Problem 71 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 71-76: Verify the identity. $$ \sin ^{-1} x=\tan ^{-1} \frac{x}{\sqrt{1-x^{2}}} $$

Step-by-Step Solution

Answer

The identity is true; \( \sin^{-1}(x) \) equals \( \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) \).

Step 1: Understand the Identity

We need to verify the identity \( \sin^{-1}(x) = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) \). This requires showing that both sides of the equation evaluate to the same value for a given \( x \).

Step 2: Consider Trigonometric Definitions

Recall the definition of \( \sin^{-1}(x) \), which is the angle whose sine is \( x \). Similarly, \( \tan^{-1}(y) \) is the angle whose tangent is \( y \). This means we need to find an expression involving \( \sin^{-1}(x) \) and evaluate the other side using trigonometric identities.

Step 3: Express Angles in Terms of a Right Triangle

Consider a right triangle where \( \theta = \sin^{-1}(x) \), meaning \( \sin(\theta) = x \). Since \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \), let the opposite side be \( x \) and the hypotenuse be \( 1 \). Then the adjacent side is \( \sqrt{1-x^2} \) by the Pythagorean theorem.

Step 4: Calculate \( \tan(\theta) \)

In the right triangle, \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{\sqrt{1-x^2}} \). Thus \( \tan\left(\sin^{-1}(x)\right) = \frac{x}{\sqrt{1-x^2}} \).

Step 5: Conclude the Verification

Since we have \( \tan\left(\sin^{-1}(x)\right) = \frac{x}{\sqrt{1-x^2}} \), and we know \( \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) \) is the angle whose tangent is \( \frac{x}{\sqrt{1-x^2}} \), it follows that \( \sin^{-1}(x) = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) \). Thus, the identity is verified.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are vital mathematical tools that help find the angle when given the value of a trigonometric function. The two inverse functions in this context are \( \sin^{-1}(x) \) and \( \tan^{-1}(x) \). For \( \sin^{-1}(x) \), the result is the angle \( \theta \) for which the sine value is \( x \). Similarly, \( \tan^{-1}(x) \) gives the angle whose tangent is \( x \). These functions are crucial when verifying identities like \( \sin^{-1} x = \tan^{-1} \left( \frac{x}{\sqrt{1-x^2}} \right) \).

• Definition: \( \sin^{-1}(x) \) gives the angle with sine value \( x \) and \( \tan^{-1}(x) \) gives the angle with tangent \( x \).
• Usage: These are used when dealing with angles rather than direct trigonometric ratios.

Inverse trigonometric functions are essential in solving triangles where only side lengths are known and you need to find angle measures.

Right Triangle Trigonometry

Right triangle trigonometry is a fundamental element in mathematics used to relate angles and sides of right triangles. In a right triangle, the sides have specific names: the hypotenuse, which is the longest side, and the opposite and adjacent sides relative to the angle in question.Using trigonometric functions like sine, cosine, and tangent, we can find unknown side lengths or angles. For example, if you have a right triangle where \( \theta = \sin^{-1}(x) \), then you know: \( \sin(\theta) = x \). This means:

• Sine: \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \)
• Cosine: \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \)
• Tangent: \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \)

Understanding the layout of a right triangle makes it easier to visualize and solve trigonometric identities by matching angles and sides correctly.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry that is applied in right triangle trigonometry. It allows you to compute an unknown side of a right triangle when the other two sides are known. The theorem is often represented as \( a^2 + b^2 = c^2 \). Here is how it's generally used:

• In the context: If \( \theta = \sin^{-1}(x) \), with opposite side \( x \) and hypotenuse 1, the remaining side (adjacent) is found using \( \sqrt{1-x^2} \).
• Importance: Knowing this side helps in calculating \( \tan(\theta) \), which is \( \frac{x}{\sqrt{1-x^2}} \).

By confirming the triangle dimensions meet the condition \( a^2 + b^2 = c^2 \), the validity of the sides' lengths is ensured, even as it applies to solving and verifying identities like \( \sin^{-1} x = \tan^{-1} \left( \frac{x}{\sqrt{1-x^2}} \right) \).

Angle Verification

Angle verification is the process of confirming if two angles are indeed the same or correspond to each other in a given trigonometric identity. In our identity \( \sin^{-1}(x) = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) \), we must prove that both side expressions predict the same angle.Steps to verify an angle include:

• Calculate one side: Start with \( \sin^{-1}(x) \) to find an angle \( \theta \) such that \( \sin(\theta) = x \).
• Use the identity: Apply Pythagorean theorem to determine the last triangle side and calculate \( \tan(\theta) \).
• Match angles: Confirm \( \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) \) gives the same angle, thereby verifying the original identity.

This method of angle verification helps to establish trust in your mathematical work by ensuring logical consistency and correctness across all parts of trigonometric calculations.