Problem 38 - A History of Mathematics: An Introduction

Question

38\. If \(y=\arctan x\), show that \(\sin y=x / \sqrt{1+x^{2}}\) and \(\cos y=\) \(1 / \sqrt{1+x^{2}}\). Then, if \(p=x / \sqrt{1+x^{2}}\), show that \(\sqrt{1-p^{2}}=\) \(1 / \sqrt{1+x^{2}}\). Since \(y=\arcsin p\), it follows that \(d y=\) \(d p / \sqrt{1-p^{2}}\) and \(d p=d x /\left(1+x^{2}\right)^{3 / 2}\). Conclude that $$ d y=\frac{d x}{1+x^{2}} \quad \text { (Euler) } $$

Step-by-Step Solution

Answer

Short Answer: We are given the function y = \(\arctan x\), and we are asked to derive relationships involving \(\sin y\), \(\cos y\), a new variable p, and their differentials. We first express y as \(\arctan x\) and find x in terms of y. Then, we express \(\sin y\) and \(\cos y\) in terms of x using the Pythagorean identity. Next, we define a new variable \(p = \frac{x}{\sqrt{1+x^2}}\) and find an expression for \(\sqrt{1-p^2}\). After relating y and p, we find the differentials dy, dp, and dx. Finally, we find the relationship between dy and dx, which is consistent with Euler's formula.

Step 1: Express y as \(\arctan x\) and find x in terms of y

Given that \(y = \arctan x\), we know that \(x = \tan y\). Now we can use this relationship to find expressions for \(\sin y\) and \(\cos y\) in terms of x.

Step 2: Express \(\sin y\) and \(\cos y\) in terms of x

Using the Pythagorean identity, we know that \(\sin^2 y + \cos^2 y = 1\). Since \(x = \tan y\), we have \(\sin y = x\cos y\). We can square both sides to get \((\sin y)^2 = x^2(\cos y)^2\). Now, substitute the identity \(1 - (\cos y)^2\) for \((\sin y)^2\) to get \(1 - (\cos y)^2 = x^2(\cos y)^2\). Rearranging terms, we find that \((1+x^2)(\cos y)^2 = 1\). Therefore, \(\cos y = \frac{1}{\sqrt{1 + x^2}}\) and \(\sin y = \frac{x}{\sqrt{1 + x^2}}\).

Step 3: Define p and find an expression for \(\sqrt{1-p^2}\)

Now, let \(p = \frac{x}{\sqrt{1+x^2}}\). Squaring both sides, we get \(p^2 = \frac{x^2}{1+x^2}\). Subtracting \(p^2\) from 1, we have \(1-p^2 = \frac{1}{1+x^2}\). Taking the square root, we get \(\sqrt{1-p^2} = \frac{1}{\sqrt{1+x^2}}\).

Step 4: Relate y and p and find the differentials

Since \(\sin y = p\), we have \(y = \arcsin p\). Now, we need to find the differentials dy, dp, and dx. Differentiating both sides of \(y = \arcsin p\) with respect to y, we get \(dy = \frac{dp}{\sqrt{1-p^2}}\). Differentiating the expression for p with respect to x, we get \(dp = \frac{dx}{(1+x^2)^{3/2}}\).

Step 5: Find the relationship between dy and dx

We need to find dy expressed as a function of dx and x. We can use the expressions found in Step 4 to do this: \(\frac{dy}{dx} = \frac{dp}{\sqrt{1-p^2}} \cdot \frac{dx}{dp} = \frac{dp}{\sqrt{1-p^2}} \cdot \frac{dx}{(1+x^2)^{3/2}}\). Using the relationship between p and x found in Step 3, we can simplify this to obtain \(\frac{dy}{dx} = \frac{dx}{1+x^2}\), which is consistent with Euler's formula.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverses of the regular trigonometric functions. They allow us to find an angle when given the value of a trigonometric function. Common inverse trigonometric functions include \( \arcsin(x) \) (inverse sine), \( \arccos(x) \) (inverse cosine), and \( \arctan(x) \) (inverse tangent). These functions are crucial when you need to determine an angle from a ratio in right-angled triangles. For example, if \( y = \arctan(x) \), you know that x represents the tangent of angle y. Consequently, you can find \( \sin(y) \) and \( \cos(y) \) in terms of x:\
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\( \sin(y) = \frac{x}{\sqrt{1+x^2}} \) and \( \cos(y) = \frac{1}{\sqrt{1+x^2}} \).
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Understanding inverse trigonometric functions is also crucial when dealing with calculus, as they often present themselves in problems involving integration and differentiation. When solving for \( dy \) with respect to \( dx \) for the function \( y = \arctan(x) \) as shown in the exercise, these functions help structure the relationship between the differential of the angle and the original function.

Trigonometric Identities

Trigonometric identities are equalities that involve trigonometric functions and are true for all values of the variables involved. One of the most widely used is the Pythagorean identity, which describes the relationship between the sine and cosine of an angle: \( \sin^2(y) + \cos^2(y) = 1 \). Such identities are essential tools for simplifying expressions involving trigonometric functions. They are particularly useful in calculus for tasks like integration, solving differential equations, and finding limits.
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In the given exercise, the identity is used to express \( \sin(y) \) and \( \cos(y) \) in terms of x. This is a fundamental step that prevents the equations from becoming too complex and allows for easier manipulation when differentiating or integrating trigonometric functions.

Pythagorean Identity

The Pythagorean identity is a specific form of trigonometric identity that is derived from the Pythagorean theorem. In its most common form, it states that for any angle y, \( \sin^2(y) + \cos^2(y) = 1 \). This identity is incredibly useful when you need to convert between sine and cosine or to simplify trigonometric expressions.
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In our exercise, we utilize this identity to find \( \cos(y) \) given \( \sin(y) \). By recognizing that \( x = \tan(y) = \frac{\sin(y)}{\cos(y)} \), we could manipulate the Pythagorean identity to solve for the cosine and sine in terms of x, which leads us to the expressions \( \cos(y) = \frac{1}{\sqrt{1 + x^2}} \) and \( \sin(y) = \frac{x}{\sqrt{1 + x^2}} \), streamlining the process of finding the differential.

Implicit Differentiation

Implicit differentiation is a technique used when a function is not defined explicitly in terms of one variable. Instead of being directly solved for one variable in terms of another, an equation defines a relationship between them. When we differentiate both sides of this equation with respect to a variable, we are implicitly differentiating the function.
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In the context of our exercise, we implicitly differentiate the function \( y = \arcsin(p) \) to find \( dy \) in terms of \( dp \) without solving for \( p \) explicitly. As \( p \) itself is a function of \( x \) defined by \( p = \frac{x}{\sqrt{1+x^2}} \), we also differentiate it with respect to \( x \) which leads us to the expression \( dp = \frac{dx}{(1+x^2)^{3/2}} \). Combining these differentials, we ultimately arrive at Euler’s formula, showing that \( dy = \frac{dx}{1+x^2} \), which succinctly describes the relationship between the differential of y and x in terms of a function we started with, \( y = \arctan(x) \).