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Chapter 6: Problem 92

For each of the following expressions, write an equivalent expression in terms of only the variable \(u\). $$ \tan \left(\sin ^{-1} u\right) $$

Short Answer

Expert verified
\( \tan(\sin^{-1}(u)) = \frac{u}{\sqrt{1-u^2}} \).

Step by step solution

01

Understand the Problem

We need to express \( \tan(\sin^{-1}(u)) \) in terms of only the variable \( u \). This involves converting the expression containing the inverse sine into an expression with just \( u \).
02

Use the Inverse Sine Identity

Recall that if \( \theta = \sin^{-1}(u) \), then \( \sin(\theta) = u \), and \( \theta \) is the angle whose sine is \( u \). We will use the right triangle relationships to express \( \tan(\theta) \) in terms of \( u \).
03

Draw and Label a Right Triangle

Consider a right triangle where the angle \( \theta \) has an opposite side of length \( u \) and a hypotenuse of length 1, because \( \sin(\theta) = \frac{u}{1} \). This means adjacent side \( \cos(\theta) = \sqrt{1-u^2} \) using the Pythagorean identity.
04

Calculate the Tangent of \( \theta \)

The identity for the tangent is \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \). Substituting the known values: \( \tan(\theta) = \frac{u}{\sqrt{1-u^2}} \).
05

Final Expression

Since \( \theta = \sin^{-1}(u) \), we have \( \tan(\sin^{-1}(u)) = \frac{u}{\sqrt{1-u^2}} \), which is the expression in terms of only \( u \).

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Key Concepts

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

Right Triangle Trigonometry
Right triangle trigonometry is a fundamental concept that involves the relationships between the angles and the sides of right triangles. In a right triangle, one angle is always 90 degrees. The other two angles, along with the sides, help establish the primary trigonometric functions: sine, cosine, and tangent. These functions relate the angles to the ratios of the lengths of two sides of the triangle.
  • Sine (\( heta\)): This function represents the ratio of the length of the opposite side to the hypotenuse. It is expressed as \( rac{opposite}{hypotenuse}\).
  • Cosine (\( heta\)): Represents the ratio of the length of the adjacent side to the hypotenuse. It is expressed as \( rac{adjacent}{hypotenuse}\).
  • Tangent (\( heta\)): Defined as the ratio of the sine to the cosine, or equivalently, the opposite side to the adjacent side. It is expressed as \( rac{opposite}{adjacent}\).
Understanding these relationships is essential to solve problems involving angles and lengths, like expressing a trigonometric function in terms of a variable, as seen in the exercise.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that hold true for all values of the variables within their domains. These identities are tools that allow us to simplify complex trigonometric expressions and solve equations. Some of the most common identities include:
  • Pythagorean Identities: These are derived from the Pythagorean theorem and are essential in right triangle trigonometry. An example is \( an( heta) = \frac{\sin( heta)}{\cos( heta)}\), which is used in the solution to simplify \(\tan(\sin^{-1}(u))\).
  • Reciprocal Identities: These identities represent the basic trigonometric functions in terms of their reciprocals, like \(\sin(\theta) = \frac{1}{\csc(\theta)}\).
  • Angle Sum and Difference Identities: These are used to express trigonometric functions of sums or differences of angles.
By leveraging these identities, one can convert inverse trigonometric functions, such as \(\sin^{-1}(u)\), into other forms where calculations or further simplifications are more manageable.
Tangent Function
The tangent function is one of the three primary trigonometric functions. It is particularly important because it connects the sine and cosine functions through the identity: \( an(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\). This relationship forms the basis for many calculations in trigonometry.
  • The tangent of an angle provides a ratio of the opposite and adjacent sides in a right triangle. It is essential in determining slopes and angles in various geometric settings.
  • In the context of inverse trigonometric functions, the tangent function \(\tan(\sin^{-1}(u))\) can be expressed entirely using a known variable, as shown in the problem. By using a right triangle representation, the tangent can be simplified to \(\frac{u}{\sqrt{1-u^2}}\).
  • This simplification is crucial in solving trigonometric expressions where direct calculation through original definitions might be cumbersome.
Understanding the tangent function and its properties allows for the solving of complex trigonometric problems using inverse functions.

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Most popular questions from this chapter

In Exercises 37-48, solve each of the trigonometric equations on the interval \(0^{\circ} \leq \theta<360^{\circ}\). Give answers in degrees and round to two decimal places. $$ \sec ^{2} x=2 \tan x+4 $$

In Exercises \(71-74\), explain the mistake that is made. Solve \(\sqrt{2+\sin \theta}=\sin \theta\) on \(0 \leq \theta<2 \pi\). Solution: Square both sides. \(2+\sin \theta=\sin ^{2} \theta\) Gather all terms to one side. \(\quad \sin ^{2} \theta-\sin \theta-2=0\) Factor. \(\quad(\sin \theta-2)(\sin \theta+1)=0\) Set each factor equal \(\quad \sin \theta-2=0\) or \(\begin{array}{lrlrl}\text { to zero. } & & \sin \theta+1=0 \\ \text { Solve for } \sin \theta . & \sin \theta=2 & \text { or } \sin \theta=-1\end{array}\) Solve \(\sin \theta=2\) for \(\theta\). no solution Solve \(\sin \theta=-1\) for \(\theta\). \(\theta=\frac{3 \pi}{2}\) This is incorrect. What mistake was made?

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