How is the phase of a sine function found?
Phase and initial phase are two important concepts in the sine function.
Phase is the phase angle in a sine function, which indicates the position of the sine function in the coordinate system.
Initial phase, on the other hand, is the phase of the sine function at t=0, which is the value of the phase angle at t=0.
To solve for the phase and initial phase of the sine function, you need to first express it in standard form as y=Asin(ωt+φ).
Where A denotes the amplitude, ω the angular frequency, and φ the phase.
Phase can be obtained by translating the sine function by a length equal to the arc length corresponding to the phase angle.
That is, if the phase angle of a sine function is φ, then translating it to the right by φ arc lengths gives a sine function in standard form.
The initial phase can be determined by taking the value of the sine function at t=0.
Specifically, if the value of the sine function at t=0 is y0, then shifting it up by y0 units gives the sine function in standard form.
In summary, we can use the following methods to find the phase and initial phase of the sine function:
Denote the sine function as the standard form y=Asin(ωt+φ);
According to the phase angle φ, translate it to the right by φ arc length, and then get the sine function with the standard form;
According to the value y0 at t=0, translate it upward by y0 units, and then get the sine function with the standard form. y0 units to get the sine function in standard form.
How to solve for phase and initial phase of a sine function?
In a sine function (sin function), phase and initialphase are concepts used to describe the position and relative starting point of the function image. Their specific meanings are as follows:
1. Phase: phase indicates the translational position of the function image in the x-axis direction. In the sine function y=A*sin(ωx+φ), the phase is φ in (ωx+φ), which determines the horizontal translation position of the sine function image.
2. Initial phase: The initial phase is the value of x corresponding to the starting point of the image of the sine function. In the sine function y=A*sin(ωx+φ), when x=0, y=A*sin(φ). The initial phase is the value of the smallest non-negative angle that makes sin(φ) = 0. It determines the value of x that corresponds to the starting point of the image of the sine function.
The method of solving for phase and initial phase can be determined by the properties of the function and by using some mathematical techniques. In general, the solution can be done in the following ways:
1. Observing the image of the function: By observing the properties of the image of the sine function, the approximate range of the phase and initial phase can be estimated. Notice that the period of the sine function is 2π and that the sine function minimizes when x + φ = 0. This can give an estimate of the initial phase.
2. Equation-based solution: the phase and initial phase can be solved by substituting some specific values of x and picking the x values that satisfy the conditions. For example, when x=π/2, sin(ωx+φ)=1, this condition can be utilized to solve the values of phase and initial phase.
3. Mathematical analysis: Using some mathematical analysis techniques, such as Taylor series expansion, Fourier series, etc., a given sine function can be analyzed to derive the exact values of phase and initial phase.
It should be noted that both phase and initial phase are expressed in radian units in the expression of the sine function. In practice, the phase and initial phase can be solved by observation, substitution, and analysis according to the specific problem and data.
How to see from the sine function image of the initial phase of the positive and negative
A while ago in the micro-signal published an article for the “trigonometric interval problem of the seconds” this article, a friend asked: in the search for the initial phase of the why to use the most valuable point to find, and can not be used to find the zero point? In fact, it is possible to use the zero point, but the amount of thinking is greater than the amount of thinking with the initial phase of the most valuable point, so the teacher in the teaching, usually emphasize the use of the most valuable point.
such as mentioned in the previous article:
We try to use the zero point to solve:
Seeking out two answers, obviously there is an incorrect, we need to go to round off an answer that does not meet the meaning of the question, why does this situation occur, in order to explain this problem, we today reacquainted with the old friend “initial phase “
We’re going to have to get rid of one of the answers that doesn’t fit!
We know from physics that simple harmonic motion explains trigonometry:
Phase is, for a wave, the position of a particular moment in its cycle: a kind of scale of whether it is at the peaks of the wave, the valleys of the wave, or a point between them. Phase describes a measure of change in the waveform of a signal, usually in degrees (angles), also known as phase angle. And the initial phase is the phase at time 0, also known as the initial phase angle or initial phase. The initial phase reflects the starting point of alternating current alternation and is related to the choice of the starting point of time.
That is to say, in telescopic (periodic) transformation, we can consider trigonometric functions as dynamic elastic steel ropes, which are obtained by fixing the point of intersection with the y-axis in telescopic (periodic) transformation, and compressing or stretching the rest toward the y-axis!
Through the graphical demonstration above, we are able to see that the monotonic tendency of the prime of the function image before and after the change does not change when the amplitude transformation is not involved, and so, with the help of monotonicity, we are able to uniquify this problem. Let’s return and look again, at the original problem:
Next, we are changing this problem a bit:
In this adapted problem, the period is not easy to find without the coordinates of M. So can we find the period by the coordinates of the point P and the point N? We have the previous telescopic translation transformations know that the relative position of these plasmas in the change is unchanged, that is to say, the proportion is the same. So, we can also find the period with the help of the relative positions of these two points.
Or, we can get it by telescopic transformation:
It is easy to find that the conclusion we used:
For sine-type functions, the distance between the two neighboring axes of symmetry (the most significant points) is half of the period, the distance between the two neighboring centers of symmetry is half of the period, and the horizontal distance between the neighboring axes of symmetry and the center of symmetry is one-fourth of the period. This is a special case of this problem.
We go back to the original problem, you can get the first phase of the function directly through the image of a complete cycle:
We are more intuitive from the perspective of the motion of the function easier to portray the image of the transformation of the function’s “past life”, the essence of the image transformation is the transformation of the point, and then from the point of the transformation of the transformation of the shape of the transformation, the last From the appearance of the shape of the transformation to realize the essence of the transformation of points and points to grasp.
How about, through this article, you are not on the trigonometric function of the “initial phase” has a new understanding, a more thorough understanding.
How do you see the initial phase of a sinusoidal AC waveform?
U=Umsin(wt+φ), when t=0, U=0, so initial phase φ=0.
If the 0 point is shifted to the left to π/2, when t=0, U=Um, so initial phase φ=π/2.
What are the phase and initial phase in a sine function? How should I find them?
1, in y=Asin(ωx+φ), A is called the amplitude; ωx+φ is called the phase; the phase at x=0 (ωx+φ=0+φ=φ) is called the initial phase.
2. There are specific functions that can be found. y is a function of x, and A, ω, and φ are fixed values.
Sine function y=sinx; cosine function y=cosx
1, monotonic interval
Sine function monotonically increasing on [-π/2+2kπ,π/2+2kπ], monotonically decreasing on [π/2+2kπ,3π/2+2kπ]
Cosine function Monotonically increasing on [-π+2kπ,2kπ] and monotonically decreasing on [2kπ,π+2kπ]
The sine function is odd
The cosine function is even
The sine function is symmetric about the axis x = π/2 + 2kπ, and centrally symmetric about (kπ,0)
The cosine function is symmetric about (kπ,0). p>
The cosine function is symmetric about x=2kπ and centrally symmetric about (π/2+kπ,0)
The period of the sine-cosine function is 2π
How to interpret the sine function y=asin(ωx+φ)
y=Asin(ωx+φ) is called the sine function.
Analytic equation of sine type function: y=Asin(ωx+φ)+h
The effect of each constant value on the function image:
φ (initial phase): determines the position of the waveform in relation to the X-axis or the distance of transverse movement (left plus right minus)
ω: determines the period (the smallest positive period T=2π/|ω|)
A: determines the peak value ( i.e. longitudinal stretching and compression times)
h: indicates the position of the waveform in the y-axis or the longitudinal movement distance (up and down)
Graphing method using the “five-point method” for graphing
The “five-point method” means that the waveform will move in the longitudinal direction when ω x+φ are 0, π/2, π, 3π/2 and 2π respectively.