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Chapter 6: Problem 9

Find the exact value of the expression whenever it is defined. (a) \(\sin ^{-1}\left(\sin \frac{\pi}{3}\right)\) (b) \(\cos ^{-1}\left[\cos \left(\frac{5 \pi}{6}\right)\right]\) (c) \(\tan ^{-1}\left[\tan \left(-\frac{\pi}{6}\right)\right]\)

Short Answer

Expert verified
(a) \(\frac{\pi}{3}\); (b) \(\frac{5\pi}{6}\); (c) \(-\frac{\pi}{6}\).

Step by step solution

01

Understanding Inverse Sine

The function \( ext{sin}^{-1}(x)\) or \( ext{arcsin}(x)\) gives us the angle \(y\) such that \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\) and \(\sin(y) = x\). For \(\text{sin}^{-1}(\sin(x))\), this simplifies to \(x\) if \(x\) is within \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
02

Evaluating Part (a)

For the expression \(\sin^{-1}(\sin \frac{\pi}{3})\), since \(\frac{\pi}{3}\) is within the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\), the function simplifies directly to \(\frac{\pi}{3}\).
03

Understanding Inverse Cosine

The function \(\text{cos}^{-1}(x)\) or \(\text{arccos}(x)\) gives the angle \(y\) such that \(0 \leq y \leq \pi\) and \(\cos(y) = x\). For \(\text{cos}^{-1}(\cos(x))\), this simplifies to \(x\) if \(x\) is within \([0, \pi]\).
04

Simplifying \(\frac{5\pi}{6}\) in Part (b)

\(\frac{5\pi}{6}\) is within the principal range \([0, \pi]\) of the inverse cosine function, so the expression \(\cos^{-1}(\cos(\frac{5\pi}{6}))\) simplifies directly to \(\frac{5\pi}{6}\).
05

Understanding Inverse Tangent

The function \(\text{tan}^{-1}(x)\) or \(\text{arctan}(x)\) yields the angle \(y\) such that \(-\frac{\pi}{2} < y < \frac{\pi}{2}\) and \(\tan(y) = x\). For \(\text{tan}^{-1}(\tan(x))\), this simplifies to \(x\) when \(x\) is already within \((-\frac{\pi}{2}, \frac{\pi}{2})\).
06

Evaluating Part (c)

Since \(-\frac{\pi}{6}\) is within the principal range \((-\frac{\pi}{2}, \frac{\pi}{2})\) of the inverse tangent function, the expression \(\tan^{-1}(\tan(-\frac{\pi}{6}))\) simplifies directly to \(-\frac{\pi}{6}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Arcsin
The inverse sine function, known as arcsin or \(\sin^{-1}\), is a fundamental aspect of trigonometry. It essentially reverses the sine function. This means it takes a value from the sine function's range and provides an angle whose sine is that value.

Some important aspects to remember about arcsin are:
  • The range of arcsin is \([-\frac{\pi}{2}, \frac{\pi}{2}]\), which ensures we get only one value of angle for each sine value.
  • When evaluating expressions like \(\sin^{-1}(\sin(x))\), remember it simplifies to \(x\) only when \(x\) is within the range of \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

For example, in the expression \(\sin^{-1}(\sin(\frac{\pi}{3}))\), the angle \(\frac{\pi}{3}\) is comfortably within the range of arcsin. Therefore, the result simplifies easily to \(\frac{\pi}{3}\) without any additional steps. This functionality stems from the fact that arcsin directly undoes the sine function within its specified range.
Arccos
Arccosine, or \(\cos^{-1}\), is the inverse function of cosine. Arccos takes a cosine value and provides the angle whose cosine is that value. This inverse function is crucial for solving trigonometric equations involving cosine.

Several key points to note about arccos are:
  • The range for \(\cos^{-1}(x)\) is \([0, \pi]\). This restricts the output to angles that can provide a unique cosine value.
  • An expression like \(\cos^{-1}(\cos(x))\) will yield \(x\) if \(x\) is situated within \([0, \pi]\).

In the case of the expression \(\cos^{-1}(\cos(\frac{5\pi}{6}))\), the value \(\frac{5\pi}{6}\) falls within the necessary range \([0, \pi]\) of arccos. Thus, the calculation straightforwardly simplifies to \(\frac{5\pi}{6}\), as the original angle is within the "acceptable" zone of the arccos function.
Arctan
The inverse tangent function, also known as arctan or \(\tan^{-1}\), is used to find an angle whose tangent is a given number. Arctan helps in determining angles corresponding to particular tangent values.

Essential features of arctan include:
  • The range of \(\tan^{-1}(x)\) is \((-\frac{\pi}{2}, \frac{\pi}{2})\); this means the possible resulting angles fall between these values.
  • For expressions like \(\tan^{-1}(\tan(x))\), the solution simplifies to \(x\) when \(x\) lies within \((-\frac{\pi}{2}, \frac{\pi}{2})\).

For \(\tan^{-1}(\tan(-\frac{\pi}{6}))\), the angle \(-\frac{\pi}{6}\) is already within the range required by arctan. This allows the expression to easily reduce to \(-\frac{\pi}{6}\), as it sits comfortably in the desired range for the arctan function, highlighting the inverse nature.

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Most popular questions from this chapter

Express in terms of the cosine function with exponent 1. $$\sin ^{4} \frac{\theta}{2}$$

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The weight \(W\) of a person on the surface of Earth is directly proportional to the force of gravity \(g\) (in \(\mathrm{m} / \mathrm{sec}^{2}\) ). Because of rotation, Earth is flattened at the poles, and as a result weight will vary at different latitudes. If \(\theta\) is the latitude, then \(g\) can be approximated by \(g=9.8066(1-0.00264 \cos 2 \theta)\) (a) At what latitude is \(g=9.8 ?\) (b) If a person weighs 150 pounds at the equator \(\left(\theta=0^{\circ}\right)\) at what latitude will the person weigh 150.5 pounds?

Use an addition or subtraction formula to find the solutions of the equation that are in the interval \([0, \pi)\). $$\cos 5 t \cos 2 t=-\sin 5 t \sin 2 t$$
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