Problem 8 - Algebra and Trigonometry

Question

Find the exact value of each expression, if it is defined. (a) \(\sin ^{-1} 0\) (b) \(\cos ^{-1} 0\) (c) \(\cos ^{-1}\left(-\frac{1}{2}\right)\)

Step-by-Step Solution

Answer

(a) 0, (b) \(\frac{\pi}{2}\), (c) \(\frac{2\pi}{3}\).

Step 1: Understanding Inverse Sine Function

The function \( ext{sin}^{-1}\) or arcsin is the inverse of the sine function. Its range is \([-\frac{\pi}{2}, \frac{\pi}{2}]\), meaning it outputs angles in this specific range. We are asked to find \(\sin^{-1}(0)\).

Step 2: Calculating \(\sin^{-1}(0)\)

To find \(\sin^{-1}(0)\), we need to determine the angle in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\) whose sine is 0. The sine of 0 is 0.

Step 3: Result for \(\sin^{-1}(0)\)

So, \(\sin^{-1}(0) = 0\). The angle 0 is within the range of \([-\frac{\pi}{2}, \frac{\pi}{2}]\) and satisfies \(\sin(0) = 0\).

Step 4: Understanding Inverse Cosine Function

The function \(\text{cos}^{-1}\) or arccos is the inverse of the cosine function. Its range is \([0, \pi]\), meaning it gives angles within this interval. We need to find \(\cos^{-1}(0)\).

Step 5: Calculating \(\cos^{-1}(0)\)

For \(\cos^{-1}(0)\), we seek an angle whose cosine is 0 within the range \([0, \pi]\). The cosine of \(\frac{\pi}{2}\) is 0.

Step 6: Result for \(\cos^{-1}(0)\)

Thus, \(\cos^{-1}(0) = \frac{\pi}{2}\). The angle \(\frac{\pi}{2}\) lies within \([0, \pi]\) and satisfies \(\cos(\frac{\pi}{2}) = 0\).

Step 7: Calculating \(\cos^{-1}ig(-\frac{1}{2}\big)\)

We need to find the angle in \([0, \pi]\) whose cosine is \(-\frac{1}{2}\). The cosine of \(\frac{2\pi}{3}\) is \(-\frac{1}{2}\).

Step 8: Result for \(\cos^{-1}ig(-\frac{1}{2}\big)\)

Hence, \(\cos^{-1}\big(-\frac{1}{2}\big) = \frac{2\pi}{3}\). The angle \(\frac{2\pi}{3}\) is within \([0, \pi]\) and satisfies \(\cos(\frac{2\pi}{3}) = -\frac{1}{2}\).

Key Concepts

Arcsin Function

The arcsin function, also known as the inverse sine function, is a crucial concept in trigonometry. It reverses the sine function, essentially answering the question: "What angle has a given sine value?" For arcsin, we denote it as \(\sin^{-1}(x)\).

The range of arcsin is \([-\frac{\pi}{2}, \frac{\pi}{2}]\). This means the angles it provides as output are always within this interval.

• For example, when you need to find \(\sin^{-1}(0)\), you are looking for the angle between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\) whose sine is 0.

The angle 0 fits this requirement, as \(\sin(0) = 0\), and it's clearly within the permitted range.

Arccos Function

The arccos function is the inverse of the cosine function, often represented as \(\cos^{-1}(x)\). When you use arccos, you're trying to find the angle that, when the cosine is applied, equals \(x\).
The range for arccos is \([0, \pi]\), delivering results within this range.

• To illustrate, finding \(\cos^{-1}(0)\) asks for the angle where cosine equals 0 within \([0, \pi]\).
• The angle \(\frac{\pi}{2}\) fulfills this, as \(\cos(\frac{\pi}{2}) = 0\).

Similarly, for \(\cos^{-1}(-\frac{1}{2})\), you seek an angle where cosine is \(-\frac{1}{2}\), which is \(\frac{2\pi}{3}\). Each of these results stays comfortably within the specified range.

Trigonometric Identities

Trigonometric identities are formulas that relate the angles and sides of a triangle using trigonometric functions.
These are powerful tools in simplifying and solving various mathematical problems, including evaluating inverse trigonometric functions.

• To appreciate their usefulness, consider the Pythagorean identity: \(\sin^2(x) + \cos^2(x) = 1\), which is fundamental when proving other identities.
• Another example is the complementary angle identity, \(\sin(\pi/2 - x) = \cos(x)\), which helps in understanding arcsin and arccos relationships.

Using trigonometric identities can simplify complex expressions and aid in finding exact values for inverse functions, like the arcsin and arccos.