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Published byJanelle Gorse Modified about 1 year ago

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1. sin (-720°) 2. tan (-180°) 3. cos (540°) 4. tan ( π ) 5. csc (4 π ) 6. sec ( π ) Undefined

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1. sin θ = sec θ = tan θ = 0 3. sin θ = 0 270° + 360k° where k is any integer 180° + 360k° where k is any integer 180k° where k is any integer 180k° where k is any integer

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1.) y = sin x [-90° ≤ x ≤ 90°], scale of 45°

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1.) y = tan x [- π /2 ≤ x ≤ 3 π /2], scale = π /4

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1.) y = cos x [-360° ≤ x ≤ 360°]

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1.) y = sec x [-360° ≤ x ≤ 360°]

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1.) y = -2sin θ 2.) y = 10sec θ 3.) y = -3sin 4 θ 4.) y = 0.5sin ( θ - ) 5.) y = 2.5 cos( θ + 180°)6.) y = -1.5sin (4 θ - ) Amp = 2 Per = 360° PS = 0° Amp = 10 Per = 360° PS = 0° Amp = 0.5 Per = 360° PS = right Amp = 3 Per = 90° PS = 0° Amp = 2.5 Per = 360° PS = 180° left Amp = 1.5 Per = 90° PS = right

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1.) Amp = 0.75, period = 360°, PS = 30° 2.) Amp = 4, period = 3°, PS = -30° y = ± 0.75 sin( θ - 30°) y = ± 4 sin(120 θ +3600°)

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1.) Amp = 0.75, period = 360°, PS = 30° 2.) Amp = 4, period = 3°, PS = -30° y = ± 3.75 cos (4 θ - 16°) y = ± 12 cos (8 θ )

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1.) y = 0.5 sin x

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1.) y = 2 cos (3x)

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1.) y = 2 cos (2x – 45°)

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1.) y = tan (x + 60°)

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Find the exact value of each expression without using a calculator. When your answer is an angle, express it in radians. Work out the answers yourself before you click.

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Answers for problems 1 – 9. Negative ratios for arccos generate angles in Quadrant II. y x 1 2 The reference angle is so the answer is

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y x x 1 2 y 15.

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1.) y = -2cos (3 θ ), scale π /4, -2 π ≤ θ ≤ 2 π

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1.) y = ½ cos(x – π /2)

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1. 2.

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1.Find the values of θ for which the equation tan θ = 1 is true. 2.State the domain and range for the function y = -csc x 3.State the amplitude, period, and phase shift of: 4.Write an equation of the cosine function with amplitude 7, period π, and phase shift 3 π /2 45°+180°k D = all reals except 180°k R = y ≤ -1 or y ≥ 1 A = ⅓ PS = - π /6 P = 2 π

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