Respuesta :
You did not include the graphs, but I can explain how the graph is and so you can recognize it.
1) for x = 0, y = sin(0) = 0, so the origin (0,0) is part of the graph.
2) The range of the function is the interval [-1, 1]. This is, y is from -1 to 1, and the graph oscilates (is a wave) between those minimum and maximum.
3) The periodicity, P, of the function is such that 0.5P = 2π = > P = 4π
That means that the values repeat every 4π interval (which is the same that 720°)
4) The zeroes of the function are 2π * n, where n is any whole number (negative, zero or positive).
That means that the function crosses the x-axis at
..., -6π, -4π, 2π, 0, 2π, 4π , 6π, ...
5) It is an odd function (which means that f(x) = sin(0.5x) = - f(-x) = - sin(-0.5x).
Now you have plenty information to identify the graph.
I also include a graph of the function in the attached pdf file.
1) for x = 0, y = sin(0) = 0, so the origin (0,0) is part of the graph.
2) The range of the function is the interval [-1, 1]. This is, y is from -1 to 1, and the graph oscilates (is a wave) between those minimum and maximum.
3) The periodicity, P, of the function is such that 0.5P = 2π = > P = 4π
That means that the values repeat every 4π interval (which is the same that 720°)
4) The zeroes of the function are 2π * n, where n is any whole number (negative, zero or positive).
That means that the function crosses the x-axis at
..., -6π, -4π, 2π, 0, 2π, 4π , 6π, ...
5) It is an odd function (which means that f(x) = sin(0.5x) = - f(-x) = - sin(-0.5x).
Now you have plenty information to identify the graph.
I also include a graph of the function in the attached pdf file.
The graph of y = sin (0.5x) can be seen in the picture in the attachment
Further explanation
Firstly , let us learn about trigonometry in mathematics.
Suppose the ΔABC is a right triangle and ∠A is 90°.
sin ∠A = opposite / hypotenuse
cos ∠A = adjacent / hypotenuse
tan ∠A = opposite / adjacent
There are several trigonometric identities that need to be recalled, i.e.
[tex]cosec ~ A = \frac{1}{sin ~ A}[/tex]
[tex]sec ~ A = \frac{1}{cos ~ A}[/tex]
[tex]cot ~ A = \frac{1}{tan ~ A}[/tex]
[tex]tan ~ A = \frac{sin ~ A}{cos ~ A}[/tex]
Let us now tackle the problem!
Given :
[tex]y = \sin (0.5x)[/tex]
Let us try to find some important parameters of this function:
To find amplitude just look at the coefficient in front of the sine function which is 1, so the amplitude for this function is 1 unit.
y = 1 sin (0.5x)
To find the period, we will divide 2π by the coefficient of the variable x that is 0.5 in this function, so the period is 2π/0.5 = 4π
y = 1 sin (0.5x)
To find the maximum and minimum values, we substitute the maximum value of the sine function which is 1 and the minimum value of the sine function which is -1, so that the equation becomes:
Maximum Values of y = sin (0.5x) → y = 1
Minimum Values of y = sin (0.5x) → y = -1
From the values obtained above, a graph of the function can be sketched as shown in the picture in the attachment.
Learn more
- Calculate Angle in Triangle : https://brainly.com/question/12438587
- Periodic Functions and Trigonometry : https://brainly.com/question/9718382
- Trigonometry Formula : https://brainly.com/question/12668178
Answer details
Grade: College
Subject: Mathematics
Chapter: Trigonometry
Keywords: Sine , Cosine , Tangent , Opposite , Adjacent , Hypotenuse

