Problem 1 - Precalculus: Functions and Graphs

Question

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

Step-by-Step Solution

Answer

(a) \(-\frac{\pi}{4}\), (b) \(\frac{2\pi}{3}\), (c) \(-\frac{\pi}{3}\).

Step 1: Understanding Inverse Sine Function

The function \( ext{sin}^{-1}(x)\) is defined as taking a value \(x\) from \(-1\) to \(1\) and returning an angle \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\) such that \(\sin(y) = x\).

Step 2: Finding the Angle for Part (a)

We are given \(x = -\frac{\sqrt{2}}{2}\). We know \(\sin(-\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}\). Thus, \(\sin^{-1}(-\frac{\sqrt{2}}{2}) = -\frac{\pi}{4}\).

Step 3: Understanding Inverse Cosine Function

The function \( ext{cos}^{-1}(x)\) is defined as taking a value \(x\) from \(-1\) to \(1\) and returning an angle \(0 \leq y \leq \pi\) such that \(\cos(y) = x\).

Step 4: Finding the Angle for Part (b)

We need to find an angle \(y\) where \(\cos(y) = -\frac{1}{2}\). The cosine function is \(-\frac{1}{2}\) at \(y = \frac{2\pi}{3}\) in this range. Therefore, \(\cos^{-1}(-\frac{1}{2}) = \frac{2\pi}{3}\).

Step 5: Understanding Inverse Tangent Function

The function \( an^{-1}(x)\) takes any real number \(x\) and returns an angle \(-\frac{\pi}{2} < y < \frac{\pi}{2}\) such that \(\tan(y) = x\).

Step 6: Finding the Angle for Part (c)

Given \(x = -\sqrt{3}\), \(\tan(-\frac{\pi}{3}) = -\sqrt{3}\). Thus, \(\tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}\).

Key Concepts

Inverse Sine Function

The inverse sine function, denoted as \( \sin^{-1}(x) \), is used to find the angle whose sine is a given number \( x \), within specific bounds. The range of \( \sin^{-1}(x) \) is between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), which means that it includes all the angles from the fourth quadrant upwards to the first quadrant. These angles are important because they represent the only possible solutions where the sine value can be equal to negative or positive numbers within the interval \(-1\) to \(1\).

To solve for the inverse sine of \(-\frac{\sqrt{2}}{2}\), we look for an angle within this range whose sine is \(-\frac{\sqrt{2}}{2}\). The angle that satisfies \( \sin(y) = -\frac{\sqrt{2}}{2} \) is \(-\frac{\pi}{4}\). This is because, in the fourth quadrant, the sine is negative, and \(\pi/4\) is the reference angle for \(\sqrt{2}/2\). When flipped to the fourth quadrant, we find \( y = -\frac{\pi}{4} \).

• Remember, sine is negative in the third and fourth quadrants.
• Inverse sine is used to "undo" the sine function to find the angle.

Inverse Cosine Function

The inverse cosine function, noted as \( \cos^{-1}(x) \), helps in determining the angle whose cosine is a specific number \( x \). Its range is from \(0\) to \(\pi\), covering angles in the first and second quadrants. The inverse cosine gives a unique angle from this range, making it easier for calculations and avoiding ambiguity.

In the exercise, for \( \cos^{-1}(-\frac{1}{2}) \), we're tasked to find an angle \( y \) within \(0\) to \(\pi\) with a cosine of \(-\frac{1}{2}\). Since cosine is negative in the second quadrant, we know \(-\frac{1}{2}\) is found at the angle \( \frac{2\pi}{3} \). This means the cosine of \( \frac{2\pi}{3} \) equals \(-\frac{1}{2}\), allowing us to conclude that \( \cos^{-1}(-\frac{1}{2}) = \frac{2\pi}{3} \).

• Cosine is positive in the first quadrant but negative in the second.
• Inverse cosine efficiently provides a definitive angle without confusion.

Inverse Tangent Function

The inverse tangent function, expressed as \( \tan^{-1}(x) \), is essential for finding the angle whose tangent equals a given real number \( x \). This function works within the range \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), meaning it only considers angles from the fourth quadrant up to the first quadrant. These constraints ensure precision and prevent overlap with other possible angle solutions.

In the problem involving \( \tan^{-1}(-\sqrt{3}) \), we're finding an angle \( y \) in the specified range that satisfies \( \tan(y) = -\sqrt{3} \). We know that \(\tan(\frac{\pi}{3}) = \sqrt{3}\), and since we need a negative value, \( y = -\frac{\pi}{3} \) is the solution. Here, \(\tan(y)\) is negative in the fourth quadrant, giving us \(-\sqrt{3}\) for \( y = -\frac{\pi}{3} \).

• Recall that tangent is positive in the first and third quadrants, but negative in the second and fourth.
• Reverse tangent function can handle all real number inputs due to its nature.