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91Ó°ÊÓ

Write expression in terms of sine and cosine, and simplify it. (The final expression does not have to be in terms of sine and cosine.) $$(1-\cos \theta)(1+\sec \theta)$$

Short Answer

Expert verified
The expression simplifies to \(-\tan \theta\).

Step by step solution

01

Expand the Expression

Start by expanding the given expression \((1 - \cos \theta)(1 + \sec \theta)\). This results in two terms: \((1)(1 + \sec \theta) - (\cos \theta)(1 + \sec \theta)\).
02

Apply the Identity for Secant

Recall the identity \(\sec \theta = \frac{1}{\cos \theta}\). Replace \(\sec \theta\) in the expanded expression:\(1 + \frac{1}{\cos \theta}\).
03

Combine and Simplify Terms

Now you have:\(1 - \cos \theta + \frac{1}{\cos \theta} + \frac{-1}{1}\).Simplify each part by finding common denominators and terms.
04

Simplify Using Common Denominators

Combine all terms over a common denominator:- For \(1 - \cos \theta\), write both terms over a common denominator:\(\frac{\cos \theta - 1}{\cos \theta}\).- Combine with the secant part: \(\frac{(1)(1 + \sec \theta) - (\cos \theta)(1 + \sec \theta))}{\cos \theta}\).After simplification, this reduces to \(-\tan \theta\).

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Key Concepts

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

Sine and Cosine
In trigonometry, sine and cosine are foundational functions that help describe angles and relationships in various geometric figures, especially circles. The sine function, denoted as \( \sin \theta \), represents the ratio of the length of the opposite side to the hypotenuse in a right triangle. Similarly, the cosine function, \( \cos \theta \), is the ratio of the adjacent side to the hypotenuse.
Understanding how these two functions interact is crucial for solving trigonometric expressions and identities. When asked to rewrite expressions in terms of sine and cosine, it can simplify the process and make the relationship between components more apparent.
  • Simplifies complex expressions.
  • Makes it easier to find common denominators.
  • Helps in understanding the trigonometric relationships.
In the given exercise, breaking down secants to their cosine components allows for a straightforward simplification.
Secant
Secant is one of the reciprocal trigonometric functions. Specifically, it is the reciprocal of the cosine function. In mathematical terms, this is expressed as \( \sec \theta = \frac{1}{\cos \theta} \). Secant can be thought of as representing how far away from the circle a point is, if the hypotenuse were extended.
When dealing with expressions involving secant, transformations to express them in terms of sine and cosine make it easier to integrate and simplify these expressions.
  • Secant helps in simplifying expressions involving reciprocal relations.
  • Transforming secant into \( \frac{1}{\cos \theta} \) reveals more relationships with other trig functions.
In our exercise, replacing \( \sec \theta \) with \( \frac{1}{\cos \theta} \) clarifies the relationship with \( 1 - \cos \theta \), enabling easier manipulation and simplification.
Simplification
Simplification in mathematics is the process of reducing a complex expression into a simpler or more understandable form. This involves combining like terms, factoring, or applying identities. For trigonometric expressions, simplification is crucial to make calculations more accurate and understandable.
In this exercise, we start by expanding the expression \((1 - \cos \theta)(1 + \sec \theta)\) to break it into simpler components involving trigonometric identities. Breaking down helps reveal opportunities to cancel out terms or combine them into more concise forms.
  • Use identities such as \( \sec \theta = \frac{1}{\cos \theta} \) to transform expressions.
  • Combine terms with a common denominator for ease of computation.
  • Be mindful of canceling opportunities that might emerge upon simplification.
Simplifying expressions is about making them ring understandable while preserving their original values or relationships.
Common Denominator
Finding a common denominator is a strategy that enables the addition or subtraction of fractions. In trigonometric manipulations, when dealing with expressions in fractional forms, identifying a common denominator simplifies operations. It aligns terms such as \( \cos \theta \) or \( \frac{1}{\cos \theta} \) under a uniform denominator and paves the way for combining them effectively.
In the original exercise, having \((1 - \cos \theta)\) and \((1 + \sec \theta)\) may turn complicated at first, but converting terms to a common denominator yields a straightforward pathway to integration and subtraction.
  • Facilitates combining different trigonometric terms.
  • Streamlines the process of computation for complex fractions.
  • Shows clearer paths toward ultimate solutions and simplifications.
Using a common denominator effectively turns an expansive expression into a simplified result, as shown when combining the terms leading to \(-\tan \theta\).

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Most popular questions from this chapter

Solve each equation in part (a) analytically over the interval \([0,2 \pi) .\) Then use a graph to solve each inequality in part (b). (a) \(\cos 2 x=\frac{\sqrt{3}}{2}\) (b) \(\cos 2 x>\frac{\sqrt{3}}{2}\)

Solve each problem. Hearing Beats in Music Musicians sometimes tune instruments by playing the same tone on two different instruments and listening for a phenomenon known as beats. When the two instruments are in tune, the beats disappear. The ear hears beats because the pressure slowly rises and falls as a result of the slight variation in the frequency. This phenomenon can be seen on a graphing calculator. (a) Consider two tones with frequencies of 220 and \(223 \mathrm{Hz}\) and pressures \(P_{1}=0.005 \sin 440 \pi t \quad\) and \(\quad P_{2}=0.005 \sin 446 \pi t\) respectively. A graph of \(P_{1}+P_{2}\) as \(Y_{3}\) felt by an eardrum over the 1 -second interval \([0.15,1.15]\) is shown here. How many beats are there in 1 second? (Graph can't copy) (b) Repeat part (a) with frequencies of 220 and 216 (c) Determine a simple way to find the number of beats per second if the frequency of each tone is given.

Use a graphical method to solve each equation over the interval \([0,2 \pi) .\) Round values to the nearest thousandth. $$\sin 3 x-\sin x=0$$

Solve each equation in part (a) analytically over the interval \([0,2 \pi) .\) Then use a graph to solve each inequality in part (b). (a) \(\sin \frac{x}{2}=\sqrt{2}-\sin \frac{x}{2}\) (b) \(\sin \frac{x}{2}>\sqrt{2}-\sin \frac{x}{2}\)

Solve each problem. Ear Pressure from a Pure Tone A pure tone has a constant frequency and amplitude, and it sounds rather dull and uninteresting. The pressures caused by pure tones on the eardrum are sinusoidal. The change in pressure \(P\) in pounds per square foot on a person's eardrum from a pure tone at time \(t\) in seconds can be modeled by the equation $$ P=A \sin (2 \pi f t+\phi) $$ where \(f\) is the frequency in cycles per second and \(\phi\) is the phase angle. When \(P\) is positive, there is an increase in pressure and the eardrum is pushed inward; when \(P\) is negative, there is a decrease in pressure and the eardrum is pushed outward. (a) Middle C has frequency 261.63 cycles per second. Graph this tone with \(A=0.004\) and \(\phi=\frac{\pi}{7}\) in the window \([0,0.005]\) by \([-0.005,0.005]\) (b) Determine analytically the values of \(t\) for which \(P=0\) on \([0,0.005],\) and support your answers graphically. (c) Determine graphically when \(P<0\) on \([0,0.005]\) (d) Would an eardrum hearing this tone be vibrating outward or inward when \(P<0 ?\)

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