/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 Write an algebraic expression th... [FREE SOLUTION] | 91Ó°ÊÓ

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Write an algebraic expression that is equivalent to the given expression. (Hint: Sketch a right triangle, as demonstrated in Example 7.). $$\cot (\arctan x)$$

Short Answer

Expert verified
The algebraic expression equivalent to \( \cot (\arctan x) \) is \( \frac{1}{x} \).

Step by step solution

01

Express Cotangent in terms of Tangent

We know that cotangent can be expressed as the reciprocal of tangent. Therefore, \( \cot x = \frac{1}{\tan x} \). So, \( \cot(\arctan x) = \frac{1}{\tan(\arctan x)} \)
02

Simplify the algebraic expression

Next, recognize that tangent of arctangent of x is simply x. This is by definition of the inverse tangent function. Therefore, \( \frac{1}{\tan(\arctan x)} = \frac{1}{x} \).
03

Present the final answer

Finally, present the solution. The equivalent algebraic expression to \( \cot (\arctan x) \) is \( \frac{1}{x} \).

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Key Concepts

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

Understanding Algebraic Expressions
Algebraic expressions can be thought of as a combination of numbers, variables, and operators that together represent a particular mathematical concept. These expressions are essential in both algebra and trigonometry because they allow us to represent quantities and relationships in a generalized form. In the given exercise, we are translating a trigonometric expression into an algebraic one.
  • The expression \( \cot(\arctan x) \) symbolizes a cotangent function combined with an inverse tangent function, both of which are types of trigonometric functions.
  • By understanding the operations and transformations of these functions, we can simplify and express them as simple algebraic fractions, such as \( \frac{1}{x} \).
To convert such trigonometric identities into algebraic expressions, it's often helpful to break them down into simpler steps, focusing on how each trigonometric and inverse trigonometric function behaves. This technique allows solutions that might not initially be intuitive but can be understood through step-by-step simplification.
Inverse Trigonometric Functions
Inverse trigonometric functions enable us to determine angles when the trigonometric values are known. They essentially 'reverse' the process of the primary trigonometric functions, like sine, cosine, and tangent.
  • The notation \( \arctan(x) \) refers to the inverse of the tangent function, meaning it finds the angle whose tangent is \( x \).
  • In our exercise, we applied this concept by recognizing that \( \tan(\arctan(x)) = x \), a principle that holds true because applying a function and then its inverse results in the original input.
Understanding these inverse functions can simplify complex expressions, eventually allowing us to manipulate equations efficiently. In this solution, this understanding was critical in converting \( \cot(\arctan(x)) \) into \( \frac{1}{x} \), as it relies on knowing how tangent and cotangent relate through their inverse actions.
Right Triangles and Their Importance
Right triangles play a vital role in understanding trigonometric functions, as these functions are originally defined based on the ratios of the sides in a right triangle. This geometrical approach provides a visual framework for comprehending how these functions work.
  • For a right triangle with an angle \( \theta \), the tangent of \( \theta \) is the ratio of the opposite side to the adjacent side.
  • Using a sketch of a right triangle can help visualize and solve expressions involving trig functions, as recommended in the hint of the exercise.
By referencing a right triangle, we can easily see how the reciprocal relationships, like that of \( \tan \) and \( \cot \), function as ratios of different sides. Visualizing these relationships often makes solving algebraic expressions involving trig identities more intuitive. In this exercise, understanding a right triangle's angle and side relationships facilitated finding \( \cot(\arctan x) = \frac{1}{x} \). This is because once the arctangent gives you \( x \), the cotangent provides the reciprocal angle relationship by default.

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Most popular questions from this chapter

Prove each identity. (a) \(\arcsin (-x)=-\arcsin x\) (b) \(\arctan (-x)=-\arctan x\) (c) \(\arctan x+\arctan \frac{1}{x}=\frac{\pi}{2}, \quad x>0\) (d) \(\arcsin x+\arccos x=\frac{\pi}{2}\) (e) \(\arcsin x=\arctan \frac{x}{\sqrt{1-x^{2}}}\)

The normal monthly high temperatures \(H\) (in degrees Fahrenheit) in Erie, Pennsylvania, are approximated by $$H(t)=56.94-20.86 \cos \left(\frac{\pi t}{6}\right)-11.58 \sin \left(\frac{\pi t}{6}\right)$$ and the normal monthly low temperatures \(L\) are approximated by $$L(t)=41.80-17.13 \cos \left(\frac{\pi t}{6}\right)-13.39 \sin \left(\frac{\pi t}{6}\right)$$ where \(t\) is the time (in months), with \(t=1\) corresponding to January (see figure). (Source: National Climatic Data Center) (a) What is the period of each function? (b) During what part of the year is the difference between the normal high and normal low temperatures greatest? When is it smallest? (c) The sun is northernmost in the sky around June \(21,\) but the graph shows the warmest temperatures at a later date. Approximate the lag time of the temperatures relative to the position of the sun.

Graph \(f\) and \(g\) in the same coordinate plane. Include two full periods. Make a conjecture about the functions. $$f(x)=\sin x, \quad g(x)=\cos \left(x-\frac{\pi}{2}\right)$$

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. $$d=\frac{1}{64} \sin 792 \pi t$$

Find two solutions of each equation. Give your answers in degrees \(\left(0^{\circ} \leq \theta<360^{\circ}\right)\) and in radians \((0 \leq \theta<2 \pi) .\) Do not use a calculator. (a) \(\sin \theta=\frac{1}{2}\) (b) \(\sin \theta=-\frac{1}{2}\)

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