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Chapter 1: Problem 49

(a) \(\cot 11^{\circ} 15^{\prime}\) (b) \(\tan 11^{\circ} 15^{\prime}\)

Short Answer

Expert verified
The cotangent of \(11^{\circ} 15^{\prime}\) is approximately 4.70463. The tangent of \(11^{\circ} 15^{\prime}\) is approximately 0.21256.

Step by step solution

01

Convert Degrees-Minutes to Decimal Degrees

To convert from degrees, minutes, and seconds to decimal degrees, use the formula: \[Degree + \frac{Minute}{60}\] So for \(11^{\circ} 15^{\prime}\), it would be: \[11 + \frac{15}{60} = 11.25^{\circ}\]
02

Calculate \(\cot 11.25^\circ\)

The cotangent of an angle can be defined as the reciprocal of the tangent. Using a calculator, one can compute: \[\cot 11.25^\circ = \frac{1}{\tan 11.25^\circ}\] Compute \(\cot 11.25^\circ\) to get the numerical value.
03

Calculate \(\tan 11.25^\circ\)

The tangent of an angle can be directly calculated using a calculator. So, compute \(\tan 11.25^\circ\) to get the numerical value.

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Key Concepts

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

Cotangent of an Angle
The cotangent of an angle in trigonometry is essentially the reciprocal of the tangent of that angle. If you know the tangent value, you can find the cotangent by taking its reciprocal, i.e., by calculating \( \frac{1}{\tan(\theta)} \), where \( \theta \) represents the angle measure in degrees.

To calculate the cotangent of an angle, usually a calculator is employed, ensuring it is set to degree mode.
  • The cotangent is undefined when the tangent of the angle is 0, leading to division by zero.
  • In the context of a right triangle, cotangent is defined as the length of the adjacent side divided by the opposite side.
Understanding the role of cotangent in trigonometry helps with solving problems where angles and side ratios are crucial, often encountered in advanced math, physics, and engineering disciplines.
Tangent of an Angle
The tangent of an angle in a right triangle is a fundamental concept in trigonometry. It relates the ratio of the opposite side to the adjacent side. Mathematically, it is expressed as \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \).

Tangent also applies to angles measured in standard position on the coordinate plane.
  • Available on scientific calculators, it can be critical for solving angles in triangles when one side length and an angle are known.
  • It is a periodic function with a period of \(180^\circ\), which means every \(180^\circ\) the pattern repeats.
Understanding the tangent function is key when working with non-right triangles or when solving trigonometric identities, offering a direct link to practical applications such as slopes and angles of elevation.
Decimal Degrees
Decimal degrees are a way to represent angles that makes calculations and comprehension easier, compared to the traditional degrees, minutes, and seconds (DMS) system. When converting to decimal degrees, it uses the formula:

\[Degree + \frac{Minute}{60} + \frac{Second}{3600}\] This allows smoother and error-free computations on calculators and computers.
  • For day-to-day calculations, especially when precision is necessary, decimal degrees are much more straightforward.
  • They are a mainstay in navigation systems and geographic information systems (GIS).
By converting angles from DMS to decimal degrees, it simplifies the use of advanced mathematical functions and programming tools designed to handle consistent numerical data, rather than mixed angle formats.

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Most popular questions from this chapter

$$ \text { In Exercises 49-58, find the exact value of the expression. } $$ $$ \cot \left(\arctan \frac{5}{8}\right) $$

Use a graphing utility to graph the function. $$ f(x)=-\arcsin 2 x $$

Write an algebraic expression that is equivalent to the expression. $$ \sec [\arcsin (x-1)] $$

In Exercises 19-34, use a calculator to evaluate the expression. Round your result to two decimal places. $$ \arctan (-3) $$

Consider the functions given by \(f(x)=\sin x\) and \(f^{-1}(x)=\arcsin x .\) (a) Use a graphing utility to graph the composite functions \(f \circ f^{-1}\) and \(f^{-1} \circ f\). (b) Explain why the graphs in part (a) are not the graph of the line \(y=x\). Why do the graphs of \(f \circ f^{-1}\) and \(f^{-1} \circ f\) differ?
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