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Chapter 8: Problem 106

Show that the functions \(f\) and \(g\) are identically equal. $$\text { Show that } \sqrt{9 \sec ^{2} \theta-9}=3 \tan \theta \text { if } \pi \leq \theta<\frac{3 \pi}{2}$$

Short Answer

Expert verified
\( \sqrt{9 \sec ^{2} \theta - 9} = 3 \tan \theta \)

Step by step solution

01

Simplify the expression inside the square root

Start by simplifying the expression inside the square root. We have \( f(\theta) = \sqrt{9 \sec^2 \theta - 9} \).
02

Factor out the common term

Factor out the common term 9 inside the square root: \[ \sqrt{9 (\sec^2 \theta - 1)} \]
03

Use trigonometric identity to simplify

Recall the trigonometric identity \(\sec^2 \theta - 1 = \tan^2 \theta\): \[ \sqrt{9 \tan^2 \theta} \]
04

Simplify the square root

Simplify the square root expression: \[ \sqrt{9 \tan^2 \theta} = 3 \tan \theta \]
05

Check the domain condition

Verify that the domain condition \( \pi \leq \theta < \frac{3\pi}{2} \) does not affect the result because \( \tan \theta \) is defined for this interval.
06

Conclude the identity

Since the steps were performed correctly, it is shown that \[ \sqrt{9 \sec ^{2} \theta - 9} = 3 \tan \theta \]

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Key Concepts

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

trigonometric simplification
Trigonometric simplification often involves reducing complex expressions by using known identities and algebraic techniques. In this exercise, the goal was to show that \( \sqrt{9 \sec ^{2} \theta - 9} = 3 \tan \theta \). Here's a brief breakdown of the trigonometric simplification:
  • First, recognize the expression inside the square root: \( 9 \sec ^{2} \theta - 9 \).

  • To simplify it, factor out the common term \( 9 \). This gives us \( \sqrt{9 ( \sec^2 \theta - 1 )} \).

  • Next, utilize the trigonometric identity \( \sec^2 \theta - 1 = \tan^2 \theta \) to replace the contents inside the parentheses: \( \sqrt{9 \tan^2 \theta} \).

  • Finally, apply the property of square roots to simplify: \( \sqrt{9 \tan^2 \theta} = 3 \tan \theta \).

Using these steps, we simplified a complex trigonometric expression into a more manageable form.
secant and tangent relationship
Understanding the relationship between secant and tangent functions is crucial in simplifying trigonometric expressions. The key identity here is \( \sec^2 \theta - 1 = \tan^2 \theta \). Let's explore why this identity holds:
  • Start from the fundamental Pythagorean identity: \( \cos^2 \theta + \sin^2 \theta = 1 \).

  • Divide every term by \( \cos^2 \theta \): \( 1 + \tan^2 \theta = \sec^2 \theta \).

  • Rearrange the equation to isolate \( \tan^2 \theta \): \( \sec^2 \theta - 1 = \tan^2 \theta \).

This identity is incredibly useful and appears frequently in trigonometric simplification problems. Here, it allowed us to transform \( 9 \sec^2 \theta - 9 \) into a simpler expression involving tangent, leading to a much easier simplification process.
domain verification
When working with trigonometric functions, it's crucial to consider the domain 鈥� the set of values for which the function is defined. In this exercise, we verified that \( \tan \theta \) is defined for the given interval \( \pi \leq \theta < \frac{3 \pi}{2} \). Let's break down why domain verification is essential:
  • Trigonometric functions like tangent can be undefined for certain values of \( \theta \). For instance, \( \tan \theta \) is undefined at \( \theta = \frac{\pi}{2} \), \( \theta = \frac{3\pi}{2} \), etc.

  • In our interval \( \pi \leq \theta < \frac{3 \pi}{2} \), \( \theta \) does not include these undefined points.

  • Therefore, \( \tan \theta \) is well-defined and real within this interval, ensuring our simplification and resulting function \( 3 \tan \theta \) are valid and correct.

Verifying the domain ensures that the functions we work with behave as expected within the specified range. It helps avoid potential pitfalls that could arise from undefined values.

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Most popular questions from this chapter

Establish each identity. $$ \cos (\theta+k \pi)=(-1)^{k} \cos \theta, k \text { any integer } $$

Establish each identity. $$ \frac{\cos (\alpha+\beta)}{\cos (\alpha-\beta)}=\frac{1-\tan \alpha \tan \beta}{1+\tan \alpha \tan \beta} $$

Use the following discussion. The formula $$ D=24\left[1-\frac{\cos ^{-1}(\tan i \tan \theta)}{\pi}\right] $$ Approximate the number of hours of daylight for any location that is \(66^{\circ} 30^{\prime}\) north latitude for the following dates: (a) Summer solstice \(\left(i=23.5^{\circ}\right)\) (b) Vernal equinox \(\left(i=0^{\circ}\right)\) (c) July \(4\left(i=22^{\circ} 48^{\prime}\right)\) (d) Thanks to the symmetry of the orbital path of Earth around the Sun, the number of hours of daylight on the winter solstice may be found by computing the number of hours of daylight on the summer solstice and subtracting this result from 24 hours. Compute the number of hours of daylight for this location on the winter solstice. What do you conclude about daylight for a location at \(66^{\circ} 30^{\prime}\) north latitude?

Area of a Dodecagon Part I A regular dodecagon is a polygon with 12 sides of equal length. See the figure. (a) The area \(A\) of a regular dodecagon is given by the formula \(A=12 r^{2} \tan \frac{\pi}{12},\) where \(r\) is the apothem, which is a line segment from the center of the polygon that is perpendicular to a side. Find the exact area of a regular dodecagon whose apothem is 10 inches. (b) The area \(A\) of a regular dodecagon is also given by the formula \(A=3 a^{2} \cot \frac{\pi}{12},\) where \(a\) is the length of a side of the polygon. Find the exact area of a regular dodecagon if the length of a side is 15 centimeters.

Challenge Problem Show that \(\sin ^{-1} v+\cos ^{-1} v=\frac{\pi}{2}\).
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