Problem 47
Question
Use fundamental identities to write the first expression in terms of the second, for any acute angle \(\theta\). $$\sin \theta, \sec \theta$$
Step-by-Step Solution
Verified Answer
\( \sin \theta = \sqrt{\frac{\sec^2 \theta - 1}{\sec^2 \theta}} \)
1Step 1: Understand the Relation between Sine and Secant
The goal is to express \( \sin \theta \) in terms of \( \sec \theta \). We know that \( \sec \theta = \frac{1}{\cos \theta} \). Also, the Pythagorean identity states that \( \sin^2 \theta + \cos^2 \theta = 1 \). These identities will help us relate sine and secant.
2Step 2: Express Cosine in Terms of Secant
We know \( \sec \theta = \frac{1}{\cos \theta} \) implies \( \cos \theta = \frac{1}{\sec \theta} \). This will allow us to replace \( \cos \theta \) with terms involving \( \sec \theta \).
3Step 3: Use Pythagorean Identity with Cosine Expression
Start with the identity \( \sin^2 \theta + \cos^2 \theta = 1 \). Substitute \( \cos \theta = \frac{1}{\sec \theta} \) to get \( \cos^2 \theta = \left( \frac{1}{\sec \theta} \right)^2 = \frac{1}{\sec^2 \theta} \). Now the identity becomes \( \sin^2 \theta = 1 - \frac{1}{\sec^2 \theta} \).
4Step 4: Simplify the Expression
Simplify \( \sin^2 \theta = 1 - \frac{1}{\sec^2 \theta} \) into a single fraction: \( \sin^2 \theta = \frac{\sec^2 \theta - 1}{\sec^2 \theta} \). This equation expresses \( \sin^2 \theta \) in terms of \( \sec \theta \).
5Step 5: Solve for Sine in Terms of Secant
To find \( \sin \theta \), take the square root: \( \sin \theta = \sqrt{\frac{\sec^2 \theta - 1}{\sec^2 \theta}} \). Since \( \theta \) is an acute angle, \( \sin \theta \) is positive.
Key Concepts
Sine and Secant RelationshipPythagorean IdentityExpressing Trigonometric FunctionsAcute Angles in Trigonometry
Sine and Secant Relationship
Understanding the relationship between sine and secant in trigonometry is helpful to bridge different trigonometric functions. The secant function, denoted as \( \sec \theta \), is expressed as \( \sec \theta = \frac{1}{\cos \theta} \). This tells us that secant is directly related to the cosine function. On the other hand, the sine function \( \sin \theta \) is concerned with the opposite side of a right triangle when an angle \( \theta \) is involved.
To express sine in terms of secant, we leverage these relationships through known identities. The cosine function \( \cos \theta \) can be rewritten using its connection with secant: \( \cos \theta = \frac{1}{\sec \theta} \). This substitution forms a bridge, allowing us to transition and express sine using secant by inserting it into relevant identities.
To express sine in terms of secant, we leverage these relationships through known identities. The cosine function \( \cos \theta \) can be rewritten using its connection with secant: \( \cos \theta = \frac{1}{\sec \theta} \). This substitution forms a bridge, allowing us to transition and express sine using secant by inserting it into relevant identities.
Pythagorean Identity
The Pythagorean identity is one of the fundamental relationships in trigonometry. It states that for any angle \( \theta \), the relationship \( \sin^2 \theta + \cos^2 \theta = 1 \) holds true. This identity essentially describes a property of right-angled triangles where the squares of the sine and cosine add up to 1.
When working with expressions, substituting known relationships can be invaluable. For instance, replacing \( \cos \theta \) with \( \frac{1}{\sec \theta} \) allows us to redefine this identity in terms of secant, transforming it to \( \sin^2 \theta = 1 - \frac{1}{\sec^2 \theta} \). This deduction offers a path to express sine in terms of secant, simply by using fundamental trigonometric relationships.
When working with expressions, substituting known relationships can be invaluable. For instance, replacing \( \cos \theta \) with \( \frac{1}{\sec \theta} \) allows us to redefine this identity in terms of secant, transforming it to \( \sin^2 \theta = 1 - \frac{1}{\sec^2 \theta} \). This deduction offers a path to express sine in terms of secant, simply by using fundamental trigonometric relationships.
Expressing Trigonometric Functions
Expressing trigonometric functions in terms of each other is a critical skill in solving geometrically based problems and simplifying calculus expressions. By using identities like the Pythagorean identity, one can manipulate expressions to bridge different functions.
Here, we aimed to express \( \sin \theta \) using \( \sec \theta \). By taking the equation \( \sin^2 \theta = \frac{\sec^2 \theta - 1}{\sec^2 \theta} \), we can then solve for \( \sin \theta \) directly. Solving this expression involves taking the square root of both sides, considering \( \sin \theta = \sqrt{\frac{\sec^2 \theta - 1}{\sec^2 \theta}} \). With the constraint that \( \theta \) is an acute angle, ensuring the positivity of \( \sin \theta \), understanding such expressions becomes intuitive.
Here, we aimed to express \( \sin \theta \) using \( \sec \theta \). By taking the equation \( \sin^2 \theta = \frac{\sec^2 \theta - 1}{\sec^2 \theta} \), we can then solve for \( \sin \theta \) directly. Solving this expression involves taking the square root of both sides, considering \( \sin \theta = \sqrt{\frac{\sec^2 \theta - 1}{\sec^2 \theta}} \). With the constraint that \( \theta \) is an acute angle, ensuring the positivity of \( \sin \theta \), understanding such expressions becomes intuitive.
Acute Angles in Trigonometry
In trigonometry, acute angles are those less than \( 90^\circ \). They are crucial as they are the foundation of trigonometric studies, especially within the confines of a right-angled triangle. Knowing that \( \theta \) is an acute angle helps ensure specific properties about trigonometric functions, such as the fact that \( \sin \theta \) is always positive for acute angles.
This positivity is vital when working with square root expressions. For example, in the expression \( \sin \theta = \sqrt{\frac{\sec^2 \theta - 1}{\sec^2 \theta}} \), knowing \( \theta \) is acute ensures that the square root yields a valid sine value. Thus, perceiving angles geometrically within a triangle aids in visualizing and confirming trigonometric calculations.
This positivity is vital when working with square root expressions. For example, in the expression \( \sin \theta = \sqrt{\frac{\sec^2 \theta - 1}{\sec^2 \theta}} \), knowing \( \theta \) is acute ensures that the square root yields a valid sine value. Thus, perceiving angles geometrically within a triangle aids in visualizing and confirming trigonometric calculations.
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