Problem 63 - A Graphical Approach to Precalculus with Limits

Question

Give solutions over the interval \([0,2 \pi)\) as approximations to the nearest hundredth when exact values cannot be determined. You may need to use the quadratic formula. Give approximate answers in Exercises \(59-64\) to the nearest tenth of a degree over the interval \(\left[0^{\circ}, 360^{\circ}\right)\) $$\cos ^{2} \theta+\sin \theta=0$$

Step-by-Step Solution

Answer

The solution is \( \theta \approx 5.62 \) radians.

Step 1: Recognize the Trigonometric Identity

We start by recognizing that the equation involves a trigonometric identity. The expression \( \cos^2 \theta + \sin \theta = 0 \) can also be written using the Pythagorean identity \( \cos^2 \theta = 1 - \sin^2 \theta \). Thus, we rewrite the equation as \( (1 - \sin^2 \theta) + \sin \theta = 0 \).

Step 2: Rearrange the Equation

Now, let's simplify the equation by rearranging it. We have \( 1 - \sin^2 \theta + \sin \theta = 0 \). Rearrange to obtain \( -\sin^2 \theta + \sin \theta + 1 = 0 \). This can be reordered to resemble a quadratic equation: \( \sin^2 \theta - \sin \theta - 1 = 0 \).

Step 3: Solve Using the Quadratic Formula

We use the quadratic formula to solve for \( \sin \theta \):\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \( a = 1 \), \( b = -1 \), and \( c = -1 \). Substitute these values:\[x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}\]Simplify this:\[x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}\].

Step 4: Calculate Numerical Solutions

Compute the approximate values for \( \sin \theta \):1. \( \sin \theta = \frac{1 + \sqrt{5}}{2} \approx 1.618 \) is not valid since sine cannot be greater than 1.2. \( \sin \theta = \frac{1 - \sqrt{5}}{2} \approx -0.618 \) which is valid.

Step 5: Find Angles Using \(\sin^{-1}\) Function

Find \( \theta \) such that \( \sin \theta = -0.618 \). Since \( \sin \theta \) is negative, the angle \( \theta \) is in the third or fourth quadrant:\(\theta = \sin^{-1}(-0.618) \approx -38.19^\circ\) (not in our interval).Add 360°: \(\theta \approx 321.81^\circ\), which lies in the specified interval. Convert it into radians: \(\theta \approx 5.62\) radians.

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool for solving equations of the form \(ax^2 + bx + c = 0\). This formula provides a general method to find solutions, or roots, for any quadratic equation, and is expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation. Using the quadratic formula involves several steps:

• Identify the coefficients \(a\), \(b\), and \(c\).
• Substitute these into the formula.
• Perform the calculations under the square root (discriminant).
• Calculate the roots using the \(\pm\) sign, which accounts for the two possible solutions.

In this exercise, the transformation of the trigonometric equation into a quadratic form allowed the use of the quadratic formula, with values \(a = 1\), \(b = -1\), and \(c = -1\). Solving this resulted in two potential solutions, where only one was valid due to the sine range constraint.

Pythagorean Identity

The Pythagorean identity is a fundamental relationship in trigonometry that connects the square of the sine and cosine functions. It is expressed as:\[\sin^2 \theta + \cos^2 \theta = 1\]This identity is incredibly useful when solving trigonometric equations because it allows you to substitute for one trigonometric function in terms of the other. In this particular exercise:

• The expression \(\cos^2 \theta\) was substituted using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\).
• This substitution simplified the trigonometric equation into a quadratic form suitable for applying the quadratic formula.

Recognizing how to apply the Pythagorean identity is key in converting trigonometric expressions and making them more manageable for solving.

Sine Function

The sine function is one of the primary trigonometric functions, commonly defined for angles in various quadrants of the unit circle. When solving trigonometric equations, knowing the properties of the sine function is crucial, especially the range and periodicity:

• The sine function oscillates between -1 and 1. Thus, any solution for \(\sin \theta\) must fall within this range.
• It repeats every \(360^\circ\) or \(2\pi\) radians, known as its periodicity.

In this task, solving \(\sin \theta = -0.618\) involved determining which quadrants this angle could be in. Since the sine is negative, the solutions lie in the third and fourth quadrants. Additionally, we make use of the inverse sine function \(\sin^{-1}\) to find precise angle measurements within the allowed range.

Radian Measure

Radian measure is an angle measurment system used in trigonometry that relates the angle to the radius of a circle. One radian is the angle formed when the arc length is equal to the radius of the circle. Here are some key points about radians:

• There are \(2\pi\) radians in a full circle, corresponding to \(360^\circ\).
• To convert degrees to radians, you multiply by \(\frac{\pi}{180}\).
• Radian measure is often more natural to use in mathematical formulas and calculations because it simplifies many expressions.

In the solution, converting from degrees into radians was the final step to ensure the answer fit within the specified interval \([0, 2\pi)\). This important conversion allowed us to not only deduce angles within the familiar system of degrees but to express them in radians, which is crucial for many applications and problem-solving situations in trigonometry.