Which of the following values of y is a solution to the equation y^3-5?
1 answer:
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The equation of the function f(x) is f(x) = 2 sin(π/2(x + 6)) - 3
<h3>How to create the sine function?</h3>A sine function is represented as:
f(x) = A sin(B(x + C)) + D
Where
A = Amplitude
Period = 2π/B
C = Phase shift
D = Vertical shift
The requirements in the question are:
- Amplitude not equal to 1
- Period not equal to 2π
- Non-zero phase and vertical shifts
So, we can use the following assumptions
A = 2
Period = 4
C = 6
D = -3
So, we have:
f(x) = 2 sin(B(x + 6)) - 3
The value of B is
4 = 2π/B
This gives
B = π/2
So, we have:
f(x) = 2 sin(π/2(x + 6)) - 3
<h3>The amplitude, vertical shift, period of f(x)and the phase shift</h3>Using the representations in (a), we have:
- Amplitude = 2
- Vertical shift = -3
- Period = 4
- Phase shift = 6
See attachment for the graph of f(x)
<h3>The value of cos θ </h3>Let θ = 3π
So, we have:
cos(3π)
This is calculated as:
cos(3π) = cos(2π + π)
Expand
cos(3π) = cos(2π) *cos(π) - sin(2π) *sin(π)
Evaluate
cos(3π) = -1
<h3>The value of sin θ </h3>Let θ = 3π
So, we have:
sin(3π)
This is calculated as:
sin(3π) = sin(2π + π)
Expand
sin(3π) = sin(2π) *cos(π) + cos(2π) *sin(π)
Evaluate
sin(3π) = 0
<h3>The value of tan 2θ </h3>Let θ = 3π
So, we have:
tan(2 * 3π)
tan(6π)
This is calculated as:
tan(6π) = tan(3π + 3π)
Evaluate
tan(3π) = 0
Read more about sinusoidal functions at:
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