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Assuming you mean that the pi is not within the sin(2pi),
its a vertical shift of +pi
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If you are given the equation of an normal sine curve how would you determine its period?
a normal sine curve exists with the formula Asin(Bx+C)+D. The formula to derive a phase shift would be such: 2pi/B (for whatever value B exists at). Thus, for a normal sine curve (sin(x) we would get 2pi/1, and arrive at 2pi for the period.
Phase shift of y equals sin x-2?
Shift = +2
What is the formula for calculating time displacement in phasors of the form Asin b plus c?
The graph of y = A sin (Bx + C) is obtained by horizontally shifting the graph of y = A sin Bx, where 1) Amplitude = |A|, the maximum of y is A, if A > 0 and -(-A) if A < 0 The minimum of y is -A if A > 0 and A if A < 0 2) Period = (2pi)/B 3)Phase shift = C/B So that the starting point of the cycle is shifted from x = 0 to x = C/B. If C/B > 0 the shift is to the right. If C/B < 0 the shift is to the left.
Determine the period of y equals -3 sin x?
the period is 2pi. period is 2pi/b and the formula is y=AsinBx.
How do you solve sin2x plus sinX equals 0?
sin(2x) + sin(x) = 0 2sin(x)cos(x) + sin(x) = 0 sin(x)[2cos(x) + 1] = 0 sin(x) = 0 OR 2cos(x) + 1 = 0 sin(x) = 0 OR cos(x) = -1/2 x = n*pi OR x = 2/3*pi + 2n*pi OR x = -2/3*pi + 2n*pi x = pi*[2n + (0 OR 2/3 OR 1 OR 4/3)] Note that n may be any integer. The solutions in [-2pi, 2pi] are: -2pi, -4/3pi, -pi, -2/3pi, 0, 2/3pi, pi, 4/3pi, 2pi
If you are given the equation of an normal sine curve how would you determine its period?
a normal sine curve exists with the formula Asin(Bx+C)+D. The formula to derive a phase shift would be such: 2pi/B (for whatever value B exists at). Thus, for a normal sine curve (sin(x) we would get 2pi/1, and arrive at 2pi for the period.
Phase shift of y equals sin x-2?
Shift = +2
What is the formula for calculating time displacement in phasors of the form Asin b plus c?
The graph of y = A sin (Bx + C) is obtained by horizontally shifting the graph of y = A sin Bx, where 1) Amplitude = |A|, the maximum of y is A, if A > 0 and -(-A) if A < 0 The minimum of y is -A if A > 0 and A if A < 0 2) Period = (2pi)/B 3)Phase shift = C/B So that the starting point of the cycle is shifted from x = 0 to x = C/B. If C/B > 0 the shift is to the right. If C/B < 0 the shift is to the left.
What is the phase shift for a sin wave with the maximum amplitude at time zero?
For a sine wave with maximum amplitude at time zero, there is no phase shift. The wave starts at its peak at time zero, and therefore, its phase angle is zero.
How do you solve sin2x plus sinX equals 0?
sin(2x) + sin(x) = 0 2sin(x)cos(x) + sin(x) = 0 sin(x)[2cos(x) + 1] = 0 sin(x) = 0 OR 2cos(x) + 1 = 0 sin(x) = 0 OR cos(x) = -1/2 x = n*pi OR x = 2/3*pi + 2n*pi OR x = -2/3*pi + 2n*pi x = pi*[2n + (0 OR 2/3 OR 1 OR 4/3)] Note that n may be any integer. The solutions in [-2pi, 2pi] are: -2pi, -4/3pi, -pi, -2/3pi, 0, 2/3pi, pi, 4/3pi, 2pi
Determine the period of y equals -3 sin x?
the period is 2pi. period is 2pi/b and the formula is y=AsinBx.
How do you find theta if 2sin theta plus 1 equals 0 and 0 is less than or equal to theta which is less than 2pi?
2 sin(x) + 1 = 0 2 sin(x) = -1 sin(x) = -1/2 x = 210° and 330°
What equation is this x equals a sin wt?
sine wave, with a period of 2pi/w
What is the integral of cosx divided by sinx plus cosx from 0 to 2pi?
The Integral diverges. It has singularities whenever sin(x)+cos(x)=0. Singularities do not necessarily imply that the integral goes to infinity, but that is the case here, since the indefinite integral is x/2 + 1/2 Log[-Cos[x] - Sin[x]]. Obviously this diverges when evaluated at zero and 2pi.
What is the difference between phase and frequency?
No matter what frequency, there are 360 degrees that can be associated with it (the phase). Here's an equation to summarize: V(t) = A sin ([w*t] + p) A is amplitude w = frequency p = phase shift
What s the amplitude and period of y cos 8x?
y = A sin Bx y = A Cos BxAmplitude = APeriod = 2pi/Bso....Amplitude = 1Period = 2pi/8
How do you find the complex cube roots of 1 plus 0i?
z = 1 + 0i So |rz| = 1 and az = 0 radians. which allows you to write z = rz*cos(az) + i*sin(az) Then, if y = z1/3 then |y| = |z1/3| = |11/3| = 1 and ay is the angle in [0, 360) such that 3*ay = 0 mod(2*pi) that is, ay = 0, 2pi/3 and 4pi/3 And therefore, Root 1 = cos(0) + i*sin(0) Root 2 = cos(2pi/3) + i*sin(2pi/3) and Root 3 = cos(4pi/3) + i*sin(4pi/3).
