Textbook Question
Solving Trigonometric Equations
For Exercises 51–54, solve for the angle θ, where 0 ≤ θ ≤ 2π.
sin² θ = cos² θ
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Ch. 1 - Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 1, Problem 1.3.68
General Sine Curves
For
f(x) = A sin ((2π/B)(x – C) +D
identify A, B, C, and D for the sine functions in Exercises 67–70 and sketch their graphs.
y = ½ sin (πx – x) + ½
Verified step by step guidance1
Step 1: Start by comparing the given sine function y = ½ sin(πx - x) + ½ to the general sine function f(x) = A sin((2π/B)(x - C)) + D. Notice that the given function has a simplified argument inside the sine function, so simplify πx - x to (π - 1)x.
Step 2: Identify the amplitude A. The amplitude is the coefficient of the sine function, which in this case is ½. Therefore, A = ½.
Step 3: Identify the period B. The period is determined by the coefficient of x inside the sine function. Here, the coefficient of x is (π - 1). The period B is calculated as B = 2π / |π - 1|.
Step 4: Identify the phase shift C. The phase shift is determined by the term (x - C) inside the sine function. Since there is no explicit constant added or subtracted from x in the argument, C = 0 (no horizontal shift).
Step 5: Identify the vertical shift D. The vertical shift is the constant added outside the sine function, which in this case is ½. Therefore, D = ½. With these values (A = ½, B = 2π / |π - 1|, C = 0, D = ½), you can now sketch the graph of the sine function.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Amplitude
The amplitude of a sine function, represented by 'A' in the equation f(x) = A sin(...), determines the height of the wave from its midline to its peak. It indicates how far the graph stretches vertically. In the given function, the amplitude is ½, meaning the maximum value of the sine wave will be ½ above and ½ below the midline.
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Introduction to Cotangent Graph Example 1
Period
The period of a sine function, determined by 'B' in the equation, indicates the length of one complete cycle of the wave. It is calculated using the formula Period = 2π/B. In the provided function, the coefficient of x inside the sine function affects how quickly the wave oscillates; a larger value of B results in a shorter period.
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Phase Shift and Vertical Shift
The parameters 'C' and 'D' in the sine function represent the phase shift and vertical shift, respectively. The phase shift, given by C, moves the graph left or right, while the vertical shift, represented by D, moves the graph up or down. In the function y = ½ sin(πx - x) + ½, the phase shift can be derived from the expression inside the sine function, and the vertical shift is +½, indicating the entire graph is shifted up by ½.
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Related Practice
Textbook Question
Even and Odd Functions
In Exercises 47–62, say whether the function is even, odd, or neither. Give reasons for your answer.
sin x²
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Textbook Question
Graph the functions in Exercises 37–56.
y = (x + 1)²/³
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Textbook Question
In Exercises 9–16, determine whether the function is even, odd, or neither.
𝔂 = x cos x
240
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Textbook Question
Finding Formulas for Functions
Express the side length of a square as a function of the length d of the square’s diagonal. Then express the area as a function of the diagonal length.
534
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Textbook Question
Shifting Graphs
Exercises 27–36 tell how many units and in what directions the graphs of the given equations are to be shifted. Give an equation for the shifted graph. Then sketch the original and shifted graphs together, labeling each graph with its equation.
y = x³ Left 1, down 1
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