Problem 21 - Precalculus: Functions and Graphs

Question

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) \( \sin \left[ \frac{1}{2} \sin^{-1}(-\frac{7}{25}) \right] = \frac{\sqrt{2}}{10} \), (b) \( \cos \left(\frac{1}{2} \tan^{-1} \frac{8}{15}\right) = \frac{4\sqrt{2}}{17} \), (c) \( \tan \left(\frac{1}{2} \cos^{-1} \frac{3}{5}\right) = \frac{1}{2} \).

Step 1: Understand Inverse Trigonometry

For all parts of this exercise, recognize that inverse trigonometric functions such as \( \sin^{-1} \), \( \cos^{-1} \), and \( \tan^{-1} \) return angles. These angles are then used in calculations involving trigonometric functions divided by 2.

Step 2: Rewrite Part (a)

For part (a), set \( \theta = \sin^{-1}\left(-\frac{7}{25}\right) \), which means \( \sin \theta = -\frac{7}{25} \). We need to find \( \sin \left(\frac{\theta}{2}\right) \).

Step 3: Use Half-Angle Identity for Sine

Use the identity \( \sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{2}} \). Since \( \sin \theta = -\frac{7}{25} \), use the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \) to find \( \cos \theta \).

Step 4: Calculate \( \cos \theta \) in (a)

\( \cos \theta = \sqrt{1 - \left(-\frac{7}{25}\right)^2} = \sqrt{1 - \frac{49}{625}} = \frac{24}{25} \).

Step 5: Apply Half-Angle Formula in (a)

Calculate \( \sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \frac{24}{25}}{2}} = \pm \sqrt{\frac{1}{50}} = \pm \frac{1}{5\sqrt{2}} \). Choose the positive root since \( \sin \left(\frac{\theta}{2}\right) \) is in the first or second quadrant. Rationalize to get \( \frac{\sqrt{2}}{10} \).

Step 6: Rewrite Part (b)

Set \( \phi = \tan^{-1}\left(\frac{8}{15}\right) \), which implies \( \tan \phi = \frac{8}{15} \). Calculate \( \cos\left(\frac{\phi}{2}\right) \).

Step 7: Use Half-Angle Identity for Cosine

The identity is \( \cos\left(\frac{\phi}{2}\right) = \pm \sqrt{\frac{1 + \cos \phi}{2}} \). Use the Pythagorean identity \( 1 + \tan^2 \phi = \sec^2 \phi \) to first find \( \cos \phi \).

Step 8: Calculate \( \cos \phi \) in (b)

\( 1 + \left(\frac{8}{15}\right)^2 = \cos^{-2} \phi \implies \cos \phi = \frac{15}{17} \).

Step 9: Apply Half-Angle Formula in (b)

\( \cos\left(\frac{\phi}{2}\right) = \pm \sqrt{\frac{1 + \frac{15}{17}}{2}} = \frac{4\sqrt{2}}{17} \). Choose the positive root as it ensures correct quadrant for the function.

Step 10: Rewrite Part (c)

Set \( \psi = \cos^{-1}\left(\frac{3}{5}\right) \) which means \( \cos \psi = \frac{3}{5} \). We need \( \tan\left(\frac{\psi}{2}\right) \).

Step 11: Use Half-Angle Identity for Tangent

The identity is \( \tan\left(\frac{\psi}{2}\right) = \pm \sqrt{\frac{1 - \cos \psi}{1 + \cos \psi}} \). Thus, \( \tan\left(\frac{\psi}{2}\right) = \pm \sqrt{\frac{1 - \frac{3}{5}}{1 + \frac{3}{5}}} = \pm \sqrt{\frac{2}{8}} = \pm \frac{1}{2} \).

Step 12: Choose Positive Roots

For all parts, given we are dealing with principal values of the inverse trigonometric functions, choose the positive sign for the half-angles since they are in the first quadrant (part (c) specifically is from 0 to \(\frac{\pi}{2}\)).

Key Concepts

Half-Angle Identities

Half-angle identities are powerful tools in trigonometry that allow us to determine the sine, cosine, or tangent of half an angle given the trigonometric values of the whole angle. For instance, if you know the cosine of an angle, you can find the cosine of half of that angle using the identity:

• \( \cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{2}} \)

This provides a way to break down complex angle calculations into simpler parts.
Another useful identity is for the sine of half an angle:

• \( \sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{2}} \)

These plus or minus signs reflect the different possible quadrants where the angle could lie. It's important to choose the sign that aligns with the angle's actual direction. In this exercise, we focus on using these identities to evaluate trigonometric expressions at half-angles, ensuring to select the correct quadrant for each calculated angle.

Trigonometric Functions

Trigonometric functions describe relationships within triangles and are essential in dealing with angles. The basic trigonometric functions include sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)).
Each function is derived from a right triangle:

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

These functions can also be extended beyond triangle relationships to describe periodic waves and rotations. Inverse trigonometric functions like \( \sin^{-1} \), \( \cos^{-1} \), and \( \tan^{-1} \) allow us to find angles when given a trigonometric value, which is crucial in calculating expressions that involve these angles.
Understanding these functions and their inverses forms the foundation for solving trigonometric identities and expressions.

Pythagorean Identity

The Pythagorean identity is a fundamental relation between the sine and cosine functions:

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

This equation arises from the Pythagorean theorem applied to a unit circle, where each point is defined by its sine and cosine values. It enables you to find one trigonometric value if you have the other.
In particular, when you know \( \sin \theta \), you can determine \( \cos \theta \) using:

• \( \cos \theta = \sqrt{1 - \sin^2 \theta} \)

And vice versa, find \( \sin \theta \) if \( \cos \theta \) is known:

• \( \sin \theta = \sqrt{1 - \cos^2 \theta} \)

The simplicity of this identity makes it a convenient tool in both theoretical and applied trigonometry. In the exercise, this identity was used to find remaining trigonometric values necessary to apply half-angle identities effectively.