Create a sine function in theta with an amplitude 1/3 , period pi/4 , and phase shift -pi/3

Question

I came up with y=(1/3)sin(((2pi)/8)x - pi/3) . Could someone verify for me? thank you!

Answer 1

Start from the general equation. f(t) = A*sin(ωt - Θ) = A*sin(2πft - Θ)

where A = Amplitude , ω = 2*π*f*t = angular frequency[rad/sec]

f = frequency[cycle/sec] , t = time[s] , Θ = angle[rad]

given A = 1/3 , T = π /4 => f = 1/T = 4/π , Θ = -π/3

hence f(t) = 1/3sin((2*π*4/π)t - (-π/3))

after simplification

f(t) = (1/3)sin(8t+π/3)