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Graphs of Trigonometric Functions Here are the graphs of the six main trigonometric functions along with their domains and ranges. The windows shown have x-values from β2π to 2π, and y-values from -3 to 3: y ο½ sin x Domain: Range: Period: All Real Numbers [-1, 1] 2π y ο½ cos x Domain: Range: Period: All Real Numbers [-1, 1] 2π y ο½ tan x Domain: All Real Numbers π except 2 ± ππ Range: Period: y ο½ csc x Domain: Range: (-β, β) π All Real Numbers except :0 ± ππ (-β, -1] and [1, β) Period: y ο½ sec x Domain: Range: Period: y ο½ cot x Domain: Range: Period: 2π All Real Numbers π except 2 ± ππ (-β, -1]and[1,β) 2π All Real Numbers Except 0 ± ππ (-β, β) π TransformationsThe general form for the sine equation is: (In our parent graph: y ο½ sin x ; ο· ο· ο· ο· y ο½ a sin(bx ο« c ) ο« d a = b = 1, c = d = 0) Changing the value of βaβ changes the amplitude of the graph. βaβ is the amplitude of the graph. Changing the value of βbβ changes the period of the graph. The new period is found by taking the original period and dividing by βbβ Changing the values of βbβ and βcβ creates phase shift. This c moves the graph horizontally. The phase shift is equal to: ο b Changing the value of βdβ creates vertical displacement. If βdβ is positive, the graph will be moved up. If βdβ is negative, the graph will be moved down. Example Set 1: Find the amplitude, period, phase shift and vertical displacement of these graphs: a) y ο½ 3sin(4 x ο 20) ο« 5 b) yο½ 1 1 cos( x ο« 4) ο 1 2 2 c) π π¦ = tanβ‘(2π₯ β 2 ) Example Set 2: Give the equation for the sine function that has the following characteristics: a) Amplitude = Period = 2 b) Amplitude = Period = 8π 2π 5 Phase Shift = 5π ½ π Phase Shift = β 4 9 Answers: Example Set 1: a 3 Amplitude Period Phase Shift 5 Vertical Displacement +5 Note: b ½ π 2 c n/a (See Note) π 2 π 4 None 4π -8 -1 We donβt talk about amplitude for tangent, as this graph goes off to infinity in both directions. Period Calculations: a) 2π 4 = π 2 b) 2π 1 2 = 4π c) π 2 = π 2 Phase Shift Calculations: a) c ο20 ο ο½ο ο½5 b 4 b) c 4 ο ο½ ο ο½ ο8 1 b 2 c) π βπ = β β π 2 2 = π 4 Example Set 2: a) π¦ = 2sinβ‘(5π₯ β 25π 9 ) 1 b) π₯ π π¦ = 2 sinβ‘(4 + 16) Calculations to find βbβ: Recall that the new period is the old period divided by b: a) 2π 5 = π= 2π π 2π b) 8π = 2π 5 2π π 2π π = 8π 1 π=5 π=4 Calculations to find βcβ: We find βcβ by using the phase shift as well as the value we just found for βbβ. c Remember that the phase shift is equal to: ο b π a) ππ = β π 5π 9 π = β5 π= β25π 9 b) π ππ = β π π π β4 = β 1 4 π π = 16