Problem 50 - Contemporary Precalculus

Question

Write the expression as an algebraic expression in \(v\). $$\tan \left(\cos ^{-1} v\right)$$

Step-by-Step Solution

Answer

Question: Write the given function of $\tan (\cos^{-1} v)$ as an algebraic expression in \(v\). Answer: The function can be written as $\tan (\cos^{-1} v) = \frac{\sqrt{1 - v^2}}{v}$.

Step 1: Recall the Pythagorean Identity

Recall the Pythagorean identity involving sine, cosine, and tangent functions: $$\sin^2 \theta + \cos^2 \theta = 1$$ We can solve for \(\sin \theta\) in terms of \(\cos \theta\): $$\sin^2 \theta = 1 - \cos^2 \theta$$ $$\sin \theta = \pm \sqrt{1 - \cos^2 \theta}$$

Step 2: Apply the identity to the given expression

In this step, we will apply the identity to the given expression, where \(\theta = \cos^{-1} v\). Since cosine is the ratio of the adjacent side to the hypotenuse in a right triangle, it has a range of \(0 \leq v \leq 1\). So we can use the positive square root. $$\tan (\cos^{-1} v) = \frac{\sin (\cos^{-1} v)}{\cos (\cos^{-1} v)}$$ We know that \(\cos (\cos^{-1} v) = v\). Also, we can use the identity from step 1 to find \(\sin (\cos^{-1} v) = \sqrt{1 - v^2}\). Substitute these values into the expression: $$\tan (\cos^{-1} v) = \frac{\sqrt{1 - v^2}}{v}$$

Step 3: Write the final expression

Therefore, the final algebraic expression of the given function in terms of \(v\) is: $$\tan (\cos^{-1} v) = \frac{\sqrt{1 - v^2}}{v}$$

Key Concepts

Pythagorean Identity

The Pythagorean Identity is a fundamental aspect of trigonometry, connecting sine, cosine, and tangent. It tells us that the square of the sine of an angle plus the square of the cosine of the same angle equals one, expressed as:

• \( \sin^2 \theta + \cos^2 \theta = 1 \)

This identity is derived from the Pythagorean theorem in a unit circle where the radius is equal to one. Each point on the circle corresponds to the angle \( \theta \) and its sine and cosine values represent the y and x coordinates, respectively.
When you need to find \( \sin \theta \) given \( \cos \theta \), rearrange the Pythagorean Identity to:

• \( \sin \theta = \pm \sqrt{1 - \cos^2 \theta} \)

In this exercise, the angle \( \theta \) is \( \cos^{-1} v \), meaning the cosine value \( v \) is between 0 and 1. Hence, the sine value is the positive square root. Using this identity helps convert expressions involving inverse trigonometric functions into algebraic terms.

Inverse Trigonometric Functions

Inverse trigonometric functions reverse the operations of regular trigonometric functions. They allow you to find an angle when given a trigonometric value.
For example, \( \cos^{-1} v \) returns the angle whose cosine is \( v \). This function is useful when working with angles in practical problems.
With \( \cos^{-1} v \), \( v \) must fall within the interval \([0, 1]\) for it to produce meaningful results because cosine values are restricted to this range for real angles between \(0\) and \(\pi\).
Utilizing the inverse cosine, we can quickly determine the angle \( \theta \), which then lets us convert the trigonometric function into an algebraic expression. This is a perfect example of how inverse trigonometric functions intertwine with other mathematical operations to simplify more complex trigonometric tasks.

Algebraic Expression Manipulation

Manipulating algebraic expressions is a core skill in advanced mathematics, crucial for simplifying complex expressions into more workable forms.
In this case, we transformed \( \tan(\cos^{-1} v) \) using identities to rewrite them as quarrel-free algebraic expressions in terms of \( v \).
Here's a step-wise breakdown:

• First, identify the inverse trigonometric function and understand the expected range and properties.
• Apply relevant trigonometric identities to exchange sine, cosine, and tangent based on known values.
• For \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), substitute \( \sin(\cos^{-1} v) = \sqrt{1 - v^2} \) and \( \cos(\cos^{-1} v) = v \) based on identity use.

These steps lead to the conclusion \( \tan (\cos^{-1} v) = \frac{\sqrt{1 - v^2}}{v} \), effectively turning the trigonometric expression into an algebraic one that's more straightforward for calculations. This manipulative technique showcases algebra's power in unveiling hidden simplicity behind complex operations.