Problem 21 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-22: Find the exact value of the expression whenever it is defined. (a) \(\sin \left[\frac{1}{2} \sin ^{-1}\left(-\frac{7}{25}\right)\right]\) (b) \(\cos \left(\frac{1}{2} \tan ^{-1} \frac{8}{15}\right)\) (c) \(\tan \left(\frac{1}{2} \cos ^{-1} \frac{3}{5}\right)\)

Step-by-Step Solution

Answer

(a) \(-\frac{1}{5\sqrt{2}}\), (b) \(\frac{4}{\sqrt{17}}\), (c) \(\frac{1}{2}\).

Step 1: Introduction to Problem

We need to find the exact values of the trigonometric expressions given as sub-expressions of arcsine, arctangent, and arccosine. Each part requires us to compute values using trigonometric identities.

Step 1: Simplifying Part (a)

For \( a = \sin \left[ \frac{1}{2} \sin^{-1} \left( -\frac{7}{25} \right) \right] \):Let \( x = \sin^{-1} \left( -\frac{7}{25} \right) \), then \( \sin x = -\frac{7}{25} \). Use the half-angle identity:\[ \sin \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{2}} \]First find \( \cos x \) using \( \sin^2 x + \cos^2 x = 1 \).So,\( \cos x = \sqrt{1 - (-\frac{7}{25})^2} = \sqrt{1 - \frac{49}{625}} = \sqrt{\frac{576}{625}} = \frac{24}{25} \).Thus,\( \sin \frac{x}{2} = \pm \sqrt{\frac{1 - \frac{24}{25}}{2}} = \pm \sqrt{\frac{1}{50}} = \pm \frac{1}{\sqrt{50}} \). Since \( x = \sin^{-1}(-\frac{7}{25}) \) is in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\), \( \sin \frac{x}{2} \) should be negative.Thus, \( \sin \frac{x}{2} = -\frac{1}{5\sqrt{2}} \).

Step 2: Simplifying Part (b)

For \( b = \cos \left( \frac{1}{2} \tan^{-1} \frac{8}{15} \right) \):Let \( y = \tan^{-1} \frac{8}{15} \), then \( \tan y = \frac{8}{15} \).Create a right triangle where opposite = 8, adjacent = 15, so \( hypotenuse = \sqrt{8^2 + 15^2} = 17 \).Thus, \( \cos y = \frac{15}{17} \).Apply half-angle identity:\[ \cos \frac{y}{2} = \pm \sqrt{\frac{1 + \cos y}{2}} \]So,\( \cos \frac{y}{2} = \pm \sqrt{\frac{1 + \frac{15}{17}}{2}} = \pm \sqrt{\frac{32}{34}} = \pm \sqrt{\frac{16}{17}} \).Since \( y = \tan^{-1} \frac{8}{15} \) is in the first quadrant, \( \cos \frac{y}{2} \) is positive, thus \( \cos \frac{y}{2} = \sqrt{\frac{16}{17}} = \frac{4}{\sqrt{17}} \).

Step 3: Simplifying Part (c)

For \( c = \tan \left( \frac{1}{2} \cos^{-1} \frac{3}{5} \right) \):Let \( z = \cos^{-1} \frac{3}{5} \), then \( \cos z = \frac{3}{5} \).Use a right triangle where adjacent = 3, hypotenuse = 5, so \( opposite = \sqrt{5^2 - 3^2} = 4 \).Thus, \( \sin z = \frac{4}{5} \).Use the half-angle formula for tangent:\[ \tan \frac{z}{2} = \frac{1 - \cos z}{\sin z} \]So,\( \tan \frac{z}{2} = \frac{1 - \frac{3}{5}}{\frac{4}{5}} = \frac{\frac{2}{5}}{\frac{4}{5}} = \frac{1}{2} \).

Step 5: Concluding the Solution

We found exact values for each part: (a) \(-\frac{1}{5\sqrt{2}}\), (b) \(\frac{4}{\sqrt{17}}\), (c) \(\frac{1}{2}\). These computations involved applying half-angle trigonometric identities and simplifying using fundamental definitions and properties of trigonometric functions.

Key Concepts

Understanding Trigonometric Identities

Trigonometric identities are equations that relate different trigonometric functions to one another. They are essential tools in mathematics, especially when simplifying complex expressions and solving trigonometric equations.
Here are some general purposes of trigonometric identities:

• They simplify trigonometric expressions by reducing their complexity.
• They allow us to find unknown angles and sides in triangles by expressing one trigonometric function in terms of another.

Half-angle identities are a specific type of trigonometric identity. They express trigonometric functions of half angles, for example, \( \sin \frac{x}{2} \) or \( \cos \frac{y}{2} \). These are extremely useful when calculating the exact values of expressions involving inverse trigonometric functions, as seen in the given exercise.
Understanding these identities enables you to transition smoothly from inverse trigonometric values to regular trigonometric values, facilitating precise computation.

Delving into Arcsine

Arcsine, denoted as \( \sin^{-1} \) or \( \text{asin} \), is the inverse function of sine. It returns the angle whose sine is a given number.
In mathematical terms, \( y = \sin^{-1} x \) means that \( \sin y = x \) where \( -1 \leq x \leq 1 \) and \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \). This specific range ensures the function is a true inverse by being one-to-one.
Here's why arcsine is beneficial:

• It helps in finding angles when the value of sine is known.
• It is crucial in defining the relationship between angles and their trigonometric values.

Applying arcsine wisely, such as in the exercise, involves transitioning from the arcsine value to a trigonometric function using identities like the half-angle identity, helping solve trigonometric problems efficiently.

Exploring Arccosine

Arccosine, usually represented as \( \cos^{-1} \) or \( \text{acos} \), serves as the inverse function of cosine. It returns the angle whose cosine is a given number, centered primarily between \( 0 \) and \( \pi \) to maintain its one-to-one nature.
Mathematically, \( y = \cos^{-1} x \) indicates that \( \cos y = x \) where \( -1 \leq x \leq 1 \) and \( 0 \leq y \leq \pi \). This calculation aids in the following:

• Finding angles provided the cosine value is known and understanding angles within specific quadrants.
• Creating expressions that allow the conversion between different trigonometric forms.

Using arccosine in conjunction with identities like the half-angle identity, as performed in the exercise, elucidates how to handle these expressions rigorously. It illustrates the process of uncovering exact trigonometric function values from inverse trigonometric values, an essential computational task.

Grasping Arctangent

Arctangent, denoted as \( \tan^{-1} \) or \( \text{atan} \), is the inverse of the tangent function. It provides the angle whose tangent is a specific value. The principal value lies between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \), aligning with the unique one-to-one nature of inverse functions.
Mathematically, it means \( y = \tan^{-1} x \) results in \( \tan y = x \), which caters to a full real range of \( x \). This brings several advantages:

• Computing angles for given tangent values in the common x-y plane applications.
• Simplifying expressions to convert inverse tangent values into calculable angles.

Arctangent aligns perfectly with trigonometric identities in simplifying and clarifying complex trigonometric expressions expressed in angular terms, as displayed in the provided exercise. This functionality makes arctangent indispensable for solving a variety of mathematical problems involving angles.