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Chapter 20: Problem 45

Find an algebraic expression for each of the given expressions. $$\sin \left(\sec ^{-1} \frac{x}{4}\right)$$

Short Answer

Expert verified
\( \sin(\sec^{-1}(\frac{x}{4})) = \frac{\sqrt{x^2 - 16}}{x} \).

Step by step solution

01

Understand the Definition

The expression involves the inverse secant function, denoted by \( \sec^{-1} z \). Let's first understand that \( \sec(y) = z \Rightarrow y = \sec^{-1} z \). This translates to \( z = \frac{x}{4} \), where \( y \) is an angle whose secant is the given expression.
02

Use Triangle Representation

Since \( y = \sec^{-1} \left( \frac{x}{4} \right) \), this implies that \( \sec(y) = \frac{x}{4} \). In a right triangle, secant is the ratio of the hypotenuse to the adjacent side, so we can construct a right triangle where the hypotenuse is \( x \) and the adjacent side is \( 4 \).
03

Find the Opposite Side

Using the Pythagorean theorem, we find the opposite side of the triangle. Let this side be \( a \). Then, \( a^2 + 4^2 = x^2 \). Solve for \( a \): \( a = \sqrt{x^2 - 16} \).
04

Express \( \sin(y) \) Using Triangle

In the right triangle representation, \( \sin(y) \) is the ratio of the opposite side to the hypotenuse. Therefore, \( \sin(y) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{x^2 - 16}}{x} \).
05

Write the Final Algebraic Expression

The algebraic expression for \( \sin(\sec^{-1}(\frac{x}{4})) \) is given by \( \frac{\sqrt{x^2 - 16}}{x} \).

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Key Concepts

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

Inverse Trigonometric Functions
Inverse trigonometric functions allow us to find angles when given a trigonometric ratio.
  • The inverse of the secant function is denoted as \( \sec^{-1} z \). This expression indicates the angle whose secant is \( z \).
  • If you see \( \sec^{-1} \left( \frac{x}{4} \right) \), it means you are dealing with an angle \( y \), such that \( \sec(y) = \frac{x}{4} \).
These function inverses are crucial in solving many trigonometry problems because they bridge the gap between trigonometric values and the angles themselves. Recognizing how to convert complex expressions involving inverses, as illustrated in the example above, is key to mastering the topic. For instance, by acknowledging that the secant of an angle is the reciprocal of its cosine, we can establish a relationship between the angle and the sides of a triangle, which becomes incredibly useful when analyzing geometric shapes.
Pythagorean Theorem
The Pythagorean Theorem is integral to relating the sides of a right-angled triangle. Simply stated, for a triangle with a 90-degree angle:
  • The square of the hypotenuse equals the sum of the squares of the other two sides. Mathematically, \( c^2 = a^2 + b^2 \).
  • In our example, this theorem helps us find the length of the opposite side once we know the hypotenuse \( x \) and the adjacent \( 4 \).
  • After applying the theorem, the length of the opposite side is \( \sqrt{x^2 - 16} \).
This step is vital as it allows us to use our triangle to compute trigonometric functions, such as the sine, from the computed side lengths. Understanding and applying the Pythagorean theorem is essential in problems that involve triangles and helps convert between angles and sides effectively.
Algebraic Expressions
Algebraic expressions represent values using symbols and constants, allowing us to formulate equations that describe real-world scenarios.
  • In our discussed problem, we transformed the trigonometric function into an algebraic expression: \( \sin(y) = \frac{\sqrt{x^2 - 16}}{x} \).
  • This form is easier to manipulate and substitute into other expressions or equations. It harnesses the power of algebra to give a tangible form to a trigonometric value.
  • By breaking down complex trigonometric forms into simpler algebraic expressions, it becomes much simpler to analyze and compute with these values, especially when integrating into broader problems and solutions.
Algebra is a fundamental component of math that ties into nearly every other field, including trigonometry. Mastering its use will help you skillfully solve and understand a broad array of mathematical challenges.

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Most popular questions from this chapter

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