/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Give the degree measure of \(\th... [FREE SOLUTION] | 91Ó°ÊÓ

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Give the degree measure of \(\theta,\) if it exists. Do not use a calculator. $$\theta=\arcsin \left(-\frac{\sqrt{2}}{2}\right)$$

Short Answer

Expert verified
\(\theta = -45^\circ\).

Step by step solution

01

Understanding the problem

The problem is asking for the degree measure of the angle \( \theta \) for which \( \sin(\theta) = -\frac{\sqrt{2}}{2} \). This is done by evaluating the \( \arcsin \) function.
02

Identify the reference angle for sine

The value \( -\frac{\sqrt{2}}{2} \) is a commonly known value for sine at specific angles. Recall that \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \) and similarly \( \sin(-45^\circ) = -\frac{\sqrt{2}}{2} \). So, the reference angle is \( 45^\circ \).
03

Determine the correct angle in the arcsin range

The range for \( \arcsin \) function is \([-90^\circ, 90^\circ]\). Since the reference angle is \( 45^\circ \) and the sine value is negative, we are in the fourth quadrant. Thus, \( \theta = -45^\circ \).
04

Verify the solution

Check if \( \sin(-45^\circ) = -\frac{\sqrt{2}}{2} \). Since the sine of \(-45^\circ\) is indeed \(-\frac{\sqrt{2}}{2}\), the calculated angle is correct.

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Key Concepts

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

Arcsine Function
The arcsine function, often denoted as \( \arcsin(x) \), is the inverse of the sine function when restricted to its principal range. This means we can find the angle whose sine is a given number within the range [-1, 1]. The arcsine function allows us to reverse the sine when we know the result but need to find the angle. In mathematical terms, if \( y = \sin(x) \), then \( x = \arcsin(y) \) for angles within the specified range. The output of the arcsine function is an angle in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\) or equivalently, in degrees,
  • \(-90^\circ\) to \(90^\circ\)
This makes \( \arcsin(x) \) particularly useful, as it gives us a unique angle for every value of x between -1 and 1 that corresponds to a possible sine value. For example, when evaluating \( \arcsin\left(-\frac{\sqrt{2}}{2}\right) \), we look for an angle within this range whose sine is \(-\frac{\sqrt{2}}{2}\).
Sine Function
The sine function, a crucial aspect of trigonometry, relates an angle in a right triangle to the ratio of the length of the opposite side over the hypotenuse. While we typically think of sine in terms of acute angles in a right triangle, the function extends into the unit circle. This circular definition means that for any angle \(\theta\), \(\sin(\theta)\) represents the y-coordinate of the corresponding point on a unit circle. Some important sine values are commonly used, such as \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\).
When we encounter angles whose sine is negative, such as \(-\frac{\sqrt{2}}{2}\), it indicates that the angle lies in either the third or fourth quadrant of the circle. This is critical when processing inverse functions like arcsin, where negative values yield results that indicate angles below the horizontal axis.
Reference Angle
A reference angle is a simplified measure used to find an angle's trigonometric functions quickly. It is defined as the acute angle formed by the terminal side of the given angle and the horizontal axis. The reference angle is always between \(0^\circ\) and \(90^\circ\), making it a powerful tool for determining trigonometric values, especially in conjunction with the unit circle.
For example, to find the sine of an angle \(\theta\) such that \(\sin(\theta) = -\frac{\sqrt{2}}{2}\), we use the reference angle \(45^\circ\), known from the positive sine value \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\). However, since we are looking for a negative sine, the original angle \(\theta\) needs to be in a quadrant where sine values are negative. The reference angle tells us the absolute size of the angle without regard to direction (positive or negative).
Angle Measures
Understanding angle measures is crucial to solving trigonometric equations and evaluating inverse trigonometric functions correctly. Angles can be measured in degrees or radians, with degrees being more intuitive in common geometric contexts. It is essential to know which quadrant an angle resides in because it affects the sine, cosine, and tangent values associated with that angle.
To solve \(\arcsin\left(-\frac{\sqrt{2}}{2}\right)\), we recognize that the sine is negative, instructing us that the angle lies either in the third or fourth quadrant when considering the unit circle's layout. However, the arcsine's range imposes that we only consider angles between \(-90^\circ\) and \(90^\circ\). Therefore, \(-45^\circ\) is the correct angle measure within this range for which the sine value is \(-\frac{\sqrt{2}}{2}\). This understanding underpins the steps required to derive the correct answer in trigonometric problems.

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Most popular questions from this chapter

Solve each problem. Inducing Voltage A coil of wire rotating in a magnetic field induces a voltage $$ V=20 \sin \left(\frac{\pi t}{4}-\frac{\pi}{2}\right) $$ where \(t\) is time in seconds. Find the least positive time required to produce each voltage. (a) 0 (b) \(10 \sqrt{3}\)

Solve each equation ( \(x\) in radians and \(\theta\) in degrees) for all exact solutions where appropriate. Round approximale values in radians to four decimal places and approximate values in degrees to the nearest tenth. Write answers using the least possible nonnegative angle measures. $$\sin 2 x=2 \cos ^{2} x$$

Verify that each equation is an identity. $$\frac{\cos (A-B)}{\cos A \sin B}=\tan A+\cot B$$

Write each expression as an algebraic expression in \(u, u>0\). $$\cos (\arcsin u)$$

Sound Waves Sound is a result of waves applying pressure to a person's eardrum. For a pure sound wave radiating outward in a spherical shape, the trigonometric function $$P=\frac{a}{r} \cos \left(\frac{2 \pi r}{\lambda}-c t\right)$$ can be used to model the sound pressure \(P\) at a radius of \(r\) feet from the source, where \(t\) is time in seconds, \(\lambda\) is length of the sound wave in feet, \(c\) is speed of sound in feet per second, and \(a\) is maximum sound pressure at the source measured in pounds per square foot. (Source: Beranek, L.., Noise and Vibration Control, Institute of Noise Control Engineering. Washington, DC.) Let \(\lambda=4.9\) feet and \(c=1026\) feet per second. (IMAGE CANNOT COPY) (a) Let \(a=0.4\) pound per square foot. Graph the sound pressure at a distance \(r=10\) feet from its source over the interval \(0 \leq t \leq 0.05 .\) Describe \(P\) at this distance. (b) Now let \(a=3\) and \(t=10 .\) Graph the sound pressure for \(0 \leq r \leq 20 .\) What happens to the pressure \(P\) as the radius \(r\) increases? (c) Suppose a person stands at a radius \(r\) so that $$r=n \lambda$$ where \(n\) is a positive integer. Use the difference identity for cosine to simplify \(P\) in this situation.

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