Problem 58 - Precalculus with Limits

Question

In Exercises 57 - 60, write the trigonometric expression as an algebraic expression. \( \sin\left(\arctan 2x - \arccos x\right) \)

Step-by-Step Solution

Answer

The algebraic expression equivalent to the given trigonometric expression is \( \frac{2x^2}{\sqrt{(1+(2x)^2)(1-x^2)}} - \frac{\sqrt{1-x^2}}{\sqrt{1+(2x)^2}} \)

Step 1: Apply the identities for arc-tan and arc-cos

Express \(2x\) and \(x\) in terms of sine and cosine. Given that \( \tan(y) = \frac{a}{b} \) and \( \cos(y) = a \), we can use the Pythagorean identity \( \sin^2(y) + \cos^2(y) = 1 \) to express these. The triangle for arctan will be opposite over adjacent (2x over 1) and the triangle for arccos will be adjacent over hypotenuse (x over 1). By using the Pythagorean theorem, the hypotenuse in each triangle can be found as \( \sqrt{1+(2x)^2} \) and \( \sqrt{1-x^2} \), respectively.

Step 2: Apply the sine of difference formula

Now we have all the necessary components to apply the difference of angles identity for sine which is \( \sin(A - B) = \sin(A) \cos(B) - \cos(A) \sin(B) \). The two angles in this case are \( \arctan(2x) \) and \( \arccos(x) \). By substituting the sine and cosine of these angles based on our constructed right triangles, we get \( \sin(\arctan 2x - \arccos x) = \left(\frac{2x}{\sqrt{1+(2x)^2}} \right) \left( \frac{x}{\sqrt{1-x^2}} \right) - \left(\frac{1}{\sqrt{1+(2x)^2}} \right) \left(\sqrt{1-x^2}\right) \).

Step 3: Simplify the expression

The last step is simplifying the obtained expression. The result is \( \frac{2x^2}{\sqrt{(1+(2x)^2)(1-x^2)}} - \frac{\sqrt{1-x^2}}{\sqrt{1+(2x)^2}} \). It is the algebraic version of the original trigonometric expression.

Key Concepts

Algebraic Expression

An algebraic expression is a combination of constants, variables, and arithmetic operations like addition, subtraction, multiplication, and division. In this exercise, we convert a trigonometric expression into an algebraic expression using mathematical identities and properties.
To start, recall that trigonometric expressions involve functions like sine, cosine, and tangent, often needing specific angles or transformations. In contrast, algebraic expressions rely on simpler operations.
When you transform a trigonometric expression into its algebraic equivalent, you simplify calculations by expressing it without trigonometric functions. To achieve this, one might derive values from a constructed right triangle using known components, such as sides or angles, then apply these to derive the desired algebraic form.

Sine of Difference Formula

The sine of difference formula is crucial for solving expressions involving the sine of two angles subtracted from each other. It states: \(\sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B)\).
This formula decomposes the sine of a difference into a combination of products of sinusoids, making complex trigonometric problems more solvable.
Applying this rule involves identifying the angles in question—in this case, the inverse trigonometric functions \({\arctan(2x)}\) and \({\arccos(x)}\)\—then expressing them as sides of a constructed right triangle.
By transforming these angles into terms of sine and cosine using the triangles, it becomes feasible to directly apply the formula, yielding a simplified algebraic expression.

Pythagorean Identity

The Pythagorean identity is a foundational concept in trigonometry stating that \( \sin^2(y) + \cos^2(y) = 1 \).
This identity helps connect trigonometric functions to algebraic forms, particularly when dealing with inverse trigonometric transformations. A common application is finding missing sides of a right triangle given one side and the trigonometric ratio. For instance, using triangles defined by \(\arctan(2x)\) and \(\arccos(x)\), we employ the Pythagorean identity to deduce their respective hypotenuses, \(\sqrt{1+(2x)^2}\) and \(\sqrt{1-x^2}\).
With these hypotenuses, we can express the sine and cosine values as ratios of triangle sides, crucial for transforming our trigonometric expression into its algebraic counterpart.

Inverse Trigonometric Functions

Inverse trigonometric functions take a trigonometric ratio and return the angle that would produce it. This transformation is key when dealing with expressions like \(\arctan(2x)\) and \(\arccos(x)\), which relate directly to specific angles in a right triangle.

• \(\arctan\) gives us an angle with tangent \(2x\), interpreted as the ratio of opposite to adjacent sides.
• \(\arccos\) involves finding an angle with cosine \(x\), reflecting the ratio of the adjacent side to the hypotenuse.

Understanding these functions allows us to convert geometric relationships into algebraic terms by constructing right triangles.
This conversion informs subsequent calculations using identities like the sine of difference formula and the Pythagorean identity, ultimately allowing for comprehensive algebraic expression derivation.