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Chapter 7: Problem 12

Find the following exactly in radians and degrees. $$\sin ^{-1} \frac{1}{2}$$

Short Answer

Expert verified
\( \text{sin}^{-1} \frac{1}{2} = \frac{\text{蟺}}{6} \text{ radians} = 30^\text{o} \).

Step by step solution

01

Understand the Inverse Sine Function

The function \(\text{sin}^{-1}(x)\) or \(\text{arcsin}(x)\) is the inverse of the sine function. It gives the angle for which the sine value is equal to \(x\) within the range \(-\frac{\text{蟺}}{2} \leq y \leq \frac{\text{蟺}}{2}\).
02

Identify the Sine Value

Notice that \(\text{sin}^{-1} \frac{1}{2}\) means we need to find the angle \(y\) such that \(\text{sin}(y) = \frac{1}{2}\).
03

Find the Corresponding Angle in Radians

The angle in radians whose sine is \(\frac{1}{2}\) is \(\frac{\text{蟺}}{6}\), since \(\text{sin}(\frac{\text{蟺}}{6}) = \frac{1}{2}\).
04

Convert the Angle to Degrees

To convert from radians to degrees, use the conversion factor \(180^\text{o} / \text{蟺}\). Thus, \(\frac{\text{蟺}}{6}\) radians is equal to \( \frac{\text{蟺}}{6} \times \frac{180^\text{o}}{\text{蟺}} = 30^\text{o} \).

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Key Concepts

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

arcsine
The arcsine function, represented as \(\text{sin}^{-1}(x)\) or \(\text{arcsin}(x)\), is the inverse of the sine function. This means it helps you find the angle when you know the sine of that angle. If you understand that the sine of an angle measures how high above or below the x-axis a certain point on the unit circle is, the arcsine reverses this process. It tells you which angle has a given sine value. The range of \(\text{arcsin}\) is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) radians (or from -90掳 to 90掳). This is because sine values repeat every 2蟺 radians (360掳), but in order to have a unique result for arcsine, we restrict it to this range.
radians to degrees conversion
Converting angles from radians to degrees is an essential skill in trigonometry. Radians and degrees are the two most common units for measuring angles. Here is a simple way to convert between the two:
  • There are \(2 \pi\) radians in a full circle, which equals 360 degrees.
  • To convert radians to degrees, multiply by \(\frac{180^{o}}{\pi}\)
  • To convert degrees to radians, multiply by \(\frac{\pi}{180^{o}}\)
For example, to convert \(\frac{\pi}{6}\) radians to degrees, you multiply \(\frac{\pi}{6} \times \frac{180^{o}}{\pi} \), which equals 30掳. Conversions like these help when switching between calculating within geometry (which often uses degrees) and trigonometry (which often uses radians).
sine function
The sine function is a fundamental trigonometric function used to relate angles to side lengths in right triangles and define the properties of waves. Here's what you need to know:
  • The sine of an angle in a right triangle is the ratio of the length of the opposite side to the hypotenuse. If 胃 is an angle in a right triangle, then \(\text{sin}(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}\)
  • The function is periodic with a period of \(2\pi\) radians (360掳), meaning its values repeat every \(2\pi\) radians.
  • Sine values range between -1 and 1, since it measures a ratio of side lengths.
  • Key angles to know are \(\theta = 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3},\ and \frac{\pi}{2}\), corresponding to sine values of 0, \(\frac{1}{2}\), \(\frac{\sqrt{2}}{2}\), \(\frac{\sqrt{3}}{2}\), and 1, respectively.
Understanding these key points can help solve many problems involving the sine function in both practical and theoretical contexts.

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Most popular questions from this chapter

Consider the following functions ( \(a\) )- ( \(f\) ). Without graphing them, answer question. a) \(f(x)=2 \sin \left(\frac{1}{2} x-\frac{\pi}{2}\right)\) b) \(f(x)=\frac{1}{2} \cos \left(2 x-\frac{\pi}{4}\right)+2\) c) \(f(x)=-\sin \left[2\left(x-\frac{\pi}{2}\right)\right]+2\) d \(f(x)=\sin (x+\pi)-\frac{1}{2}\) e) \(f(x)=-2 \cos (4 x-\pi)\) f) \((x)=-\cos \left[2\left(x-\frac{\pi}{8}\right)\right]\) Which functions have a graph with a phase shift of \(\frac{\pi}{4} ?\)

Simplify. $$\sin (u-v) \cos v+\cos (u-v) \sin v$$

First write each of the following as a trigonometric function of a single angle. Then evaluate. $$\frac{\tan 35^{\circ}-\tan 12^{\circ}}{1+\tan 35^{\circ} \tan 12^{\circ}}$$

Show that each of the following is not an identity by finding a replacement or replacements for which the sides of the equation do not name the same number. Then use a graphing calculator to show that the equation is not an identity. $$\tan ^{2} \theta+\cot ^{2} \theta=1$$

Solve, finding all solutions in \([0,2 \pi)\). $$\frac{\sin ^{2} x-1}{\cos \left(\frac{\pi}{2}-x\right)+1}=\frac{\sqrt{2}}{2}-1$$
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