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Chapter 4: Problem 80

Complete the equation. $$\arccos \frac{x-2}{2}=\arctan (\square), \quad 2

Short Answer

Expert verified
\(\sqrt{ \frac{4 - (x-2)^2}{(x-2)^2} }\)

Step by step solution

01

Write down the formula for tan(arccos(x))

The formula required is \( \tan(\arccos(x)) = \sqrt{\frac{1 - x^2}{x^2}} \)
02

Apply the formula for tan(arccos(x)) on given equation

The given equation to complete is \( \arccos \frac{x-2}{2}=\arctan (\square) \), \quad 2<x<4. \n \n Thus applying the formula, we get \n \n \( \arctan \left( \sqrt{ \frac{1 - \left(\frac{x-2}{2}\right)^2}{\left(\frac{x-2}{2}\right)^2} } \right) \)
03

Simplifying the expression

\n\n Simplify the expression to the form where it can be filled into the \(\square\) of the original equation: \n \n \(\sqrt{ \frac{4 - (x-2)^2}{(x-2)^2} } \)

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Key Concepts

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

Arccosine Function
The arccosine function, often denoted as \( \arccos \), is the inverse function of the cosine function. This means that it gives the angle whose cosine is a given number. For instance, if \( \cos(\theta) = x \), then \( \arccos(x) = \theta \). This function is essential in trigonometry for finding angles from cosine values.

### Properties of the Arccosine Function
  • **Range:** The arccosine function returns values primarily between \(0\) and \(\pi\) radians (or \(0\)掳 and \(180\)掳).
  • **Domain:** Its domain is restricted to \([-1, 1]\) since cosine values range between these numbers.
  • **Periodic Nature:** Unlike the cosine function, the arccosine function is not periodic and does not repeat its values.

When working with equations involving the arccosine function, like \( \arccos \frac{x-2}{2} \), it's vital to remember its domain and range constraints. This ensures the solutions we derive are valid within the set parameters.
Arctangent Function
The arctangent function, which can be represented as \( \arctan \), is the inverse of the tangent function. It is used to find the angle whose tangent is a given value, such that if \( \tan(\theta) = y \), then \( \arctan(y) = \theta \). Understanding this function helps in solving trigonometric equations that involve the tangent function.

### Properties of the Arctangent Function
  • **Range:** The values of arctangent lie within \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) radians (or \(-90\)掳 to \(90\)掳).
  • **Domain:** It is defined for all real numbers because tangent can take on any value.
  • **Continuous and Smooth:** This function is continuous and has no breaks in its range or domain.

The arctangent function comes into play when completing equations such as \( \arctan(\square) \) by providing a link between quadratic forms and angular solutions. This aids in redefining trigonometric identities in inverse computations.
Trigonometric Identities
Trigonometric identities are equations that are true for all values within the domain of the involved trigonometric functions. They play a critical role in simplifying complex trigonometric expressions and solving equations.

### Important Trigonometric Identities
  • **Pythagorean Identity:** One of the fundamental identities is \( \sin^2(\theta) + \cos^2(\theta) = 1 \). This is often used to derive other identities.
  • **Multiple Angle Identities:** Such as \( \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) \).
  • **Inverse Identities:** For example, combining relationships like \( \tan(\arccos(x)) = \sqrt{\frac{1-x^2}{x^2}} \) helps in solving arcs and angles-related problems.

Understanding these identities is crucial when you encounter expressions that seem complicated at first glance. By applying identities, you can simplify expressions into forms that are easier to manage, such as reducing \( \sqrt{ \frac{1 - \,x^2}{x^2} } \) into more straightforward terms.

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Most popular questions from this chapter

The numbers of hours \(H\) of daylight in Denver, Colorado, on the 15 th of each month are: \(1(9.67), 2(10.72), 3(11.92), 4(13.25)\) \(5(14.37), \quad 6(14.97), \quad 7(14.72), \quad 8(13.77), \quad 9(12.48)\) \(10(11.18), \quad 11(10.00), \quad 12(9.38) . \quad\) The month is represented by \(t,\) with \(t=1\) corresponding to January. A model for the data is $$H(t)=12.13+2.77 \sin \left(\frac{\pi t}{6}-1.60\right)$$. (a) Use a graphing utility to graph the data points and the model in the same viewing window. (b) What is the period of the model? Is it what you expected? Explain. (c) What is the amplitude of the model? What does it represent in the context of the problem? Explain.

When tuning a piano, a technician strikes a tuning fork for the A above middle \(\mathrm{C}\) and sets up a wave motion that can be approximated by \(y=0.001 \sin 880 \pi t,\) where \(t\) is the time (in seconds). (a) What is the period of the function? (b) The frequency \(f\) is given by \(f=1 / p .\) What is the frequency of the note?

Determine whether the statement is true or false. Justify your answer. $$\cot ^{2} 10^{\circ}-\csc ^{2} 10^{\circ}=-1$$

Write an equation for the function that is described by the given characteristics. A cosine curve with a period of \(4 \pi,\) an amplitude of 3 a right phase shift of \(\pi / 2,\) and a vertical translation up 2 units

For the simple harmonic motion described by the trigonometric function, find (a) the maximum displacement, (b) the frequency, (c) the value of \(d\) when \(t=5,\) and (d) the least positive value of \(t\) for which \(d=0 .\) Use a graphing utility to verify your results. $$d=\frac{1}{64} \sin 792 \pi t$$
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