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Jaycee says that the trigonometric equation \(\cos \theta=\frac{1}{2}\) has an infinite number of solutions. Do you agree? Explain.

Short Answer

Expert verified
Yes, the equation \( \cos \theta = \frac{1}{2} \) has an infinite number of solutions because it occurs periodically.

Step by step solution

01

Understand the Trigonometric Equation

The given equation is \(\cos \theta = \frac{1}{2}\). This means we need to find angles \(\theta\) where the cosine of \(\theta\) equals \frac{1}{2}.
02

Identify Principal Solutions

Cosine is positive in the first and fourth quadrants. The principal solutions for \(\theta\) with \(\cos \theta = \frac{1}{2}\) are \(\theta = \frac{\pi}{3}\) and \(\theta = -\frac{\pi}{3}\).
03

Generalize for All Solutions

The cosine function is periodic with period \(2\pi\). Therefore, the general solutions can be written as \(\theta = 2n\pi \pm \frac{\pi}{3}\), where \(n\) is any integer.
04

Conclude on Infinite Solutions

Since \(n\) can be any integer (positive, negative, or zero), there are infinitely many solutions to the equation \(\cos \theta = \frac{1}{2}\).

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Key Concepts

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

Cosine Function
The cosine function, often denoted as \( \cos \theta \), is a fundamental trigonometric function that relates angles to side lengths in a right-angled triangle. It's particularly useful in various fields like physics, engineering, and computer graphics.
The function gives the ratio of the length of the adjacent side to the hypotenuse of a right-angled triangle. In mathematical terms, if \( \text{adj} \) is the length of the adjacent side and \( \text{hyp} \) is the hypotenuse, then: \[ \cos \theta = \frac{\text{adj}}{\text{hyp}} \]
The cosine function is also one of the primary functions in the unit circle, which simplifies the study of trigonometry. The value of \( \cos \theta \) ranges from -1 to 1 for angles between 0 and 360 degrees (or 0 and 2\pi radians). This range is crucial for solving trigonometric equations like \( \cos \theta = \frac{1}{2} \).
Periodic Functions
Periodic functions are functions that repeat their values at regular intervals. The cosine function is a classic example of a periodic function.
Its periodicity comes from the unit circle, where angles are measured in radians, and each complete cycle around the circle brings the function back to the same value.
For the cosine function, the period is \(2\pi \), meaning: \[ \cos( \theta + 2\pi) = \cos \theta \]
Understanding the periodic nature of cosine is essential when solving trigonometric equations. It tells us that once we find a solution within one period, we can find infinitely many solutions by simply adding or subtracting multiples of \(2\pi \). This concept is what leads to the 'general solutions' for equations like \( \cos \theta = \frac{1}{2} \).
Principal Solutions
Principal solutions are the primary solutions to a trigonometric equation within one period, usually between 0 and \(2\pi \) or -\pi and \pi. These solutions give us a starting point to find all other solutions.
For the equation \( \cos \theta = \frac{1}{2} \), we first determine the angles within the range of 0 to 2\pi where the cosine equals \( \frac{1}{2} \). These angles are \( \frac{\pi}{3} \) (in the first quadrant) and \( -\frac{\pi}{3} \) (equivalently, \frac{5\pi}{3} in the fourth quadrant).
These principal solutions are essential because they allow us to generalize the solutions to cover the entire set of possible angles by using the periodic property of cosine.
General Solutions
General solutions expand on the principal solutions to account for the infinite nature of trigonometric functions. Since cosine is periodic, we can find an infinite number of solutions by adding or subtracting multiples of its period (2\pi).
For the equation \( \cos \theta = \frac{1}{2} \), the principal solutions give us \( \theta = \frac{\pi}{3} \) and \( -\frac{\pi}{3}. \) To find all possible solutions, we use these principal solutions and add multiples of 2\pi, the period of the cosine function.
Therefore, the general solutions are: \[ \theta = 2n\pi \pm \frac{\pi}{3}, \] where n is any integer (positive, negative, or zero).
This notation indicates that by changing the value of n, we can generate infinitely many solutions, in agreement with the periodic nature of the cosine function.

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Most popular questions from this chapter

Sketch each angle in standard position. In which quadrant does each angle terminate? a) 1 b) \(-225^{\circ}\) c) \(\frac{17 \pi}{6}\) d) \(650^{\circ}\) e) \(-\frac{2 \pi}{3}\) f) \(-42^{\circ}\)

Solve each equation for \(0 \leq \theta < 2 \pi\) Give solutions to the nearest hundredth of a radian. a) \(\tan \theta=4.36\) b) \(\cos \theta=-0.19\) c) \(\sin \theta=0.91\) d) cot \(\theta=12.3\) e) \(\sec \theta=2.77\) f) \(\csc \theta=-1.57\)

Identify a measure for the central angle \(\theta\) in the interval \(0 \leq \theta<2 \pi\) such that \(P(\theta)\) is the given point. a) (0,-1) b) (1,0) c) \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\) d) \(\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\) e) \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\) f) \(\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)\) g) \(\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\) h) \(\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\) i) \(\left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)\) j) (-1,0)

Express the given quantity using the same trigonometric ratio and its reference angle. For example, \(\cos 110^{\circ}=-\cos 70^{\circ}\) For angle measures in radians, give exact answers. For example, \(\cos 3=-\cos (\pi-3)\) a) \(\sin 250^{\circ}\) b) tan \(290^{\circ}\) c) sec \(135^{\circ}\) d) \(\cos 4\) e) csc 3 f) cot 4.95

You can locate a point in a plane using Cartesian coordinates \((x, y),\) where \(|x|\) is the distance from the \(y\) -axis and \(|y|\) is the distance from the \(x\) -axis. You can also locate a point in a plane using \((r, \theta)\) where \(r, r \geq 0,\) is the distance from the origin and \(\theta\) is the angle of rotation from the positive \(x\) -axis. These are known as polar coordinates. Determine the polar coordinates for each point. a) \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\) b) \(\left(-\frac{\sqrt{3}}{2},-\frac{1}{3}\right)\) c) (2,2) d) (4,-3)

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