Ch. 2 - Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

Blitzer3rd EditionTrigonometryISBN: 9780137316601Not the one you use?Change textbook

Blitzer 3rd Edition

Ch. 2 - Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

Problem 49

Chapter 2, Problem 49

In Exercises 43–52, determine the amplitude, period, and phase shift of each function. Then graph one period of the function. y = −3 cos (2x − π/2)

Verified step by step guidance

Identify the general form of the cosine function: \(y = A \cos(Bx - C)\), where \(A\) is the amplitude, \(B\) affects the period, and \(C\) relates to the phase shift.

Determine the amplitude by taking the absolute value of the coefficient in front of the cosine: \(\text{Amplitude} = |A| = |-3|\).

Calculate the period using the formula \(\text{Period} = \frac{2\pi}{|B|}\), where \(B\) is the coefficient of \(x\) inside the cosine function. Here, \(B = 2\).

Find the phase shift by solving \(Bx - C = 0\) for \(x\), which gives \(x = \frac{C}{B}\). In this case, \(C = \frac{\pi}{2}\), so the phase shift is \(\frac{\pi/2}{2}\).

Use the amplitude, period, and phase shift to sketch one full cycle of the function, starting at the phase shift and extending one period length along the \(x\)-axis.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Amplitude of a Trigonometric Function

Amplitude is the maximum absolute value of the function's output, representing the height from the midline to the peak. For functions like y = a cos(bx + c), the amplitude is |a|. In this question, the amplitude is the absolute value of -3, which is 3.

Period of a Trigonometric Function

The period is the length of one complete cycle of the function. For y = cos(bx + c), the period is calculated as 2π divided by |b|. Here, with b = 2, the period is 2π/2 = π, meaning the function repeats every π units.

Phase Shift of a Trigonometric Function

Phase shift is the horizontal translation of the function, determined by solving bx + c = 0 for x. It is given by -c/b. In this function, with c = -π/2 and b = 2, the phase shift is (π/2)/2 = π/4 to the right.

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In Exercises 45–52, graph two periods of each function. y = csc|x|

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