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Chapter 6: Problem 70

In Exercises 69-88, evaluate each expression exactly. $$ \sin \left[\cos ^{-1}\left(\frac{2}{3}\right)\right] $$

Short Answer

Expert verified
\( \frac{\sqrt{5}}{3} \)

Step by step solution

01

Understanding the Problem

We need to evaluate the expression \( \sin \left[ \cos^{-1} \left(\frac{2}{3}\right) \right] \). This means we need to find the sine of an angle whose cosine is \( \frac{2}{3} \).
02

Use Right Triangle Definitions

Consider a right triangle where the angle \( \theta \) has a cosine of \( \frac{2}{3} \). According to the definition of cosine, this means \( \cos \theta = \frac{2}{3} \), or \( \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{2}{3} \). This yields an adjacent side of length 2 and a hypotenuse of length 3.
03

Determine the Opposite Side

Use the Pythagorean theorem, which is \( a^2 + b^2 = c^2 \), to find the length of the opposite side. Let \( a = 2 \) (adjacent side) and \( c = 3 \) (hypotenuse). The equation becomes:\[ (\text{opposite})^2 + 2^2 = 3^2 \] \[ (\text{opposite})^2 + 4 = 9 \] \[ (\text{opposite})^2 = 5 \] Thus, \( \text{opposite} = \sqrt{5} \).
04

Find the Sine Value

Now that we know the opposite side is \( \sqrt{5} \) and the hypotenuse is 3, we can find \( \sin \theta \) using the definition \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \). Therefore, \( \sin \theta = \frac{\sqrt{5}}{3} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cosine Inverse
The cosine inverse function, denoted as \( \cos^{-1} \), is a way to find the angle whose cosine value is a given number. It's essentially the reverse of the cosine function, telling us which angle we used to arrive at a specific cosine. Remember, cosine is the ratio of the length of the adjacent side to the hypotenuse in a right triangle. For instance, in the problem \( \cos^{-1} \left(\frac{2}{3}\right) \):
  • The given ratio \( \frac{2}{3} \) indicates the relation between the adjacent side and the hypotenuse.
  • The angle corresponding to this ratio is the angle, \( \theta \), whose cosine is \( \frac{2}{3} \).
This inverse function helps us bridge from ratio values back to angles, which is crucial in connecting various trigonometric relationships.
Right Triangle Definitions
A right triangle has one angle exactly 90 degrees. The sides have specific names based on their positions relative to the right angle:
  • Adjacent Side: The side adjacent to the angle of interest and not the hypotenuse.
  • Opposite Side: The side opposite the angle of interest.
  • Hypotenuse: The longest side, opposite the right angle.
These definitions form the foundation of trigonometric ratios. In the context of cosine, such as \( \cos \theta = \frac{2}{3} \):
  • Adjacent side = 2
  • Hypotenuse = 3
Clearly identifying these sides according to the angle enables the application of trigonometric functions like sine and cosine effectively.
Pythagorean Theorem
This fundamental principle in geometry relates the sides of a right triangle:\[ a^2 + b^2 = c^2 \]Where \( a \) and \( b \) are the triangle's legs, and \( c \) is the hypotenuse. In the exercise, to find the length of the opposite side:
  • Substitute the values into the Pythagorean theorem: \( 2^2 + \text{opposite}^2 = 3^2 \).
  • Solve to find \( \text{opposite} = \sqrt{5} \).
This calculation is vital to determining other trigonometric values, like sine, by providing the length of the remaining side of the triangle.
Sine Calculation
Sine of an angle in a right triangle is the ratio of the length of the opposite side to the hypotenuse. Using the information from our triangle:
  • The opposite side = \( \sqrt{5} \)
  • The hypotenuse = 3
Thus, the sine of the angle \( \theta \):\[ \sin \theta = \frac{\sqrt{5}}{3} \] Understanding this ratio helps in resolving the given problem and calculating trigonometric values. Sine computations are crucial in different applications, such as waves, physics, and engineering problems, where angles need to be resolved into their vertical components.

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Most popular questions from this chapter

In Exercises 1-36, solve each of the trigonometric equations exactly on the interval \(0 \leq x<2 \pi\). $$ \sin (2 x)=\sqrt{3} \sin x $$

For Exercises 87-92, refer to the following: Graphing calculators can be used to find approximate solutions to trigonometric equations. For the equation \(f(x)=g(x)\), let \(Y_{1}=f(x)\) and \(Y_{2}=g(x)\). The \(x\) values that correspond to points of intersections represent solutions. Use a graphing utility to solve the equation \(\cos \theta=\csc \theta\) on \(0 \leq \theta<\pi\).

In Exercises 37-48, solve each of the trigonometric equations on the interval \(0^{\circ} \leq \theta<360^{\circ}\). Give answers in degrees and round to two decimal places. $$ \frac{\sin ^{2} x-2 \sin x}{\tan x}=0 $$

In Exercises \(37-54\), solve each of the trigonometric equations on \(0^{\circ} \leq \theta<360^{\circ}\) and express answers in degrees to two decimal places. $$ 3 \cot ^{2} \theta+2 \cot \theta-4=0 $$

For Exercises 87-92, refer to the following: Graphing calculators can be used to find approximate solutions to trigonometric equations. For the equation \(f(x)=g(x)\), let \(Y_{1}=f(x)\) and \(Y_{2}=g(x)\). The \(x\) values that correspond to points of intersections represent solutions. Use a graphing utility to find all solutions to the equation \(\sin \theta=e^{\theta}\) for \(\theta \geq 0\).
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