Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that each equation is an identity. $$ \sin \left(\tan ^{-1} x\right)=\frac{x}{\sqrt{1+x^{2}}} $$

Step-by-Step Solution

Answer

The equation is an identity: \( \sin(\tan^{-1}(x)) = \frac{x}{\sqrt{1+x^2}} \).

Step 1: Understand Inverse Trigonometric Functions

We start by recalling that the inverse tangent function, \( \tan^{-1}(x) \), returns an angle \( \theta \) whose tangent is \( x \). Thus, we have \( \tan(\theta) = x \).

Step 2: Use Trigonometric Identity

We use the identity \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \). For a right triangle where \( \tan(\theta) = \frac{x}{1} \), the opposite side is \( x \) and the adjacent side is \( 1 \).

Step 3: Determine Hypotenuse

We find the hypotenuse using the Pythagorean Theorem: hypotenuse = \( \sqrt{x^2 + 1^2} = \sqrt{1 + x^2} \).

Step 4: Calculate Sine of Angle

Using \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \), we find \( \sin(\theta) = \frac{x}{\sqrt{1 + x^2}} \). Therefore, \( \sin(\tan^{-1}(x)) = \frac{x}{\sqrt{1 + x^2}} \), confirming the identity.

Key Concepts

Inverse Trigonometric Functions

Understanding inverse trigonometric functions is a key aspect of solving the given identity problem. These functions are essentially the opposites of the regular trigonometric functions like sine, cosine, and tangent. When we talk about the inverse tangent function, denoted as \( \tan^{-1}(x) \), what we're dealing with is finding the angle \( \theta \) where the tangent is equal to \( x \). Therefore, if \( \tan(\theta) = x \), then \( \theta = \tan^{-1}(x) \). This angle is important because knowing \( \theta \) allows us to use other trigonometric identities, like sine and cosine, to find relationships involving \( x \).

These inverse functions come in very handy, especially in situations where we need to determine angles from given trigonometric ratios. So here, \( \tan^{-1}(x) \) isn't just a number; it's expressing an angular relationship that helps us derive the sine of that same angle. Remember, the range of \( \tan^{-1}(x) \) is typically between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), making this angle a familiar friend when analyzing right triangles.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental concept in geometry and trigonometry that relates the lengths of the sides of a right triangle. It states that the square of the hypotenuse (the longest side opposite the right angle) is equal to the sum of the squares of the other two sides. Algebraically, this is expressed as \( a^2 + b^2 = c^2 \), where \( c \) is the hypotenuse.

In the context of the given identity, knowing that \( \tan(\theta) = \frac{x}{1} \) allows us to express the legs of a right triangle in terms of opposite and adjacent sides. Here, the opposite side is \( x \) and the adjacent is \( 1 \). To find the hypotenuse, we apply the formula: hypotenuse = \( \sqrt{x^2 + 1^2} = \sqrt{1 + x^2} \).

This relationship is crucial when solving trigonometric identities because it provides us the actual lengths needed to calculate other ratios, like sine, as used in the exercise to confirm \( \sin(\tan^{-1}(x)) \). The Pythagorean Theorem ensures everything aligns perfectly in our calculations.

Sine Function

The sine function is one of the basic trigonometric functions that relates the angle of a triangle to the ratio of the opposite side to the hypotenuse. It is particularly useful for solving trigonometric problems, as it often appears in identities and equations. The formula for the sine of an angle \( \theta \), in a right triangle, is given by \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \).

In the problem, after converting \( \tan^{-1}(x) \) into an angle \( \theta \), we translate that into a scenario involving a right triangle. With \( \tan(\theta) = \frac{x}{1} \), it follows that the sine of this angle is \( \sin(\theta) = \frac{x}{\sqrt{1+x^2}} \), using the Pythagorean Theorem to find the hypotenuse as explained earlier.

This identity showcase that the sine function can elegantly connect with inverse functions, simplifying what might initially seem like complex mathematical expressions into more straightforward relationships.