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Magazine Chapter 2: Problem 67
Evaluate (a) \(\sin ^{-1}(0.5)\) (b) \(\sin ^{-1}(-0.5)\) (c) \(\cos ^{-1}(0.5)\) (d) \(\cos ^{-1}(-0.5)\) (e) \(\tan ^{-1}(\sqrt{3})\) (f) \(\tan ^{-1}(-\sqrt{3})\)
Short Answer
Expert verified
(a) \( \frac{\pi}{6} \), (b) \( -\frac{\pi}{6} \), (c) \( \frac{\pi}{3} \), (d) \( \frac{2\pi}{3} \), (e) \( \frac{\pi}{3} \), (f) \( -\frac{\pi}{3} \).
Step by step solution
01
Understanding the Arcsin Function
The function \( \sin^{-1}(x) \) represents the inverse sine function and gives the angle whose sine is \( x \). The range of \( \sin^{-1} \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
02
Solving Part (a)
To evaluate \( \sin^{-1}(0.5) \), we need to find an angle \( \theta \) such that \( \sin(\theta) = 0.5 \). In the given range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), \( \theta = \frac{\pi}{6} \) or 30掳. Therefore, \( \sin^{-1}(0.5) = \frac{\pi}{6} \).
03
Solving Part (b)
For \( \sin^{-1}(-0.5) \), we find \( \theta \) such that \( \sin(\theta) = -0.5 \). The angle in the range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) that satisfies this is \( \theta = -\frac{\pi}{6} \). Thus, \( \sin^{-1}(-0.5) = -\frac{\pi}{6} \).
04
Understanding the Arccos Function
The function \( \cos^{-1}(x) \) is the inverse cosine function and provides the angle whose cosine is \( x \). It has a range of \( [0, \pi] \).
05
Solving Part (c)
For \( \cos^{-1}(0.5) \), find \( \theta \) with \( \cos(\theta) = 0.5 \). In the range \( [0, \pi] \), this angle is \( \theta = \frac{\pi}{3} \) or 60掳. Thus, \( \cos^{-1}(0.5) = \frac{\pi}{3} \).
06
Solving Part (d)
To solve \( \cos^{-1}(-0.5) \), find \( \theta \) such that \( \cos(\theta) = -0.5 \) within the range \( [0, \pi] \). The angle is \( \theta = \frac{2\pi}{3} \). Therefore, \( \cos^{-1}(-0.5) = \frac{2\pi}{3} \).
07
Understanding the Arctan Function
The function \( \tan^{-1}(x) \) represents the inverse tangent, giving the angle whose tangent is \( x \). Its range is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).
08
Solving Part (e)
For \( \tan^{-1}(\sqrt{3}) \), we need the angle \( \theta \) such that \( \tan(\theta) = \sqrt{3} \). Angle \( \theta = \frac{\pi}{3} \) satisfies this requirement because \( \tan(\frac{\pi}{3}) = \sqrt{3} \). Thus, \( \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \).
09
Solving Part (f)
To evaluate \( \tan^{-1}(-\sqrt{3}) \), identify the angle \( \theta \) that satisfies \( \tan(\theta) = -\sqrt{3} \). This angle is \( \theta = -\frac{\pi}{3} \) within the range \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). Therefore, \( \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Arcsin
The arcsine function, denoted as \( \sin^{-1}(x) \), is the inverse of the sine function and it's used to determine the angle whose sine is a given number. This function is restricted to the principal values, so it only outputs angles in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\). This means any angle found using the arcsine function will always lie between \(-90掳\) and \(90掳\).
For example:
For example:
- To find \( \sin^{-1}(0.5) \), look for an angle whose sine value is 0.5. That angle is \( \frac{\pi}{6} \) or 30掳, as \( \sin(30掳) = 0.5 \).
- Similarly, when you evaluate \( \sin^{-1}(-0.5) \), you find the angle whose sine is -0.5. Within our range, the angle is \(-\frac{\pi}{6} \), as \( \sin(-30掳) = -0.5 \).
Understanding Arccos
Arccos, denoted \( \cos^{-1}(x) \), is the inverse function of cosine. It helps to find the angle whose cosine is equal to a given value, commonly used in geometry and trigonometry problems. Unlike the arcsine function, arccos returns angles in the range \([0, \pi]\) or \([0掳, 180掳]\). This ensures you always find the principal angle which is the simplest angle measured counterclockwise from the x-axis to the line segment.
Consider these calculations:
Consider these calculations:
- To compute \( \cos^{-1}(0.5) \), the angle with a cosine of 0.5 is \( \frac{\pi}{3} \) or 60掳, because \( \cos(60掳) = 0.5 \).
- For \( \cos^{-1}(-0.5) \), the angle whose cosine equals -0.5 falls in the range and is \( \frac{2\pi}{3} \) or 120掳. This is because \( \cos(120掳) = -0.5 \).
Understanding Arctan
The arctangent function \( \tan^{-1}(x) \), offers the angle whose tangent is a particular value. It is particularly indispensable for problems involving slopes and angles, notably in calculus and physics. The range of \( \tan^{-1}(x) \) is \((-\frac{\pi}{2}, \frac{\pi}{2})\), which means the output angle is always between \(-90掳\) and \(90掳\) exclusive.
Let's examine these example calculations:
Let's examine these example calculations:
- When evaluating \( \tan^{-1}(\sqrt{3}) \), the angle you seek is \( \frac{\pi}{3} \) or 60掳, to satisfy \( \tan(60掳) = \sqrt{3} \).
- For \( \tan^{-1}(-\sqrt{3}) \), locate the angle where tangent equals \(-\sqrt{3}\). This comes out to \(-\frac{\pi}{3} \) or -60掳, fulfilling \( \tan(-60掳) = -\sqrt{3} \).
