Problem 4 - Algebra and Trigonometry

Question

Find the exact value of each expression, if it is defined. (a) \(\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right) \quad\) (b) \(\cos ^{-1}\left(-\frac{\sqrt{2}}{2}\right) \quad\) (c) \(\tan ^{-1}(-\sqrt{3})\)

Step-by-Step Solution

Answer

(a) \\( -\frac{\pi}{3} \\); (b) \\( \frac{3\pi}{4} \\); (c) \\( -\frac{\pi}{3} \\).

Step 1: Understanding the Problem

We are given three inverse trigonometric problems to solve. We need to find the angles whose sine, cosine, and tangent values are the numbers given. The angles should be within the principal range of each inverse function.

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

The principal value range for \( \sin^{-1}(x) \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). We need an angle \( \theta \) such that \( \sin(\theta) = -\frac{\sqrt{3}}{2} \). The angle in this range where the sine value is \( -\frac{\sqrt{3}}{2} \) is \( -\frac{\pi}{3} \).

Step 3: Solve Part (b) \\( \cos^{-1}\left(-\frac{\sqrt{2}}{2}\right) \\\)

The principal value range for \( \cos^{-1}(x) \) is \( [0, \pi] \). We need an angle \( \theta \) such that \( \cos(\theta) = -\frac{\sqrt{2}}{2} \). The angle in this range where the cosine value is \( -\frac{\sqrt{2}}{2} \) is \( \frac{3\pi}{4} \).

Step 4: Solve Part (c) \\( \tan^{-1}(-\sqrt{3}) \\\)

The principal value range for \( \tan^{-1}(x) \) is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). We need an angle \( \theta \) such that \( \tan(\theta) = -\sqrt{3} \). The angle in this range where the tangent value is \( -\sqrt{3} \) is \( -\frac{\pi}{3} \).

Key Concepts

Sine Function

The sine function maps angles to their opposite side ratios in a right triangle. In inverse sine, denoted as \( \sin^{-1}(x) \) or \( \arcsin(x) \), we seek an angle given its sine. The principal value range for \( \sin^{-1}(x) \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), meaning any angle outside this is adjusted back into the range.

• Given: \( \sin^{-1}(-\frac{\sqrt{3}}{2}) \)
• We know \( \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \)
• Because sine is odd, \( \sin(-\theta) = -\sin(\theta) \)
• Thus, \( \sin^{-1}(-\frac{\sqrt{3}}{2}) = -\frac{\pi}{3} \)

Sine values tell us about vertical positioning on the unit circle. For inverse sine, a negative input indicates the angle below the x-axis or reflective across the origin within the permissible interval.

Cosine Function

The cosine function returns the ratio of the adjacent side to the hypotenuse. In inverse cosine, represented as \( \cos^{-1}(x) \) or \( \arccos(x) \), one finds an angle whose cosine is \( x \). It ranges from \( [0, \pi] \), covering the first and second quadrants of the unit circle.

• Given: \( \cos^{-1}(-\frac{\sqrt{2}}{2}) \)
• Recognize that \( \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \)
• Hence, \( \cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2} \)
• Therefore, \( \cos^{-1}(-\frac{\sqrt{2}}{2}) = \frac{3\pi}{4} \)

Cosine reflects horizontal positioning. A negative cosine value implies the angle sits in the second quadrant. The principal range ensures the output is always an angle from these two quadrants.

Tangent Function

The tangent function is the ratio of the sine and cosine values. Its inverse, \( \tan^{-1}(x) \) or \( \arctan(x) \), finds the angle such that the tangent is \( x \). The principal range is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), ensuring results are in the first and fourth quadrants.

• Given: \( \tan^{-1}(-\sqrt{3}) \)
• We know \( \tan(\frac{\pi}{3}) = \sqrt{3} \)
• Tangent being odd implies \( \tan(-\theta) = -\tan(\theta) \)
• Thus, \( \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \)

The tangent spans all real numbers vertically, which is why the principal range is asymmetric, allowing a negative value to fall in the fourth quadrant, showing angles below the x-axis.