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Chapter 8: Problem 111

Challenge Problem Show that \(\tan ^{-1}\left(\frac{1}{v}\right)=\frac{\pi}{2}-\tan ^{-1} v\) if \(v>0\).

Short Answer

Expert verified
\(\tan^{-1}\bigg(\frac{1}{v}\bigg) = \frac{\pi}{2} - \tan^{-1}(v)\)

Step by step solution

01

Define Inverse Tangent

Recall that the inverse tangent function, \(\tan^{-1}(x)\), provides an angle whose tangent is x. Hence, we can write:\[\tan(\theta) = x \Rightarrow \theta = \tan^{-1}(x)\]
02

Set Up Equation

To prove the given identity, start by setting \(\theta = \tan^{-1}\bigg(\frac{1}{v}\bigg)\). Then by definition, \(\tan(\theta) = \frac{1}{v}\).
03

Tangent of Complementary Angles

Recall the complementary angle identity for tangent: \[\tan\bigg( \frac{\pi}{2} - \theta \bigg) = \frac{1}{\tan(\theta)} \]Using \( \tan(\theta) = \frac{1}{v}\):\[\tan\bigg( \frac{\pi}{2} - \theta \bigg) = v\]
04

Define New Angle

Let \(\beta = \frac{\pi}{2} - \theta\) . Then, \( \tan(\beta) = v \Rightarrow \beta = \tan^{-1}(v)\).
05

Solve for \(\theta\)

Since \( \beta = \frac{\pi}{2} - \theta\):\[\tan^{-1}(v) = \frac{\pi}{2} - \tan^{-1}\bigg(\frac{1}{v}\bigg)\]Rewriting it, we obtain:\[\tan^{-1}\bigg(\frac{1}{v}\bigg) = \frac{\pi}{2} - \tan^{-1}(v)\]

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Key Concepts

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

inverse tangent
Inverse tangent, also known as arctangent, is a fundamental function in trigonometry. The inverse tangent function, denoted as \(\tan^{-1}(x)\), provides the angle whose tangent is x. For instance, if \(\tan(\theta) = x\), then \(\theta = \tan^{-1}(x)\). This concept is critical because it allows us to switch from knowing the ratio of the sides of a right triangle to knowing the actual angle. In solving problems, the notation \(\tan^{-1}(x)\) is used to signify the angle and has a range of \(\frac{-\frac{\pi}{2}}{\frac{\pi}{2}}\). This function is helpful in various mathematical and physical contexts, especially when dealing with angles and rotations.
complementary angles
Complementary angles are two angles whose measures add up to 90 degrees, or \(\frac{\pi}{2}\) radians. A key property of complementary angles is their trigonometric functions' relationships. For example, the tangent of an angle is the reciprocal of the tangent of its complementary angle. In mathematical terms, if \(\theta\) and \(\beta\) are complementary angles, then \(\theta + \beta = \frac{\pi}{2}\). This means: \[ \tan\bigg( \frac{\pi}{2} - \theta \bigg) = \frac{1}{\tan(\theta)} \]. This property is particularly useful when converting a complex trigonometric expression into something simpler to solve. In the given problem, this complementary angle identity aids in transforming and proving the given trigonometric identity.
tangent
The tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side. It is one of the basic trigonometric functions and is written as \(\tan(\theta)\). The tangent function has several unique properties and is central to trigonometric calculations. Its range is all real numbers, since the tangent value can extend to positive or negative infinity. The formula \(\tan(\theta + \frac{\pi}{2}) = - \frac{1}{\tan(\theta)}\) shows that the tangent of an angle plus 90 degrees is the negative reciprocal of the original tangent. Understanding tangent functions is essential when dealing with inverse functions, as the relationship between them helps simplify and solve trigonometric equations. For example, in the given problem, knowing the properties of tangent and how they relate to their inverses helps in proving the identity with complementary angles.

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Most popular questions from this chapter

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?

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

Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Convert \(\frac{17 \pi}{6}\) to degrees.

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.

Discuss the following derivation: $$ \tan \left(\theta+\frac{\pi}{2}\right)=\frac{\tan \theta+\tan \frac{\pi}{2}}{1-\tan \theta \tan \frac{\pi}{2}}=\frac{\frac{\tan \theta}{\tan \frac{\pi}{2}}+1}{\frac{1}{\tan \frac{\pi}{2}}-\tan \theta}=\frac{0+1}{0-\tan \theta}=\frac{1}{-\tan \theta}=-\cot \theta $$ Can you justify each step?
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