Problem 10 - Algebra and Trigonometry

Question

\(3-10=\) Find the exact value of each expression, if it is defined. $$ \begin{array}{llll}{\text { (a) } \cos ^{-1} 0} & {} & {\text { (b) } \sin ^{-1} 0} & {} & {\text { (e) } \sin ^{-1}\left(-\frac{1}{2}\right)}\end{array} $$

Step-by-Step Solution

Answer

(a) \( \frac{\pi}{2} \); (b) 0; (e) \(-\frac{\pi}{6}\).

Step 1: Determine the value of (a)

We need to find \( \cos^{-1}(0) \). The cosine inverse of 0 corresponds to the angle whose cosine is 0. Cosine is 0 at angles \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \). However, \( \cos^{-1} \) is defined in the range \([0, \pi]\), so the correct value is \( \frac{\pi}{2} \).

Step 2: Determine the value of (b)

For \( \sin^{-1}(0) \), we look for the angle whose sine is 0. Sine is 0 at angles 0, \( \pi \), 2\( \pi \), etc. However, the \( \sin^{-1} \) function is defined in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\). Therefore, \( \sin^{-1}(0) \) is 0.

Step 3: Determine the value of (e)

We find \( \sin^{-1}\left(-\frac{1}{2}\right) \). This represents the angle whose sine is \(-\frac{1}{2}\). Sine of \(-\frac{1}{2}\) is found at \(-\frac{\pi}{6}\) when considering the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\) for the inverse function. Thus, \( \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \).

Key Concepts

Understanding Cosine Inverse

The cosine inverse function, denoted as \( \cos^{-1} \) or \( \text{arccos} \), helps us find the angle whose cosine value is given. In simple terms, if you know the cosine of an angle, cosine inverse allows you to discover the actual angle. It's like looking at a footprint and figuring out who made it.
In mathematical terms, if \( y = \cos^{-1}(x) \), then \( \cos(y) = x \). Here, \( y \) is the angle in radians that ranges between \( 0 \) and \( \pi \). This is crucial because:

• Cosine inverse has its defined range from 0 to \( \pi \), which means any angle outside of this will not be considered by \( \cos^{-1} \).
• For instance, when you see \( \cos^{-1}(0) \), think about angles where cosine equals 0. These are \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \). However, only \( \frac{\pi}{2} \) is within our range of \([0, \pi]\).

Cosine inverse is a handy function when working with right triangles or problems needing angle determination. While use can seem complex, it simplifies finding the correct angle when dealing with cosine.

Exploring Sine Inverse

The sine inverse function, written as \( \sin^{-1} \) or \( \text{arcsin} \), helps determine the angle from a given sine value. If the sine of an angle is known, sine inverse will reveal that angle.

• Mathematically, if \( y = \sin^{-1}(x) \), then we have \( \sin(y) = x \).
• The sine inverse function's range is from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \). This means possible angles are limited between \(-90^\circ\) and \(90^\circ\).

This predefined range is key. For example, in the problem \( \sin^{-1}(0) \), you seek angles where sine equals 0. Commonly, these are angles like 0, \( \pi \), 2\( \pi \), and etc., but only 0 is within our specified range.
Similarly, if \( \sin^{-1}\left(-\frac{1}{2}\right) \) is needed, observe that its solution must lie within \([ -\frac{\pi}{2}, \frac{\pi}{2}] \) and correctly equates to \(-\frac{\pi}{6} \). Sine inverse ensures that only these limited angles are picked, making it especially useful for equations and trigonometric applications.

The Range of Inverse Functions

Understanding the range of inverse functions is essential when analyzing them, as it dictates which angles qualify as solutions.

• The range indicates the set of all possible outputs for the inverse function. Each trigonometric function has a specific range to ensure unique outputs.
• For \( \cos^{-1} \), the range is \([0, \pi] \). This is because cosine is an even function, symmetrically repeating, and thus the defined range limits angles to only upper semicircular values.
• The \( \sin^{-1} \) function operates within the range \([-\frac{\pi}{2}, \frac{\pi}{2}] \), allowing only angles in the lower and upper quadrants, matching sine's odd nature. This ensures distinctness.

By maintaining these specific ranges, inverse functions avoid ambiguity seen when dealing with periodic trigonometric functions. These ranges are guideposts that ensure each function delivers a single, meaningful result. Mastering these ranges not only helps in solving equations correctly but also in recognizing the necessary angle transformations in various trigonometric scenarios.