Problem 37 - Thomas Calculus

Question

Evaluate the expressions in Exercises \(29-40\) $$ \cos \left(\sin ^{-1} \frac{2 y}{3}\right) $$

Step-by-Step Solution

Answer

The value is \( \frac{\sqrt{9 - 4y^2}}{3} \).

Step 1: Understand the Inverse Sine Function

The expression involves \( \sin^{-1} (x) \), which is the inverse sine function. It gives an angle \( \theta \) such that \( \sin(\theta) = x \). In this case, \( \sin^{-1} \left( \frac{2y}{3} \right) = \theta \) which implies that \( \sin(\theta) = \frac{2y}{3} \).

Step 2: Set Up the Right Triangle

Imagine a right triangle where \( \theta \) is an angle such that the side opposite \( \theta \) has length \( 2y \) and the hypotenuse has length \( 3 \). This setup satisfies the equation \( \sin(\theta) = \frac{2y}{3} \).

Step 3: Use the Pythagorean Theorem

To find \( \cos(\theta) \), we need the length of the adjacent side of the triangle. Use the Pythagorean theorem: \( a^2 + (2y)^2 = 3^2 \) to solve for \( a \). Simplify to get \( a^2 + 4y^2 = 9 \), hence \( a^2 = 9 - 4y^2 \), and finally \( a = \sqrt{9 - 4y^2} \).

Step 4: Find the Cosine of Theta

Now use the definition of cosine, \( \cos(\theta) = \frac{\text{adjacent side}}{\text{hypotenuse}} \). With \( a = \sqrt{9 - 4y^2} \) and the hypotenuse being \( 3 \), the expression becomes \( \cos(\theta) = \frac{\sqrt{9 - 4y^2}}{3} \).

Step 5: Write the Final Expression

Therefore, the expression for \( \cos \left( \sin^{-1} \frac{2y}{3} \right) \) simplifies to \( \frac{\sqrt{9 - 4y^2}}{3} \).

Key Concepts

Inverse Sine

The inverse sine function, denoted as \( \sin^{-1} \), is a trigonometric function that retrieves an angle when given a sine value. This process essentially 'reverses' the sine function. For example, if \( \sin(\theta) = x \), applying the inverse sine function results in \( \sin^{-1}(x) = \theta \).

• The function is restricted to angles between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\) to remain classified as a function, providing a unique angle for each sine value.
• In our problem, this function helps determine the angle \( \theta \) given the value \( \frac{2y}{3} \).

By using the inverse sine, we transitioned from a sine value to an angle. This crucial step sets the stage for the subsequent geometrical interpretation.

Right Triangle

A right triangle is a triangle with one of its angles measuring exactly 90 degrees, or \( \pi/2 \) radians. It provides a solid groundwork for connecting trigonometric values to side lengths.

• The sides are traditionally labeled as the opposite, adjacent, and hypotenuse relative to an angle of interest.
• In the context of \( \sin(\theta) = \frac{2y}{3} \), a right triangle is constructed where the side opposite angle \( \theta \) has length \(2y\) and the hypotenuse is \(3\).

This setup allows us to visually represent and work with the given sine value, forming a springboard into further calculations like using the Pythagorean theorem.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry, particularly applicable to right triangles. It states that in such a triangle, the sum of the squares of the lengths of the two shorter sides equals the square of the length of the hypotenuse: \( a^2 + b^2 = c^2 \).

• In this exercise, the theorem facilitates the computation of the triangle's unknown side, labeled \( a \).
• Given the sides \( 2y \) and \( 3 \) (the hypotenuse), we apply the theorem: \( a^2 + (2y)^2 = 3^2 \).
• Solving this yields \( a^2 = 9 - 4y^2 \) and consequently \( a = \sqrt{9 - 4y^2} \).

This process equips us with the adjacent side's length necessary for calculating the cosine function.

Cosine Function

The cosine function measures the ratio of the length of the adjacent side to the hypotenuse in a right triangle, for a given angle \( \theta \). Mathematically, it is defined as \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \).

• In our scenario, having determined the adjacent side as \( \sqrt{9 - 4y^2} \) and with the hypotenuse as \(3\), we compute \( \cos(\theta) = \frac{\sqrt{9 - 4y^2}}{3} \).
• This calculation provides the solution to the exercise, embodying the elegant bridging of inverse sine and cosine through the geometry of triangles.

The cosine function effectively allows us to express an angle's distinguishing traits in terms of side ratios, completing the trigonometric translation task at hand.