Problem 103 - Precalculus Essentials

Question

Write each trigonometric expression as an algebraic expression (that is, without any trigonometric fienctions). Assume that \(x\) and \(y\) are positive and in the domain of the given inverse trigonometric function. $$\cos \left(\sin ^{-1} x-\cos ^{-1} y\right)$$

Step-by-Step Solution

Answer

The algebraic expression equivalent to \(\cos(\sin^{-1}(x) - \cos^{-1}(y))\) is \(\sqrt{1 - x^2} * y + x * \sqrt{1 - y^2}\).

Step 1: Understand Inverse Trigonometric Functions

Consider the inverse sine function as an angle, i.e. \(a = \sin^{-1}(x)\). Thus, from the definition of sine function, we know the opposite side over hypotenuse is \(x\). We can calculate the adjacent side using the Pythagorean theorem, i.e. \(\sqrt{1 - x^2}\). Now we consider the inverse cosine function as an angle, i.e. \(b = \cos^{-1}(y)\). Here the adjacent side over the hypotenuse is \(y\), and thus using Pythagorean theorem, we calculate the value of the opposite side as \(\sqrt{1 - y^2}\).

Step 2: Use the Cosine Difference Formula

Then, the given equation can be solved using the formula for the cosine of the difference of two angles. \(a\) and \(b\) are two angles such that cos(\(a - b\)) can be written as \(\cos(a)\cos(b) + \sin(a)\sin(b)\). Substituting for \(\cos(a)\), \(\sin(a)\), \(\cos(b)\) and \(\sin(b)\) from step 1, we have \(\cos(\sin^{-1}(x) - \cos^{-1}(y)) = \sqrt{1 - x^2} * y + x * \sqrt{1 - y^2}\).

Step 3: Simplify the Expression

Simplifying the right side of the equation gives the final algebraic version of the given trigonometric expression. This doesn't involve any trigonometric functions.

Key Concepts

Pythagorean theorem

The Pythagorean theorem is a fundamental principle in geometry, used to relate the sides of a right triangle. This theorem is often expressed as \(a^2 + b^2 = c^2\), where \(c\) represents the hypotenuse, or the longest side, of the right triangle, and \(a\) and \(b\) are the two other sides. It is crucial in this exercise because we need to find the missing sides of right triangles formed by the angles involving inverse trigonometric functions.
When we take \(\sin^{-1}(x)\) and view it as an angle in a right triangle, the side opposite this angle is \(x\) and the hypotenuse is 1. The Pythagorean theorem helps us determine that the adjacent side is \(\sqrt{1 - x^2}\). Similarly, for \(\cos^{-1}(y)\), the adjacent side is \(y\) and the hypotenuse is 1, which leads us to find the opposite side as \(\sqrt{1 - y^2}\).
These calculations allow us to use trigonometric identities effectively by replacing trigonometric ratios with their algebraic counterparts.

Cosine of the difference of two angles

One of the key trigonometric identities used to solve this problem is the formula for the cosine of the difference of two angles. This formula is represented as:

• \( \cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b) \)

Understanding and applying this formula is crucial when dealing with composite trigonometric expressions where two angles are subtracted. In this context, \(a = \sin^{-1}(x)\) and \(b = \cos^{-1}(y)\). By substituting the previously found values for \(\cos(a)\), \(\sin(a)\), \(\cos(b)\), and \(\sin(b)\), we can transform the original trigonometric expression into its algebraic form.
In this exercise:

• \( \cos(a) = \sqrt{1 - x^2} \)
• \( \sin(a) = x \)
• \( \cos(b) = y \)
• \( \sin(b) = \sqrt{1 - y^2} \)

Substituting these into the cosine difference formula provides us with the expression \( \sqrt{1 - x^2} \cdot y + x \cdot \sqrt{1 - y^2} \), which is free of trigonometric functions.

Trigonometric identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables where both sides of the equality are defined. These tools are indispensable when manipulating and simplifying trigonometric expressions.
In this exercise, the identity for the cosine of the difference of two angles plays a pivotal role. It transforms the expression \( \cos(\sin^{-1}(x) - \cos^{-1}(y)) \) into an algebraic expression using the identities for cosine and sine for respective angles:

• \( \cos(a) = \sqrt{1 - x^2} \)
• \( \sin(a) = x \)
• \( \cos(b) = y \)
• \( \sin(b) = \sqrt{1 - y^2} \)

By using these trigonometric identities, we replace trigonometric terms with algebraic terms, making it easier for us to carry out algebraic operations. Such transformations are crucial, especially in cases where working directly with trigonometric functions can complicate the problem due to domain considerations or in simplifying expressions for further calculations.