Problem 82 - Algebra and Trigonometry Real Mathematics, Real People

Question

Complete the equation $$\arcsin \frac{\sqrt{36-x^{2}}}{6}=\arccos (\square), \quad 0 \leq x \leq 6$$

Step-by-Step Solution

Answer

The value required to complete the equation is \( \frac{x^2}{36} \)

Step 1: Simplify the Expression under the Arcsine Function

Start by simplifying the fraction under the arcsine function: \(\frac{\sqrt{36-x^{2}}}{6}\). Then, make it the square of a sine function: This is written as \(\sin^2y\), where \(y = \frac{x}{6}\). This gives us \(\arcsin (\sin^2y) = \arccos (\square) .\)

Step 2: Recall the Property of Arcsine and Arccosine

There's a formula related to the values of arcsine and arccosine: \(\arccos A = \frac{\pi}{2} - \arcsin A\). We can express the arcsine function in terms of arccosine, and simplify. So, the equation will become \(\frac{\pi}{2}- \arccos (\sin^2 y) = \arccos (\square) .\)

Step 3: Find the Value of the box

According to step 2, the box is equivalent to \(\sin^2 y\). That is, \(\frac{\pi}{2}- \arccos (\sin^2 y) = \arccos (\sin^2y)\) by the properties of sine and cosine functions. Thus, we can see that the missing box is \(\sin^2 y\). This can also be written as \(\frac{x^2}{36}\) by back substituting the value of \(y\).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions, often called arc-functions, are the inverses of the trigonometric functions. For instance, the arcsine (\texttt{arcsin}) of a value is the angle whose sine is that value. Similarly, the arccosine (\texttt{arccos}) gives us the angle whose cosine equals the provided value.

These functions are crucial because they allow us to find angles when we know the ratios of the sides of a right triangle. The output of \texttt{arcsin} and \texttt{arccos} is an angle, typically measured in radians. In a formula, if you see \texttt{arcsin}(x), it means you're looking for an angle whose sine is x.

The domains of these functions are limited to ensure they are functions in the mathematical sense (passing the vertical line test). That implies for \texttt{arcsin}, the domain is \texttt{[-1, 1]}, and the range is \texttt{[-\(\frac{\texttt{\textpi}}{2}\), \(\frac{\texttt{\textpi}}{2}\)]}. For \texttt{arccos}, the domain is also \texttt{[-1, 1]} but the range is \texttt{[0, \(\texttt{\textpi}\)]} to maintain a one-to-one relationship and ensure the output is a unique angle.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables. These identities are indispensable tools for solving trigonometric equations.

One of the most fundamental identities is the Pythagorean identity, which states that \texttt{sin\textsuperscript{2}x + cos\textsuperscript{2}x = 1} for any angle x. This identity is a consequence of the Pythagorean theorem and the definitions of sine and cosine in terms of the sides of a right triangle.

Complementary Angle Identities

When discussing \texttt{arcsin} and \texttt{arccos}, a complementary identity becomes particularly useful: \texttt{sin(\(\frac{\texttt{\textpi}}{2}\) - x) = cos(x)} and \texttt{cos(\(\frac{\texttt{\textpi}}{2}\) - x) = sin(x)}. These show the relationship between sine and cosine functions for complementary angles. Likewise, based on these identities, we can establish the link between \texttt{arcsin} and \texttt{arccos}: \texttt{\texttt{arccos}(x) = \(\frac{\texttt{\textpi}}{2}\) - \texttt{arcsin}(x)}. This is especially relevant when converting between the two inverse functions as demonstrated in the given exercise.

Solving Trigonometric Equations

Solving trigonometric equations involves finding all angles that satisfy the equation. The process often requires a blend of algebraic manipulation and the application of trigonometric identities.

When approaching a trigonometric equation, the first step is usually to express the equation in terms of a single trigonometric function if possible. Then, we use the appropriate inverse trigonometric function to find the angle that satisfies the equation. It is also important to consider the principal domain of inverse functions to find all possible solutions within the given interval.

In the presented exercise, we began by simplifing the expression under the arcsine function and then we leveraged trigonometric identities to express arcsine in terms of arccosine. This process is typical when solving trigonometric equations because it simplifies the equation into a form where the solution can be more easily identified.

The ultimate goal is to isolate the variable, which often involves back substituting the expressions we have for trigonometric functions in terms of the variable, as shown in the last step where we went from \texttt{sin\textsuperscript{2}y} back to \texttt{\(\frac{x\textsuperscript{2}}{36}\)}. Understanding these concepts creates a solid foundation that proves beneficial when tackling a wide array of trigonometric problems.