Find an equation for a sinusoidal function that has period 360°, amplitude 1, and contains the point (180°,0). Write your answer
in the form f(x)=Asin(Bx+C)+D, where A, B, C, and D are real numbers.
1 answer:
The answer is:
f(x) = 1*Sin(1*x + k*π).
It must be remembered that:
360°= 2π.
180° = π.
Therefore we see that:
A = 1, where A represents the amplitude.
B is equal to (2π / T) and T is the period of oscillation.
If B = 1 then T = 2pi = 360 ° as requested.
C is the phase.
In the required equation C = kπ, where k is any whole number.
D = 0
Below is a graph of the equation:
f (x) = 1sin (x + kπ) with k = 2 for this case.
It can be seen that indeed the equation satisfied all the requirements of the problem
f(x) = 1*Sin(1*x + k*π).
It must be remembered that:
360°= 2π.
180° = π.
Therefore we see that:
A = 1, where A represents the amplitude.
B is equal to (2π / T) and T is the period of oscillation.
If B = 1 then T = 2pi = 360 ° as requested.
C is the phase.
In the required equation C = kπ, where k is any whole number.
D = 0
Below is a graph of the equation:
f (x) = 1sin (x + kπ) with k = 2 for this case.
It can be seen that indeed the equation satisfied all the requirements of the problem
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