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Chapter 7: Problem 51

Use a graphing calculator to determine which expression \((A)-(F)\) on the right can be used to complete the identity. Then try to prove that identity algebraically. $$\frac{\cos x+\cot x}{1+\csc x}$$ A. \(\frac{\sin ^{3} x-\cos ^{3} x}{\sin x-\cos x}\) B. \(\cos x\) C. \(\tan x+\cot x\) D. \(\cos ^{3} x+\sin ^{3} x\) E. \(\frac{\sin x}{1-\cos x}\) F. \(\cos ^{4} x-\sin ^{4} x\)

Short Answer

Expert verified
B. \(\cos x\)

Step by step solution

01

Analyze the given expression

Consider the expression \(\frac{\cos x+\cot x}{1+\csc x}\) and analyze its components. Rewrite major trigonometric functions in terms of easier functions, such as sine and cosine.
02

Rewrite in terms of sine and cosine

Replace the functions with their equivalents: \(\cot x = \frac{\cos x}{\sin x} \) and \(\csc x = \frac{1}{\sin x}\). This gives: \[ \frac{\cos x + \frac{\cos x}{\sin x}}{1 + \frac{1}{\sin x}} = \frac{\cos x \sin x + \cos x}{\sin x + 1} \].
03

Simplify the expression

Combine like terms in the numerator: \[ \frac{\cos x (1+\sin x)}{1+\sin x} \]. Since the numerator and denominator have the common factor \(1+\sin x\), cancel it out: \[ \cos x \].
04

Compare to the given options

From the simplified expression, conclude that \(\cos x\) is the resulting expression. Compare this with the provided options A-F. Option B matches the simplified expression.
05

Algebraic proof

To confirm the identity algebraically, note that simplifying \(\frac{\cos x + \cot x}{1 + \csc x}\) yields \(\cos x\), which matches Option B without discrepancy.

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Key Concepts

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

Trigonometric Functions
Trigonometric functions are fundamental in math for relating the angles of a triangle to the lengths of its sides. These functions include sine (\(\text{sin}\)), cosine (\(\text{cos}\)), and tangent (\(\text{tan}\)), among others. They are especially useful for analyzing periodic phenomena, such as sound waves and light waves.
In this exercise, we mainly dealt with three key trigonometric functions: cosine (\( \text{cos} x \)), cotangent (\( \text{cot} x \)), and cosecant (\( \text{csc} x \)).
  • Cosine: \( \text{cos} x = \frac{\text{adjacent}}{\text{hypotenuse}} \)
  • Cotangent: \( \text{cot} x = \frac{\text{cos} x}{\text{sin} x} = \frac{\text{adjacent}}{\text{opposite}} \)
  • Cosecant: \( \text{csc} x = \frac{1}{\text{sin} x} = \frac{\text{hypotenuse}}{\text{opposite}} \)
Understanding how to transform and manipulate these functions is crucial for solving trigonometric identities.
For instance, converting cotangent and cosecant in terms of sine and cosine was vital in the step-by-step solution. This made the expression simpler and more straightforward to work with.
Simplifying Expressions
Simplifying expressions in trigonometry often involves rewriting them using fundamental identities. This can make complex equations easier to handle. In the given exercise, we started by rewriting \( \frac{\text{cos} x + \text{cot} x}{1 + \text{csc} x} \) in terms of sine (\text\text{sin} x) and cosine (\text\text{cos} x).
This step helped to streamline the expression:
\[ \frac{\text{cos} x + \frac{\text{cos} x}{\text{sin} x}}{1 + \frac{1}{\text{sin} x}} = \frac{\text{cos} x \text{sin} x + \text{cos} x}{\text{sin} x + 1} \]
After combining like terms, we noticed a common factor in both the numerator and denominator, which allowed us to cancel out terms.
  • Numerator: \( \text{cos} x (1 + \text{sin} x) \)
  • Denominator: \( \text{sin} x + 1 \)
  • Simplified: \( \frac{\text{cos} x (1 + \text{sin} x)}{1 + \text{sin} x} \to \text{cos} x \)

This simplification allowed us to see that the expression simplifies neatly to a basic trigonometric function: \( \text{cos} x \).
Algebraic Manipulation
Algebraic manipulation is the process of rearranging and simplifying algebraic expressions using mathematical operations and properties. In trigonometry, this skill is essential for proving identities and simplifying complex functions.
During this exercise, algebraic manipulation played a key role in transforming the given expression into the simpler form of \( \text{cos} x \). Here are key steps to remember:
  • Substitute complex functions with their simpler equivalents (e.g., \(\text{cot} x \) and \(\text{csc} x \)).
  • Combine like terms to simplify the numerator and denominator.
  • Identify and cancel out common factors.

These steps allowed us to reduce the complex fraction \(\frac{\text{cos} x + \text{cot} x}{1 + \text{csc} x} \) to the simpler \(\text{cos} x \), showing how algebraic manipulations help simplify expressions and make calculations easier.
Mastering these manipulations will improve solving trigonometric problems and proving identities effortlessly.

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Most popular questions from this chapter

The acceleration due to gravity is often denoted by \(g\) in a formula such as \(S=\frac{1}{2} g t^{2},\) where \(S\) is the distance that an object falls in time \(t .\) The number \(g\) relates to motion near the earth's surface and is generally considered constant. In fact, however, \(g\) is not constant, but varies slightly with latitude. Latitude is used to measure north-south location on the earth between the equator and the poles. If \(\phi\) stands for latitude, in degrees, \(g\) is given with good approximation by the formula \(g=9.78049\left(1+0.005288 \sin ^{2} \phi-0.000006 \sin ^{2} 2 \phi\right)\), where \(g\) is measured in meters per second per second at sea level. a) Chicago has latitude \(42^{\circ} \mathrm{N}\). Find \(g .\) b) Philadelphia has latitude \(40^{\circ} \mathrm{N}\). Find \(g\). c) Express \(g\) in terms of \(\sin \phi\) only. That is, eliminate the double angle.

Angles Between Lines. One of the identities gives an easy way to find an angle formed by two lines. Consider two lines with equations \(l_{1}: y=m_{1} x+b_{1}\) and \(l_{2}: y=m_{2} x+b_{2}\) (GRAPH CANNOT COPY) The slopes \(m_{1}\) and \(m_{2}\) are the tangents of the angles \(\theta_{1}\) and \(\theta_{2}\) that the lines form with the positive direction of the \(x\) -axis. Thus we have \(m_{1}=\tan \theta_{1}\) and \(m_{2}=\tan \theta_{2} .\) To find the measure of \(\theta_{2}-\theta_{1},\) or \(\phi,\) we proceed as follows: This formula also holds when the lines are taken in the reverse order. When \(\phi\) is acute, tan \(\phi\) will be positive. When \(\phi\) is obtuse, tan \(\phi\) will be negative. Find the measure of the angle from \(l_{1}\) to \(l_{2}\) $$\begin{aligned} &l_{1}: 2 x+y-4=0\\\ &l_{2}: y-2 x+5=0 \end{aligned}$$

SOLVE. $$\sin ^{-1} x=\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{2}$$

Derive the identity. Check using a graphing calculator. $$\tan \left(x+\frac{\pi}{4}\right)=\frac{1+\tan x}{1-\tan x}$$

Nautical Mile. (See Exercise 60 in Exercise Set 7.2 ) In Great Britain, the nautical mile is defined as the length of a minute of arc of the earth's radius. since the earth is flattened at the poles, a British nautical mile varies with latitude. In fact, it is given, in feet, by the function $$N(\phi)=6066-31 \cos 2 \phi$$ where \(\phi\) is the latitude in degrees. At what latitude north is the length of a British nautical mile found to be \(6040 \mathrm{ft} ?\)
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