Question

Write \(\cos \left(\csc ^{-1} u\right)\) as an algebraic expression in \(u\)

Short answer

\( \cos \left(\csc ^{-1} u\right) = \sqrt{1 - \frac{1}{u^2}} \)

Step-by-step solution

Define the inverse cosecant

Recall that \(\text{csc}^{-1}(u)\) represents the angle \(\theta \) such that \( \text{csc}(\theta) = u \).

Express cosecant in terms of sine

We know that cosecant is the reciprocal of sine, i.e., \(\text{csc}(\theta) = \frac{1}{\text{sin}(\theta)} \). Therefore, \( \text{sin}(\theta) = \frac{1}{u} \).

Apply trigonometric identity

Recall the Pythagorean identity: \( \text{sin}^2(\theta) + \text{cos}^2(\theta) = 1 \). We can solve for \( \text{cos}(\theta) \).

Substitute for sine

Substitute \( \text{sin}(\theta) = \frac{1}{u} \) into the identity to get \( \frac{1}{u^2} + \text{cos}^2(\theta) = 1 \).

Solve for \( \text{cos}^2(\theta) \)

Rearrange the equation to solve for \( \text{cos}^2(\theta) \): \( \text{cos}^2(\theta) = 1 - \frac{1}{u^2} \).

Take the square root

Take the square root of both sides: \( \text{cos}(\theta) = \pm \sqrt{1 - \frac{1}{u^2}} \). Since \( \theta \) lies in the range of \( \text{csc}^{-1} \), we take the positive value: \( \text{cos}(\theta) = \sqrt{1 - \frac{1}{u^2}} \).

Key concepts

Inverse Trigonometric Functions

Inverse trigonometric functions can be thought of as the reverse process of the standard trigonometric functions. When we use an inverse trigonometric function, we are starting with the value of the ratio and finding the corresponding angle. For example, \(\text{csc}^{-1}(u)\) denotes the angle \(\theta\) such that \(\text{csc}(\theta) = u\). Each inverse trigonometric function has a specific range, defined to ensure that every input corresponds to a single output. This is what makes them 'functions'.

For the inverse cosecant, \(\text{csc}^{-1}(x)\), the range is typically \([\frac{\pi}{2}, \pi] \cup [0,\frac{\pi}{2}]\) (excluding \(\frac{\pi}{2}\)), meaning that the angle we obtain from \(\text{csc}^{-1}(u)\) will fall within this interval. Understanding the inverse trigonometric functions is crucial for converting between angle measures and their associated reciprocal trigonometric values like cosecant, secant, and cotangent.

Pythagorean Identity

The Pythagorean identity is a fundamental relationship in trigonometry that connects sine and cosine functions. It states that for any angle \(\theta\), the sum of the squares of sine and cosine of that angle is always 1: \[\text{sin}^2(\theta) + \text{cos}^2(\theta) = 1 \]. This identity is extremely useful for transforming between trigonometric functions and their reciprocals.
Let's use an example to see how it works. Assume you have \(\text{sin}(\theta)\) and need to find \(\text{cos}(\theta) \). According to the Pythagorean identity, \[\text{cos}(\theta) = \sqrt{1 - \text{sin}^2(\theta)} \]. If \(\text{sin}(\theta) = \frac{1}{u} \), substituting gives us: \[\text{cos}(\theta) = \sqrt{1 - \left( \frac{1}{u} \right) ^2} = \sqrt{1 - \frac{1}{u^2}} \].
This transformation allows us to express one trigonometric function in terms of another, which can be crucial in solving various mathematical problems involving trigonometric identities.

Algebraic Expressions

An algebraic expression is a mathematical phrase that can contain numbers, variables, and operators. They are used to represent real-world quantities and to solve equations. In trigonometry, we often use algebraic expressions to simplify complex trigonometric forms into more manageable terms.

For example, to express \(\cos(\text{csc}^{-1}(u))\) in an algebraic form, we performed a series of algebraic manipulations using trigonometric identities. We started with the understanding that \(\text{csc}(\theta) = u\), which means \(\text{sin}(\theta) = \frac{1}{u}\). By applying the Pythagorean identity, we derived that \(\text{cos}(\theta) = \sqrt{1 - \frac{1}{u^2}}\).

This process is essential in mathematics because it allows us to translate trigonometric and geometric relationships into forms that are easier to analyze and compute, bridging different mathematical concepts seamlessly.