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Magazine Chapter 7: Problem 37
Evaluate the expressions in Exercises \(29-40\) $$ \cos \left(\sin ^{-1} \frac{2 y}{3}\right) $$
Short Answer
Expert verified
The value is \( \frac{\sqrt{9 - 4y^2}}{3} \).
Step by step solution
01
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 \( \sin(\theta) = \frac{2y}{3} \).
02
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 \( \sin(\theta) = \frac{2y}{3} \).
03
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 \( a = \sqrt{9 - 4y^2} \).
04
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 \( \cos(\theta) = \frac{\sqrt{9 - 4y^2}}{3} \).
05
Write the Final Expression
Therefore, the expression for \( \cos \left( \sin^{-1} \frac{2y}{3} \right) \) simplifies to \( \frac{\sqrt{9 - 4y^2}}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( \frac{2y}{3} \).
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 \(3\).
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 \( a = \sqrt{9 - 4y^2} \).
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.
