Graphs of Circular Functions PDF
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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.
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GRAPHS OF CIRCULAR FUNCTIONS Graph: π¦ = sin π₯ Graph: π¦ = cos π₯ The domain of both π¦ = asin π(π₯ β π) + π and π¦ = π cos π(π₯ β π) + π is the set of real numbers or (ββ, β). The range of both π¦ = asin π(π₯ β π) + π and π¦ = π cos π(π₯ β π) + π is π β π , π + π. The number π is the amplitude of...
GRAPHS OF CIRCULAR FUNCTIONS Graph: π¦ = sin π₯ Graph: π¦ = cos π₯ The domain of both π¦ = asin π(π₯ β π) + π and π¦ = π cos π(π₯ β π) + π is the set of real numbers or (ββ, β). The range of both π¦ = asin π(π₯ β π) + π and π¦ = π cos π(π₯ β π) + π is π β π , π + π. The number π is the amplitude of the graphs of π¦ = π sin π(π₯ β π) + π and π¦ = π cos(π(π₯ β π)) + π. The period of π¦ = π sin π(π₯ β π) + π and 2π π¦ = π cos π(π₯ β π) + π is. π Find the domain, range, amplitude, and period of π¦ = β 2 sin(3π₯ + π). Solution: Note that π¦ = β2 sin 3π₯ + π can be written as π¦ = π π β 2 sin 3 π₯ +. This means that π = β2, π = 3, π = β , and 3 3 π = 0. Domain: (ββ, β) Range: π β π , π + π = 0 β β2 , 0 + β2 = β2,2 Amplitude: π = β2 = 2 2π 2π 2π Period: = = π 3 3 Find the domain, range, amplitude, and period of π π¦ = β2 cos β2π₯ + β1 4 Solution: π Note that π¦ = β2 cos β2π₯ + β 1 can be written as π 4 π¦ = β2 cos β2 π₯ β β 1. This means that π = β2, π = β2, π 8 π = , and π = β1. 8 Domain: (ββ, β) Range: π β π , π + π = β1 β β2 , β1 + β2 = β3,1 Amplitude: π = β2 = 2 2π 2π 2π Period: = = =π π β2 2 Graph: π¦ = sec π₯ Graph: π¦ = csc π₯ The domain of π¦ = π csc π(π₯ β π) + π is the set of π₯ β β: sin π π₯ β π β 0 = ππ π₯ β β: π₯ β π + , π β β€. π The domain of π¦ = π sec π(π₯ β π) + π is the set of π₯ β β: cos π π₯ β π β 0 = ππ π₯ β β: π₯ β π + , π is an odd integer. 2π The range of both π¦ = π csc π(π₯ β π) + π and π¦ = π sec π(π₯ β π) + π is (ββ, π β π αΏ βͺ π + π ,β. The period of both π¦ = π csc(π(π₯ β 2π π)) + π and π¦ = π sec π(π₯ β π) + π is. π The asymptotes of π¦ = π csc π(π₯ β π) + π ππ are π₯ = π + , π β β€. π The asymptotes of π¦ = π sec π(π₯ β π) + π ππ are π₯ = π + , π is an odd integer. 2π Find the domain, range, amplitude, and period of π¦ = β2 sec(3π₯ + π) Solution: Note that π¦ = β2 sec(3π₯ + π) can be written as π π π¦ = β2 sec 3 π₯ +. This means that π = β2, π = 3, π = β , 3 3 and π = 0. ππ Domain: π₯ β β: π₯ β π + 2π , π is an odd integer = π ππ π₯ β β: π₯ β β + ,π is an odd integer 3 6 Range: ββ, π β π βͺ π + π , β = ββ, 0 β β2 βͺ αΎ0 + β2 , β) = ββ, β2 βͺ 2, β Solution: 2π 2π 2π Period: = = π 3 3 ππ Asymptotes: π₯ = π + 2π , π is an odd integer; thus, the asymptotes π ππ are π₯ = β + ,π is an odd integer 3 6 Find the domain, range, amplitude, and period π of π¦ = β2 csc(β2π₯ β ) β 1 4 Solution: π Note that π¦ = β2 csc(β2π₯ β )β 1 4 π can be written as π¦ = β2 csc β2 π₯ + β 1. This means that 8 π π = β2, π = β2, π = β , and π = β1. 8 ππ π ππ Domain: π₯ β β: π₯ β π + π ,π β β€ = π₯ β β: π₯ β β 8 + 2 ,π ββ€ Range: ββ, π β π βͺ π + π , β = ββ, β1 β β2 βͺ αΎβ1 + β2 , β) = ββ, β3 βͺ 1, β Solution: 2π 2π Period: = =π π β2 ππ Asymptotes: π₯ = π + ,π π β β€; thus, the asymptotes are π ππ π₯= β + ,π ββ€ 8 2 Graph: π¦ = tan π₯ Graph: π¦ = cot π₯ The domain of π¦ = π cot π(π₯ β π) + π is the set of π₯ β β: sin π π₯ β π β 0 = ππ π₯ β β: π₯ β π + , π β β€. π The domain of π¦ = π tan π(π₯ β π) + π is the set of π₯ β β: cos π π₯ β π β 0 = ππ π₯ β β: π₯ β π + , π is an odd integer. 2π The range of both π¦ = π cot π(π₯ β π) + π and π¦ = π tan π(π₯ β π) + π is (ββ, β). The period of both π¦ = π cot(π(π₯ β π π)) + π and π¦ = π tan π(π₯ β π) + π is. π The asymptotes of π¦ = π cot π(π₯ β π) + π ππ are π₯ = π + , π β β€. π The asymptotes of π¦ = π tan π(π₯ β π) + π ππ are π₯ = π + , π is an odd integer. 2π 1 Graph: π¦ = tan 2π₯ 2 Solution: 1 1 In π¦ = tan 2π₯ , π = , π = 2, π = 0, and π = 0. 2 2 ππ Domain: π₯ β β: π₯ β π + , π is an odd integer = 2π ππ π₯ β β: π₯ β , π is an odd integer 4 Range: ββ, β Solution: π π π Period: = = π 2 2 ππ Asymptote: π₯ = π + , π is an odd integer; thus, the 2π ππ asymptotes are π₯ = , π is an odd integer 4 π₯ Graph: π¦ = 2 cot 3 Solution: π₯ 1 In π¦ = 2 cot , π = 2, π = , π = 0, and π = 0. 3 3 ππ Domain: π₯ β β: π₯ β π + π ,πββ€ = π₯ β β: π₯ β 3ππ, π β β€ Range: ββ, β Solution: π π Period: = 1 = 3π π 3 ππ Asymptote: π₯ = π + ,πβ β€; thus, the asymptotes are π π₯ = 3ππ, π β β€