Chapter 7: Problem 52
Use a Double- or Half-Angle Formula to solve the equation in the interval \([0,2 \pi)\) $$\sin \theta-\cos \theta=\frac{1}{2}$$
Short Answer
Expert verified
The solution in \([0, 2\pi)\) is given by solving \( \theta = \alpha + \frac{\pi}{4} \) and \( \theta = \pi - \alpha + \frac{\pi}{4} \).
Step by step solution
01
Use identity to transform the equation
We start with the equation: \(\sin \theta - \cos \theta = \frac{1}{2}\). To simplify this equation, we can use the identity \(\sin(\theta - \frac{\pi}{4}) \). This is derived from: \[ \sin \theta = \cos \frac{\pi}{4} \sin \theta + \sin \frac{\pi}{4} \cos \theta, \]\[ \cos \theta = \sin \frac{\pi}{4} \sin \theta + \cos \frac{\pi}{4} \cos \theta. \]Substituting into the equation, \[ \sin \theta - \cos \theta = \sqrt{2} \sin\left(\theta - \frac{\pi}{4}\right), \]we have:\[ \sqrt{2} \sin\left(\theta - \frac{\pi}{4}\right) = \frac{1}{2}. \]
02
Solve for the trigonometric function
Now, we solve for \(\sin\left(\theta - \frac{\pi}{4}\right)\): \[ \sqrt{2} \sin\left(\theta - \frac{\pi}{4}\right) = \frac{1}{2} \]\[ \sin\left(\theta - \frac{\pi}{4}\right) = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4} \]Therefore, we need to solve the equation: \[ \sin\left(\theta - \frac{\pi}{4}\right) = \frac{\sqrt{2}}{4}. \]
03
Determine the solutions for the sine equation
The general solution for \( \sin x = a \) is:\[ x = \sin^{-1}(a) + 2\pi k \quad \text{or} \quad x = \pi - \sin^{-1}(a) + 2\pi k, \]where \( k \) is an integer.Replacing \( x \) with \( \theta - \frac{\pi}{4} \) and \( a \) with \( \frac{\sqrt{2}}{4} \), we find: \[ \theta - \frac{\pi}{4} = \sin^{-1}\left(\frac{\sqrt{2}}{4}\right) + 2\pi k, \]or\[ \theta - \frac{\pi}{4} = \pi - \sin^{-1}\left(\frac{\sqrt{2}}{4}\right) + 2\pi k. \]
04
Solve for \( \theta \)
Rearrange the equations to solve for \( \theta \):\[ \theta = \sin^{-1}\left(\frac{\sqrt{2}}{4}\right) + \frac{\pi}{4} + 2\pi k, \]and\[ \theta = \pi - \sin^{-1}\left(\frac{\sqrt{2}}{4}\right) + \frac{\pi}{4} + 2\pi k. \]Evaluate \( \sin^{-1}\left(\frac{\sqrt{2}}{4}\right) \). Suppose \( \sin^{-1}\left(\frac{\sqrt{2}}{4}\right) = \alpha \), then substitute to determine \( \theta \) in the specified interval \([0, 2\pi)\).
05
Find specific solutions in the interval \([0,2\pi)\)
Compute the specific values for \( \theta \) in the interval \([0, 2\pi)\):- Calculate \( \sin^{-1}\left(\frac{\sqrt{2}}{4}\right) = \alpha \).- The solutions based on this value are: - \( \theta = \alpha + \frac{\pi}{4} \) and \( \theta = \alpha + \frac{\pi}{4} + 2\pi m \) where \( m \) is an integer. - \( \theta = \pi - \alpha + \frac{\pi}{4} \) and \( \theta = \pi - \alpha + \frac{\pi}{4} + 2\pi n \) where \( n \) is an integer.- Validate these solutions by ensuring they lie within \([0, 2\pi)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Double-angle formulas
Double-angle formulas are special trigonometric identities used to simplify expressions and solve equations. These formulas involve trigonometric functions of angles that are double a given angle. They are incredibly useful when dealing with equations involving sine, cosine, or tangent. For sine, the double-angle formula is:
- \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \)
- \( \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) \)
- \( \cos(2\theta) = 2\cos^2(\theta) - 1 \)
- \( \cos(2\theta) = 1 - 2\sin^2(\theta) \)
Half-angle formulas
Half-angle formulas are another set of trigonometric identities that are derived from double-angle formulas. They are useful when you need to solve equations or evaluate trigonometric functions of angles that are half of a known angle. These formulas include:
- \( \sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos(\theta)}{2}} \)
- \( \cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos(\theta)}{2}} \)
- \( \tan\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}} \)
Inverse trigonometric functions
Inverse trigonometric functions are functions that undo the effect of the original trigonometric functions. They allow you to determine an angle whose trigonometric ratio equals a given value. For sine, cosine, and tangent, these inverses are written as \( \sin^{-1}, \cos^{-1}, \) and \( \tan^{-1} \) respectively. For example, if \( \sin(\theta) = a \), then \( \theta = \sin^{-1}(a) \).
- \( \sin^{-1}(a) \) returns an angle \( \theta \) between \( [-\frac{\pi}{2}, \frac{\pi}{2}] \)
- \( \cos^{-1}(a) \) outputs an angle in \( [0, \pi] \)
- \( \tan^{-1}(a) \) results in an angle in \( (-\frac{\pi}{2}, \frac{\pi}{2}) \)