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Chapter 6: Problem 23

Write the expression as an algebraic expression in \(x\) for \(x>0\). $$\sin \left(\tan ^{-1} x\right)$$

Short Answer

Expert verified
\( \frac{x}{\sqrt{x^2 + 1}} \)

Step by step solution

01

Understanding the Relationship

First, let's consider the inverse tangent function, \( \tan^{-1}(x) \), which is also known as arctan. This function returns the angle \( \theta \) such that \( \tan(\theta) = x \).
02

Setting Up the Right Triangle

Visualize this situation as a right triangle where \( \theta \) is one of the angles. Recall that \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \). In this scenario, \( \tan(\theta) = x = \frac{x}{1} \), so the opposite side is \( x \) and the adjacent side is \( 1 \).
03

Finding the Hypotenuse

To find the hypotenuse of the triangle, use the Pythagorean Theorem: \( c = \sqrt{x^2 + 1^2} = \sqrt{x^2 + 1} \). Thus, the hypotenuse of the triangle is \( \sqrt{x^2 + 1} \).
04

Expressing \(\sin(\theta)\) in Terms of \(x\)

The sine function is given by \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \). So, \( \sin(\tan^{-1}(x)) = \frac{x}{\sqrt{x^2 + 1}} \). This is the algebraic expression in \( x \) that we were asked to find.

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Key Concepts

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

Inverse Trigonometric Functions
Inverse trigonometric functions, like \( \tan^{-1}(x) \), are crucial in connecting geometric angles to specific ratios. They take a ratio and return the corresponding angle. For example, \( \tan^{-1}(x) \) gives the angle \( \theta \) whose tangent is \( x \). These functions are often called arctangent, arcsine, and arccosine in various contexts.
  • Arctangent: \( \tan^{-1}(x) \) relates to the angle of a right triangle, such that \( \tan(\theta) = x \).
  • Range: Inverse trigonometric functions have restricted ranges to be defined properly. For \( \tan^{-1}(x) \), the range is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).
  • Application: These functions are vital in calculus and analytical geometry, assisting in integration, differentiation, and solving equations involving angles.
Understanding inverse trigonometric functions not only helps in trigonometry problems but also enhances the grasp of mathematical modeling involving angles and real-world situations.
Pythagorean Theorem
The Pythagorean Theorem is a fundamental principle in geometry that connects the sides of a right-angled triangle. It is expressed as \( a^2 + b^2 = c^2 \), where \( c \) is the hypotenuse, the longest side of the triangle, and \( a \) and \( b \) are the other two sides.
  • Hypotenuse Calculation: In the context of \( \tan^{-1}(x) \), if we form a right triangle with legs \( x \) and \( 1 \), the hypotenuse becomes \( \sqrt{x^2 + 1} \) through this theorem.
  • Applications: Used in various fields like physics, engineering, and architecture, it forms the basis for trigonometric identities and relationships.
This theorem is not only pivotal to solving geometric problems but also underpins key trigonometric identities and transformations in advanced mathematics. It aids in visualizing and computing relationships in right triangles and understanding spatial dimensions.
Sine Function
The sine function, noted as \( \sin(\theta) \), is one primary trigonometric function used to relate an angle of a right triangle to the lengths of its sides. Specifically, it is defined as the ratio of the length of the side opposite the angle to the hypotenuse.
  • Basic Formula: \( \sin(\theta) = \frac{\text{opposite side}}{\text{hypotenuse}} \).
  • In Context: In the example, \( \sin(\tan^{-1}(x)) = \frac{x}{\sqrt{x^2 + 1}} \), derived from the constructed triangle with opposite \( x \) and hypotenuse \( \sqrt{x^2 + 1} \).
  • Sine Wave: The sine function is also known for creating periodic sine waves, which are essential in studying waves, oscillations, and electrical circuits.
Understanding sine is crucial for comprehending wave mechanics, harmonic motion in physics, and solving problems involving periodic phenomena.

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Most popular questions from this chapter

Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. \(15 \cos ^{4} x-14 \cos ^{2} x+3=0\) \([0, \pi]\)

Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. $$3 \tan ^{4} \theta-19 \tan ^{2} \theta+2=0 ; \quad\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$

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 the graph of \(f\) to find the simplest expression \(g(x)\) such that the equation \(f(x)=g(x)\) is an Identity. Verify this identity. $$f(x)=\sec x\left(\sin x \cos x+\cos ^{2} x\right)-\sin x$$

Many calculators have viewing sereens that are wider than they are high. The approximate ratio of the height to the width is often \(2: 3 .\) Let the actual height of the calculator screen along the \(y\) -axis be 2 units, the actual width of the calculator screen along the \(x\) -axis be 3 units, and Xscl \(=\mathbf{Y s c}=1 .\) since the line \(y=x\) must pass through the point \((1,1),\) the actual slope \(m_{A}\) of this line on the calcuIator screen is given by $$m_{A}=\frac{\text { actual distance between tick marks on } y \text { -axis }}{\text { actual distance between tick marks on } x \text { -axis }}$$ Using this information, graph \(y=x\) in the given viewing rectangle and predict the actual angle \(\boldsymbol{\theta}\) that the graph makes with the \(x\) -axis on the viewing screen. \([0,3]\) by \([0,2]\)
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