Problem 93
Question
Make the indicated trigonometric substitution in the given algebraic expression and simplify. Assume that \(0 \leq \theta<\pi / 2\) $$\sqrt{x^{2}-1}, \quad x=\sec \theta$$
Step-by-Step Solution
Verified Answer
\( \tan \theta \)
1Step 1: Understand the Substitution
We start with the expression \( \sqrt{x^2 - 1} \) and know that \( x = \sec \theta \). Therefore, \( x^2 = (\sec \theta)^2 \). Since \( \sec \theta = \frac{1}{\cos \theta} \), then \( \sec^2 \theta = 1 + \tan^2 \theta \) by the identity \( \sec^2 \theta = 1 + \tan^2 \theta \).
2Step 2: Substitute and Simplify
Substitute \( x^2 \) in the expression: \( \sqrt{x^2 - 1} = \sqrt{\sec^2 \theta - 1} \). Using the identity \( \sec^2 \theta = 1 + \tan^2 \theta \), we have \( \sqrt{\tan^2 \theta} \). Since \( \tan \theta \geq 0 \) for \( 0 \leq \theta < \pi/2 \), this simplifies to \( \tan \theta \).
3Step 3: Final Expression
The simplified form of the given expression is \( \tan \theta \) as derived from the trigonometric identity and given substitution.
Key Concepts
Trigonometric IdentitiesSecant FunctionTangent Function
Trigonometric Identities
Trigonometric identities are key tools in simplifying expressions and solving equations in trigonometry. They are equations that relate the trigonometric functions, such as sine, cosine, and tangent, to one another. These identities are derived based on the right triangle definitions or the unit circle approach. They allow us to transform complex trigonometric expressions into simpler forms. For example, one crucial identity often used is
- the Pythagorean identity: \(\sin^2 \theta + \cos^2 \theta = 1\)
- \(\sec^2 \theta = 1 + \tan^2 \theta\)
Secant Function
The secant function, denoted as \(\sec \theta\), is one of the six primary trigonometric functions. It is defined as the reciprocal of the cosine function:
In the original exercise, the substitution \(x = \sec \theta\) simplifies solving methods by transforming the square root expression \(\sqrt{x^2 - 1}\) into a more manageable trigonometric form. This use of the secant function aligns with its property in trigonometric identities, especially with one involving tangent function \(\sec^2 \theta = 1 + \tan^2 \theta\).
- \(\sec \theta = \frac{1}{\cos \theta}\)
In the original exercise, the substitution \(x = \sec \theta\) simplifies solving methods by transforming the square root expression \(\sqrt{x^2 - 1}\) into a more manageable trigonometric form. This use of the secant function aligns with its property in trigonometric identities, especially with one involving tangent function \(\sec^2 \theta = 1 + \tan^2 \theta\).
Tangent Function
The tangent function, expressed as \( \tan \theta \), is another primary trigonometric function, known for the ratio it offers between sine and cosine:
In the range \(0 \leq \theta < \pi/2\), the tangent function is positive and increases in value as \(\theta\) approaches \(\pi/2\). This makes it a stable choice for many substitutions involving expressions like \(\sqrt{x^2 - 1}\), often simplifying into \(\tan \theta\).
The substitution \(x = \sec \theta\) allows us to utilize the identity \(\sec^2 \theta = 1 + \tan^2 \theta\), transforming \(\sqrt{x^2 - 1}\) into \(\sqrt{\tan^2 \theta}\). Because \(\tan \theta\) is non-negative in this specific domain, it directly simplifies to \(\tan \theta\), showcasing the practical side of trigonometric identities when efficiently substituting into algebraic expressions.
- \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)
In the range \(0 \leq \theta < \pi/2\), the tangent function is positive and increases in value as \(\theta\) approaches \(\pi/2\). This makes it a stable choice for many substitutions involving expressions like \(\sqrt{x^2 - 1}\), often simplifying into \(\tan \theta\).
The substitution \(x = \sec \theta\) allows us to utilize the identity \(\sec^2 \theta = 1 + \tan^2 \theta\), transforming \(\sqrt{x^2 - 1}\) into \(\sqrt{\tan^2 \theta}\). Because \(\tan \theta\) is non-negative in this specific domain, it directly simplifies to \(\tan \theta\), showcasing the practical side of trigonometric identities when efficiently substituting into algebraic expressions.
Other exercises in this chapter
Problem 92
Show that \(\cos 100^{\circ}-\cos 200^{\circ}=\sin 50^{\circ}\)
View solution Problem 92
Make the indicated trigonometric substitution in the given algebraic expression and simplify. Assume that \(0 \leq \theta
View solution Problem 94
Show that \(\cos 87^{\circ}+\cos 33^{\circ}=\sin 63^{\circ}\)
View solution Problem 94
Make the indicated trigonometric substitution in the given algebraic expression and simplify. Assume that \(0 \leq \theta
View solution