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Chapter 8: Problem 7

Determine all solutions of the given equations. Express your answers using radian measure. $$\sin \theta=-1 / 2$$

Short Answer

Expert verified
\( \theta = \frac{7\pi}{6} + 2k\pi \) or \( \theta = \frac{11\pi}{6} + 2k\pi \), \( k \in \mathbb{Z} \).

Step by step solution

01

Identify Reference Angle

First, we recognize that the equation involves the sine of an angle. We know that \( \sin \theta = -\frac{1}{2} \) occurs at specific angles. The reference angle for \( \sin \theta = \frac{1}{2} \) is \( \theta = \frac{\pi}{6} \).
02

Determine Quadrants

Since \( \sin \theta = -\frac{1}{2} \), and sine is negative in the third and fourth quadrants, we need to find angles in these quadrants where the sine is equal to \(-\frac{1}{2}\).
03

Find Solutions in Third Quadrant

In the third quadrant, the angle corresponding to the sine value of \(-\frac{1}{2}\) can be found by adding the reference angle to \(\pi\). Thus, \( \theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \).
04

Find Solutions in Fourth Quadrant

In the fourth quadrant, the angle related to \(-\frac{1}{2}\) is found by subtracting the reference angle from \(2\pi\). Therefore, \( \theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \).
05

Generalize the Solution

Because sine is periodic, with period \(2\pi\), any general solution can be expressed as \( \theta = \frac{7\pi}{6} + 2k\pi \) or \( \theta = \frac{11\pi}{6} + 2k\pi \), where \( k \) is any integer.

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Key Concepts

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

Sine Function
The sine function is one of the fundamental trigonometric functions used in mathematics. It relates an angle in a right triangle to the ratio of the length of the side opposite the angle to the hypotenuse. This relationship can be remembered using the acronym "SOH" from SOHCAHTOA, where S stands for sine, O for opposite, and H for hypotenuse. The sine function is periodic with a period of \(2\pi\), meaning it repeats its values every \(2\pi\) radians. This periodic nature is crucial for solving trigonometric equations, as it means that solutions can occur at regular intervals. The range of the sine function is from -1 to 1, which indicates the lowest and highest values the sine can reach. Understanding where these values occur based on the angle is important for solving equations like \( \sin \theta = -\frac{1}{2} \). Here, we're interested in finding all possible angles for which the sine value is \(-\frac{1}{2}\).
Reference Angle
The reference angle is the acute angle formed by the terminal side of an angle in standard position and the x-axis. These angles are always positive and fall between \(0\) and \(\frac{\pi}{2}\) radians (90 degrees). When solving trigonometric equations where the trigonometric value is negative or positive, the reference angle helps us determine possible angles in different quadrants. For instance, in the equation \( \sin \theta = -\frac{1}{2} \), we first determine the reference angle for \( \sin \theta = \frac{1}{2} \), which is \( \frac{\pi}{6} \) radians. By knowing this reference angle, we can then find angles in other quadrants where the sine value is negative. The reference angle helps identify the exact positions along the unit circle where the desired trigonometric function takes on the specified value. It's important for students to remember that the reference angle is independent of the actual sign of the sine function.
Angle Measurement in Radians
Angles can be measured in degrees or radians, with radians being the standard unit in mathematics. A radian measures how far around a circle's circumference the angle sweeps. Specifically, a full circle is \(2\pi\) radians, corresponding to 360 degrees. Therefore, \(\pi\) radians equal 180 degrees. Using radians simplifies the calculation in trigonometry, calculus, and advanced mathematics, as they naturally correspond to real numbers and are more conducive to algebraic manipulation. When solving trigonometric equations, expressing your answers in radians is typically the preferred approach. By understanding the radian system, you can easily convert between radians and degrees when necessary, and you can apply radian measures directly to find solutions that align with the circular nature of trigonometric functions.
Quadrantal Angles
Quadrantal angles are angles that lie along the axes of the Cartesian plane: 0, \(\frac{\pi}{2}\), \(\pi\), \(\frac{3\pi}{2}\), and \(2\pi\) radians. At these angles, trigonometric functions have their values completely defined due to their positioning on the unit circle. For example, \(\sin(\pi)\) and \(\sin(2\pi)\) both equal 0 because the sine function represents the y-coordinate of a point on the unit circle, and these angles lie on the x-axis where the y-coordinate is 0. When solving equations like \(\sin \theta = -\frac{1}{2}\), understanding the concept of quadrantal angles is important because solutions often require identifying which quadrants (or locations on the circle) yield the necessary values for the sine function. The angles that provide specific values for the sine function are non-quadrantal, but recognizing quadrantal angles helps frame where solutions not lying exactly on these axes can reside on the unit circle.

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Most popular questions from this chapter

In this exercise you will see how certain cubic equations can be solved by using the following identity (which we proved in Example 3 in Section 8.2 ): $$ 4 \cos ^{3} \theta-3 \cos \theta=\cos 3 \theta $$ For example, suppose that we wish to solve the equation $$ 8 x^{3}-6 x-1=0 $$ (2) To transform this equation into a form in which the stated identity is useful, we make the substitution \(x=a \cos \theta,\) where \(a\) is a constant to be determined. With this substitution, equation ( 2 ) can be written $$ 8 a^{3} \cos ^{3} \theta-6 a \cos \theta=1 $$ In equation (3) the coefficient of \(\cos ^{3} \theta\) is \(8 a^{3} .\) since we want this coefficient to be \(4 \text { [as it is in equation }(1)]\), we divide both sides of equation (3) by \(2 a^{3}\) to obtain $$ 4 \cos ^{3} \theta-\frac{3}{a^{2}} \cos \theta=\frac{1}{2 a^{3}} $$ Next, a comparison of equations (4) and (1) leads us to require that \(3 / a^{2}=3 .\) Thus \(a=\pm 1 .\) For convenience we choose \(a=1 ;\) equation (4) then becomes $$ 4 \cos ^{3} \theta-3 \cos \theta=\frac{1}{2} $$ Comparing equation (5) with the identity in (1) leads us to the equation $$ \cos 3 \theta=\frac{1}{2} $$ As you can check, the solutions here are of the form $$ \theta=20^{\circ}+120 k^{\circ} \quad \text { and } \quad \theta=100^{\circ}+120 k^{\circ} $$ Thus \(x=\cos \left(20^{\circ}+120 k^{\circ}\right) \quad\) and \(\quad x=\cos \left(100^{\circ}+120 k^{\circ}\right)\) Now, however, as you can again check, only three of the angles yield distinct values for \(\cos \theta,\) namely, \(\theta=20^{\circ}\) \(\theta=140^{\circ},\) and \(\theta=260^{\circ} .\) Thus the solutions of the equation \(8 x^{3}-6 x-1=0\) are given by \(x=\cos 20^{\circ}, x=\cos 140^{\circ}\) and \(x=\cos 260^{\circ} .\) Note: If you choose \(a=-1,\) your solutions will be equivalent to those we found with \(a=1\) Use the method just described to solve the following equations. (a) \(x^{3}-3 x+1=0\) Answers: \(2 \cos 40^{\circ},-2 \cos 20^{\circ}, 2 \cos 80^{\circ}\) (b) \(x^{3}-36 x-72=0\) (c) \(x^{3}-6 x+4=0 \quad\) Answers: \(2,-1 \pm \sqrt{3}\) (d) \(x^{3}-7 x-7=0\) (Round your answers to three decimal places.)

Solve the equations on the interval \([0,2 \pi]\) as follows. Graph the expression on each side of the equation and then zoom in on the intersection points until you are certain of the first three decimal places in each answer. For instance, for Exercise \(53,\) when you graph the two equations \(y=\cos x\)and \(y=0.623\) on the interval \([0,2 \pi],\) you 'll see that there are two intersection points. The \(x\) -coordinates of these points are roots of the equation \(\cos x=0.623\). $$\cos 2 x+1=\cos (2 x+1)$$

Use graphs to determine whether there are solutions for each equation in the interval \([0,1] .\) If there are solutions, use the graphing utility to find them accurately to two decimal places. (a) \(1 /\left(\sin ^{-1} x+\cos ^{-1} x\right)=4 x^{3}\) (b) \(1 /\left(\sin ^{-1} x+\cos ^{-1} x\right)=5 x^{3}\)

Consider the equation \(\sin x \cos x=1\) (a) Square both sides and then replace \(\cos ^{2} x\) by \(1-\sin ^{2} x\) Show that the resulting equation can be written \(\sin ^{4} x-\sin ^{2} x+1=0\). (b) Show that the equation \(\sin ^{4} x-\sin ^{2} x+1=0\) has no solutions. Conclude from this that the original equation has no real-number solutions.

If triangle \(A B C\) is not a right triangle, and \(\cos A=\cos B \cos C,\) show that \(\tan B \tan C=2\)
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