Problem 112
Question
In Exercises 109 - 112, use a graphing utility to solve the equation for \( \theta \), where \( 0 \le \theta < 2\pi \). \( \csc \theta = \sqrt{1 + \cot^2 \theta} \)
Step-by-Step Solution
Verified Answer
The solutions to the equation are \( \theta = \pi/2, 3\pi/2 \)
1Step 1: Convert to sine and cosine
First, convert the equation from cotangent and cosecant to cosine and sine, so it becomes easy for plotting. For this we can use the identities: \( \csc \theta = 1/\sin \theta \) and \( \cot \theta = \cos \theta/\sin \theta \). The equation becomes \(1/\sin \theta = \sqrt{1 + (\cos \theta/\sin \theta)^2}\)
2Step 2: Simplify the equation
Next, simplify the equation. This can be done by squaring both sides of the equation (to eliminate the square root) and then multiplying each term by \(\sin^2 \theta\) to get rid of the fractions. This will give the equation in a more simple form of \( \sin^2 \theta = 1 + \cos^2 \theta\) which rearranges to \( \sin^2 \theta - \cos^2 \theta = 1 \).
3Step 3: Express in the form of known trigonometric identity
Considering the identity \( \sin^2 \theta = 1 - \cos^2 \theta \), substitute this in above equation. This gives \( 1 - \cos^2 \theta - \cos^2 \theta = 1 \), which simplifies to \( 2\cos^2 \theta = 0 \) and finally to \( \cos \theta = 0 \).
4Step 4: Solve for theta
Now solve the equation \( \cos \theta = 0 \) for values of \(\theta\) within the prescribed range. You will have \( \theta = \pi/2, \ 3\pi/2 \).
Key Concepts
Trigonometric IdentitiesSolving Equations Using GraphsSine and Cosine FunctionsCosecant and Cotangent Identities
Trigonometric Identities
Trigonometric identities are essential tools in simplifying equations involving trigonometric functions. They allow us to transform complex expressions into simpler forms that are easier to work with. For example, in this exercise, we utilize two important identities:
- The identity for cosecant: \( \csc \theta = \frac{1}{\sin \theta} \)
- The identity for cotangent: \( \cot \theta = \frac{\cos \theta}{\sin \theta} \)
Solving Equations Using Graphs
Solving trigonometric equations using graphs is a visual method that can illuminate the solutions clearly. When graphed, trigonometric functions reveal their periodic nature and the points where they intersect. In this exercise, where we solve \( \cos \theta = 0 \) within the interval \( 0 \leq \theta < 2\pi \), graphing provides a handy approach.When plotting \( y = \cos \theta \), the function impacts the horizontal axis at \( \theta = \pi/2 \) and \( \theta = 3\pi/2 \). These intercepts represent the solutions to our equation, showcasing how graphical methods verify analytical findings. This graphical approach complements algebraic methods, offering a double-checking mechanism.By utilizing a graphing tool or calculator, one can easily view these relationships and determine exact or approximate solutions, reinforcing comprehension of trigonometric equation behaviors.
Sine and Cosine Functions
Sine and cosine are fundamental trigonometric functions that depict periodic oscillations. Each function represents specific relationships in a right-angled triangle:
- **Sine**: Represents the ratio of the length of the opposite side to the hypotenuse. It's given by \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \).
- **Cosine**: Represents the ratio of the length of the adjacent side to the hypotenuse. It's calculated as \( \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \).
Cosecant and Cotangent Identities
Cosecant and cotangent are reciprocal trigonometric functions that extend the use of sine and cosine.
- **Cosecant**: Reciprocal of sine, defined as \( \csc \theta = \frac{1}{\sin \theta} \).
- **Cotangent**: Reciprocal of tangent, defined as \( \cot \theta = \frac{\cos \theta}{\sin \theta} \).
Other exercises in this chapter
Problem 111
In Exercises 109 - 112, use a graphing utility to solve the equation for \( \theta \), where \( 0 \le \theta
View solution Problem 112
In Exercises 111 - 124, verify the identity. \( \sec 2 \theta = \dfrac{\sec^2 \theta}{2 - \sec^2 \theta} \)
View solution Problem 113
In Exercises 111 - 124, verify the identity. \( \sin \dfrac{\alpha}{3} \cos \dfrac{\alpha}{3} = \dfrac{1}{2} \sin \dfrac{2\alpha}{3} \)
View solution Problem 113
In Exercises 113 - 118, rewrite the expression as a single logarithm and simplify the result. \( \ln\mid\cos x\mid - \ln\mid\sin x\mid \)
View solution