{225}} = \frac{-\frac{16}{15}}{\frac{161}{225}} = \frac{-16 \times 225}{15 \times 161} = \frac{-240}{161} \]

Key Concepts

Double Angle Formulas

Double angle formulas are crucial in trigonometry for expressing trigonometric functions of double angles in terms of single angles. By knowing these formulas, we can simplify expressions and solve various trigonometric problems more easily. Here are the three main double angle formulas:

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

These identities allow us to work with angles that are twice as large by using known values of sine, cosine, and tangent of simpler angles. They are particularly helpful when dealing with expressions like \( \sin(2\tan^{-1} x) \) or \( \cos(2\arccos y) \), as seen in our original exercise. Understanding how to derive and apply these formulas will greatly enhance your ability to tackle more complex trigonometric problems efficiently.

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions and are true for every value of the involved variables. They are vital tools for simplifying expressions and solving trigonometric equations. Some common identities include the Pythagorean identities, such as

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

These identities are often combined with double angle formulas to simplify expressions further. For example, in finding the cosine of a double angle, you might use \( \cos(2\theta) = 2\cos^2(\theta) - 1 \) if you have \( \cos(\theta) \), or \( \cos(2\theta) = 1 - 2\sin^2(\theta) \) if you have \( \sin(\theta) \). Mastering these identities will make it easier to derive exact values and find solutions for trigonometric expressions in various scenarios.

Right Triangle Trigonometry

Right triangle trigonometry is a foundational concept that helps in understanding trigonometric relationships within right-angled triangles. The relationships between the sides of a right triangle are often expressed using the primary trigonometric functions: sine, cosine, and tangent. Specifically:

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

When you know two sides of a right triangle, you can easily calculate the third side using the Pythagorean theorem, \( a^2 + b^2 = c^2 \). This is invaluable in exercises where you need to find specific values of trigonometric functions based on given ratios, like in the inverse functions \( \tan^{-1} \), \( \arccos \), and \( \arcsin \). By conceptualizing these terms using right triangles, many of these problems become more intuitive and manageable.

Exact Values in Trigonometry

Exact values in trigonometry refer to those trigonometric function values that can be expressed without decimals or approximations. These values often arise from familiar angles such as 30°, 45°, and 60°, and more complex expressions can sometimes be resolved into these exact values using identities or by breaking down the expressions.
In the context of the exercise, you're tasked with finding exact values using inverse trigonometric functions in combination with double angle formulas. For example, by using a known function like \( \tan^{-1}(\frac{5}{12}) \) with a distinct angle, you can find the sine and cosine of \( \theta \) precisely. Once these are established, applying the double angle formula results in an exact value for \( \sin(2\theta) \), \( \cos(2\theta) \), or \( \tan(2\theta) \).

• This approach avoids rounding errors and gives mathematically precise solutions.
• Using these methods consistently aids in validating calculations.

By engaging with exact values, students can better understand connections between angles and their trigonometric functions, leading to a deeper comprehension of trigonometry as a whole.