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Chapter 8: Problem 97

Establish each identity. $$\ln |\sec \theta|=-\ln |\cos \theta|$$

Short Answer

Expert verified
\(\text{ln} |\text{sec} 胃| = -\text{ln} |\text{cos} 胃|\) follows from \(\text{sec} 胃 = \frac{1}{\text{cos} 胃}\) and \(\text{ln} \frac{1}{x} = -\text{ln} x\).

Step by step solution

01

Understand the identity

The given identity is \(\text{ln} |\text{sec} 胃| = -\text{ln} |\text{cos} 胃|\). To establish this identity, we need to use the definitions of the logarithm and the secant function.
02

Recall the definition of secant

Recall that the secant function is the reciprocal of the cosine function. Thus, \(\text{sec} 胃 = \frac{1}{\text{cos} 胃}\).
03

Substitute the definition into the logarithm

Substitute \(\text{sec} 胃\) with \(\frac{1}{\text{cos} 胃}\) in the logarithm: \(\text{ln} |\text{sec} 胃| = \text{ln} \bigg|\frac{1}{\text{cos} 胃}\bigg| \).
04

Use the property of logarithms

Recall the property of logarithms: \(\text{ln} \frac{1}{x} = -\text{ln} x\). Apply this property to the expression: \(\text{ln} \bigg|\frac{1}{\text{cos} 胃}\bigg| = -\text{ln} |\text{cos} 胃|\).
05

Verify the identity

By substituting and applying the property of logarithms, we have shown that \(\text{ln} |\text{sec} 胃| = -\text{ln} |\text{cos} 胃|\). Thus, the identity is established.

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Key Concepts

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

secant function
The secant function, often written as \text{sec} 胃\, is a trigonometric function that is the reciprocal of the cosine function. This means it is defined as:
  • \( \text{sec} 胃 = \frac{1}{\text{cos} 胃} \)
The secant function is useful in many areas of mathematics and engineering. Since it is the reciprocal of the cosine, it is undefined where the cosine is zero, leading to vertical asymptotes in its graph.
For example, if \( \text{cos} 胃 = 0 \), then \( \text{sec} 胃 \) would be undefined. Understanding how the secant function works in relation to cosine can help solve complex trigonometric problems.
reciprocal functions
Reciprocal functions are functions where each output value is the reciprocal (or multiplicative inverse) of the input value. For trigonometric functions, the main reciprocal identities include:
  • \( \text{sec} 胃 = \frac{1}{\text{cos} 胃} \)
  • \( \text{csc} 胃 = \frac{1}{\text{sin} 胃} \)
  • \( \text{cot} 胃 = \frac{1}{\text{tan} 胃} \)
In our context, the secant function is the focus, which we can see as \( \text{sec} 胃 \). This makes our math work simpler by turning division into multiplication by reciprocals. Utilizing reciprocal functions, we can simplify many trigonometric identities and solve equations more efficiently.
properties of logarithms
Logarithms come with several properties that make manipulating logarithmic expressions much easier. Some of the key properties include:
  • \( \text{ln}(xy) = \text{ln}(x) + \text{ln}(y) \)
  • \( \text{ln}\big(\frac{x}{y}\big) = \text{ln}(x) - \text{ln}(y) \)
  • \( \text{ln}(x^a) = a \text{ln}(x) \)
  • \( \text{ln} \big(\frac{1}{x}\big) = -\text{ln}(x) \)
In the given exercise, the property \( \text{ln} \big(\frac{1}{x}\big) = -\text{ln}(x) \) was used. This helped us transform the expression \( \text{ln} \big|\frac{1}{\text{cos} 胃}\big| \) into \( -\text{ln}|\text{cos} 胃| \), eventually leading to the final identity \( \text{ln} |\text{sec} 胃| = -\text{ln} |\text{cos} 胃| \). Mastery of these properties allows for simplifying complex logarithmic equations and expressions effortlessly.

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Most popular questions from this chapter

If the angle of incidence and the angle of refraction are complementary angles, the angle of incidence is referred to as the Brewster angle \(\theta_{B}\). The Brewster angle is related to the indices of refraction of the two media, \(n_{1}\) and \(n_{2},\) by the equation \(n_{1} \sin \theta_{B}=n_{2} \cos \theta_{B},\) where \(n_{1}\) is the index of refraction of the incident medium and \(n_{2}\) is the index of refraction of the refractive medium. Determine the Brewster angle for a light beam traveling through water (at \(20^{\circ} \mathrm{C}\) ) that makes an angle of incidence with a smooth, flat slab of crown glass.

Find the exact value of \(\sin \frac{\pi}{3} \cos \frac{\pi}{3}\)

Write each trigonometric expression as an algebraic expression containing u and \(v .\) Give the restrictions required on \(u\) and \(v\). $$ \cos \left(\cos ^{-1} u+\sin ^{-1} v\right) $$

Establish each identity. $$ \sin (\alpha-\beta) \sin (\alpha+\beta)=\sin ^{2} \alpha-\sin ^{2} \beta $$

Write each trigonometric expression as an algebraic expression containing u and \(v .\) Give the restrictions required on \(u\) and \(v\). $$ \tan \left(\sin ^{-1} u-\cos ^{-1} v\right) $$
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