Graphs of Circular Functions PDF

Summary

This document provides detailed explanations and solutions for various trigonometric functions, including sine, cosine, secant, cosecant, tangent, and cotangent. It covers topics such as domain, range, amplitude, and period for these functions.

Full Transcript

GRAPHS OF CIRCULAR FUNCTIONS Graph: 𝑦 = sin π‘₯ Graph: 𝑦 = cos π‘₯ The domain of both 𝑦 = asin 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 and 𝑦 = π‘Ž cos 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 is the set of real numbers or (βˆ’βˆž, ∞). The range of both 𝑦 = asin 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 and 𝑦 = π‘Ž cos 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 is 𝑑 βˆ’ π‘Ž , 𝑑 + π‘Ž. The number π‘Ž is the amplitude of...

GRAPHS OF CIRCULAR FUNCTIONS Graph: 𝑦 = sin π‘₯ Graph: 𝑦 = cos π‘₯ The domain of both 𝑦 = asin 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 and 𝑦 = π‘Ž cos 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 is the set of real numbers or (βˆ’βˆž, ∞). The range of both 𝑦 = asin 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 and 𝑦 = π‘Ž cos 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 is 𝑑 βˆ’ π‘Ž , 𝑑 + π‘Ž. The number π‘Ž is the amplitude of the graphs of 𝑦 = π‘Ž sin 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 and 𝑦 = π‘Ž cos(𝑏(π‘₯ βˆ’ 𝑐)) + 𝑑. The period of 𝑦 = π‘Ž sin 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 and 2πœ‹ 𝑦 = π‘Ž cos 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 is. 𝑏 Find the domain, range, amplitude, and period of 𝑦 = βˆ’ 2 sin(3π‘₯ + πœ‹). Solution: Note that 𝑦 = βˆ’2 sin 3π‘₯ + πœ‹ can be written as 𝑦 = πœ‹ πœ‹ βˆ’ 2 sin 3 π‘₯ +. This means that π‘Ž = βˆ’2, 𝑏 = 3, 𝑐 = βˆ’ , and 3 3 𝑑 = 0. Domain: (βˆ’βˆž, ∞) Range: 𝑑 βˆ’ π‘Ž , 𝑑 + π‘Ž = 0 βˆ’ βˆ’2 , 0 + βˆ’2 = βˆ’2,2 Amplitude: π‘Ž = βˆ’2 = 2 2πœ‹ 2πœ‹ 2πœ‹ Period: = = 𝑏 3 3 Find the domain, range, amplitude, and period of πœ‹ 𝑦 = βˆ’2 cos βˆ’2π‘₯ + βˆ’1 4 Solution: πœ‹ Note that 𝑦 = βˆ’2 cos βˆ’2π‘₯ + βˆ’ 1 can be written as πœ‹ 4 𝑦 = βˆ’2 cos βˆ’2 π‘₯ βˆ’ βˆ’ 1. This means that π‘Ž = βˆ’2, 𝑏 = βˆ’2, πœ‹ 8 𝑐 = , and 𝑑 = βˆ’1. 8 Domain: (βˆ’βˆž, ∞) Range: 𝑑 βˆ’ π‘Ž , 𝑑 + π‘Ž = βˆ’1 βˆ’ βˆ’2 , βˆ’1 + βˆ’2 = βˆ’3,1 Amplitude: π‘Ž = βˆ’2 = 2 2πœ‹ 2πœ‹ 2πœ‹ Period: = = =πœ‹ 𝑏 βˆ’2 2 Graph: 𝑦 = sec π‘₯ Graph: 𝑦 = csc π‘₯ The domain of 𝑦 = π‘Ž csc 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 is the set of π‘₯ ∈ ℝ: sin 𝑏 π‘₯ βˆ’ 𝑐 β‰  0 = π‘˜πœ‹ π‘₯ ∈ ℝ: π‘₯ β‰  𝑐 + , π‘˜ ∈ β„€. 𝑏 The domain of 𝑦 = π‘Ž sec 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 is the set of π‘₯ ∈ ℝ: cos 𝑏 π‘₯ βˆ’ 𝑐 β‰  0 = π‘˜πœ‹ π‘₯ ∈ ℝ: π‘₯ β‰  𝑐 + , π‘˜ is an odd integer. 2𝑏 The range of both 𝑦 = π‘Ž csc 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 and 𝑦 = π‘Ž sec 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 is (βˆ’βˆž, 𝑑 βˆ’ π‘Ž ሿ βˆͺ 𝑑 + π‘Ž ,∞. The period of both 𝑦 = π‘Ž csc(𝑏(π‘₯ βˆ’ 2πœ‹ 𝑐)) + 𝑑 and 𝑦 = π‘Ž sec 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 is. 𝑏 The asymptotes of 𝑦 = π‘Ž csc 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 π‘˜πœ‹ are π‘₯ = 𝑐 + , π‘˜ ∈ β„€. 𝑏 The asymptotes of 𝑦 = π‘Ž sec 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 π‘˜πœ‹ are π‘₯ = 𝑐 + , π‘˜ is an odd integer. 2𝑏 Find the domain, range, amplitude, and period of 𝑦 = βˆ’2 sec(3π‘₯ + πœ‹) Solution: Note that 𝑦 = βˆ’2 sec(3π‘₯ + πœ‹) can be written as πœ‹ πœ‹ 𝑦 = βˆ’2 sec 3 π‘₯ +. This means that π‘Ž = βˆ’2, 𝑏 = 3, 𝑐 = βˆ’ , 3 3 and 𝑑 = 0. π‘˜πœ‹ Domain: π‘₯ ∈ ℝ: π‘₯ β‰  𝑐 + 2𝑏 , π‘˜ is an odd integer = πœ‹ π‘˜πœ‹ π‘₯ ∈ ℝ: π‘₯ β‰  βˆ’ + ,π‘˜ is an odd integer 3 6 Range: βˆ’βˆž, 𝑑 βˆ’ π‘Ž βˆͺ 𝑑 + π‘Ž , ∞ = βˆ’βˆž, 0 βˆ’ βˆ’2 βˆͺ ሾ0 + βˆ’2 , ∞) = βˆ’βˆž, βˆ’2 βˆͺ 2, ∞ Solution: 2πœ‹ 2πœ‹ 2πœ‹ Period: = = 𝑏 3 3 π‘˜πœ‹ Asymptotes: π‘₯ = 𝑐 + 2𝑏 , π‘˜ is an odd integer; thus, the asymptotes πœ‹ π‘˜πœ‹ are π‘₯ = βˆ’ + ,π‘˜ is an odd integer 3 6 Find the domain, range, amplitude, and period πœ‹ of 𝑦 = βˆ’2 csc(βˆ’2π‘₯ βˆ’ ) βˆ’ 1 4 Solution: πœ‹ Note that 𝑦 = βˆ’2 csc(βˆ’2π‘₯ βˆ’ )βˆ’ 1 4 πœ‹ can be written as 𝑦 = βˆ’2 csc βˆ’2 π‘₯ + βˆ’ 1. This means that 8 πœ‹ π‘Ž = βˆ’2, 𝑏 = βˆ’2, 𝑐 = βˆ’ , and 𝑑 = βˆ’1. 8 π‘˜πœ‹ πœ‹ π‘˜πœ‹ Domain: π‘₯ ∈ ℝ: π‘₯ β‰  𝑐 + 𝑏 ,π‘˜ ∈ β„€ = π‘₯ ∈ ℝ: π‘₯ β‰  βˆ’ 8 + 2 ,π‘˜ βˆˆβ„€ Range: βˆ’βˆž, 𝑑 βˆ’ π‘Ž βˆͺ 𝑑 + π‘Ž , ∞ = βˆ’βˆž, βˆ’1 βˆ’ βˆ’2 βˆͺ αˆΎβˆ’1 + βˆ’2 , ∞) = βˆ’βˆž, βˆ’3 βˆͺ 1, ∞ Solution: 2πœ‹ 2πœ‹ Period: = =πœ‹ 𝑏 βˆ’2 π‘˜πœ‹ Asymptotes: π‘₯ = 𝑐 + ,π‘˜ 𝑏 ∈ β„€; thus, the asymptotes are πœ‹ π‘˜πœ‹ π‘₯= βˆ’ + ,π‘˜ βˆˆβ„€ 8 2 Graph: 𝑦 = tan π‘₯ Graph: 𝑦 = cot π‘₯ The domain of 𝑦 = π‘Ž cot 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 is the set of π‘₯ ∈ ℝ: sin 𝑏 π‘₯ βˆ’ 𝑐 β‰  0 = π‘˜πœ‹ π‘₯ ∈ ℝ: π‘₯ β‰  𝑐 + , π‘˜ ∈ β„€. 𝑏 The domain of 𝑦 = π‘Ž tan 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 is the set of π‘₯ ∈ ℝ: cos 𝑏 π‘₯ βˆ’ 𝑐 β‰  0 = π‘˜πœ‹ π‘₯ ∈ ℝ: π‘₯ β‰  𝑐 + , π‘˜ is an odd integer. 2𝑏 The range of both 𝑦 = π‘Ž cot 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 and 𝑦 = π‘Ž tan 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 is (βˆ’βˆž, ∞). The period of both 𝑦 = π‘Ž cot(𝑏(π‘₯ βˆ’ πœ‹ 𝑐)) + 𝑑 and 𝑦 = π‘Ž tan 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 is. 𝑏 The asymptotes of 𝑦 = π‘Ž cot 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 π‘˜πœ‹ are π‘₯ = 𝑐 + , π‘˜ ∈ β„€. 𝑏 The asymptotes of 𝑦 = π‘Ž tan 𝑏(π‘₯ βˆ’ 𝑐) + 𝑑 π‘˜πœ‹ are π‘₯ = 𝑐 + , π‘˜ is an odd integer. 2𝑏 1 Graph: 𝑦 = tan 2π‘₯ 2 Solution: 1 1 In 𝑦 = tan 2π‘₯ , π‘Ž = , 𝑏 = 2, 𝑐 = 0, and 𝑑 = 0. 2 2 π‘˜πœ‹ Domain: π‘₯ ∈ ℝ: π‘₯ β‰  𝑐 + , π‘˜ is an odd integer = 2𝑏 π‘˜πœ‹ π‘₯ ∈ ℝ: π‘₯ β‰  , π‘˜ is an odd integer 4 Range: βˆ’βˆž, ∞ Solution: πœ‹ πœ‹ πœ‹ Period: = = 𝑏 2 2 π‘˜πœ‹ Asymptote: π‘₯ = 𝑐 + , π‘˜ is an odd integer; thus, the 2𝑏 π‘˜πœ‹ asymptotes are π‘₯ = , π‘˜ is an odd integer 4 π‘₯ Graph: 𝑦 = 2 cot 3 Solution: π‘₯ 1 In 𝑦 = 2 cot , π‘Ž = 2, 𝑏 = , 𝑐 = 0, and 𝑑 = 0. 3 3 π‘˜πœ‹ Domain: π‘₯ ∈ ℝ: π‘₯ β‰  𝑐 + 𝑏 ,π‘˜βˆˆβ„€ = π‘₯ ∈ ℝ: π‘₯ β‰  3π‘˜πœ‹, π‘˜ ∈ β„€ Range: βˆ’βˆž, ∞ Solution: πœ‹ πœ‹ Period: = 1 = 3πœ‹ 𝑏 3 π‘˜πœ‹ Asymptote: π‘₯ = 𝑐 + ,π‘˜βˆˆ β„€; thus, the asymptotes are 𝑏 π‘₯ = 3π‘˜πœ‹, π‘˜ ∈ β„€

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