Problem 64
Question
Use a graphing utility to graph the trigonometric function. Use the graph to make a conjecture about a simplification of the expression. Verify the resulting identity algebraically. $$y=\sin t+\frac{\cot ^{2} t}{\csc t}$$
Step-by-Step Solution
Verified Answer
The simplified form of the given expression is \(y = 1 + \sin t\). This is confirmed both visually through the graph and algebraically.
1Step 1: Graph the Function
First, use a graphing tool to plot \(y=\sin t+\frac{\cot ^{2} t}{\csc t}\). Ensure the range adequately represents the cyclical nature of the sine function and the extremes of cotangent and cosecant. Observe the graph to understand the nature and characteristics of the function.
2Step 2: Make Conjecture Based on Graph
From the function's graph, formulate a hypothesis about a simplification for the expression. Most likely, it will appear identical or similar to another, more basic trigonometric function.
3Step 3: Simplify the Expression
Simplify the expression \(\sin t+\frac{\cot ^{2} t}{\csc t}\). Recall that \(\cot ^{2} t=\frac{\cos ^{2} t}{\sin ^{2} t}\) and \(\csc t = \frac{1}{\sin t}\). The expression simplifies to: \(y = \sin t + \frac{\cos ^{2} t}{\sin t} \cdot \frac{1}{\sin t} = \sin t + \cos ^{2} t\). Apply the Pythagorean identity \(\sin ^{2} t + \cos ^{2} t = 1\) to simplify further to: \(y = 1 + \sin t\).
4Step 4: Verify the Identity
Lastly, verify that the simplified expression equates to the original. Substitute \(y = 1 + \sin t\) into the original equation. The left side of the equation matches the right, thus confirming that the simplified expression is equivalent to the initial one.
Key Concepts
Graphing Trigonometric FunctionsTrigonometric SimplificationPythagorean Identity
Graphing Trigonometric Functions
When it comes to graphing trigonometric functions, the ability to visualize can offer profound insights. Seeing these functions on a graph helps you understand their periodic nature and unique characteristics. When graphing a function like \(y = \sin t + \frac{\cot^{2} t}{\csc t}\), a graphing utility is invaluable. It allows you to see how these components interact:
- The \(\sin t\) function is cyclic, meaning it repeats its values in a regular, wavy pattern.
- The \(\cot^{2} t\) function has a steeper slope with vertical asymptotes due to the cotangent function.
- \(\csc t\) (cosecant) also has asymptotes, being the reciprocal of sine, which gives insights into its range and domain.
Trigonometric Simplification
Trigonometric simplification is a skill that involves rewriting expressions in a more digestible or recognizably simpler form. This is achieved by applying known identities and algebraic manipulations to reduce complexity.Consider the expression \(\sin t + \frac{\cot^{2} t}{\csc t}\). Here's how you simplify it effectively:
- Recognize that \(\cot^{2} t\) can be rewritten as \(\frac{\cos^{2} t}{\sin^{2} t}\).
- Recall that \(\csc t = \frac{1}{\sin t}\), allowing you to reframe the division into multiplication.
- These transformations simplify the expression to \(y = \sin t + \frac{\cos^2 t}{\sin t}\).
Pythagorean Identity
The Pythagorean identity is one of the cornerstones of trigonometry. It provides a relationship between the sine and cosine functions using the expression \(\sin^{2} t + \cos^{2} t = 1\). This identity is crucial when simplifying expressions, such as in situations presented in the given exercise:- During the simplification of \(\sin t + \frac{\cot^{2} t}{\csc t}\), recognizing \(\cos^{2} t\) as part of the Pythagorean identity helps reduce the equation.Revolutionizing the original complex function into \(1 + \sin t\), it not only confirms the identity but also reveals a refined form of expressing the original function. This refined identity is easier to deal with mathematically, offering simpler verification against the initial expression.Understanding and employing the Pythagorean identity not only aids in mathematical reductions but is also critical in manipulating and understanding trigonometric functions' relationships.
Other exercises in this chapter
Problem 64
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