Assuming you mean that the pi is not within the sin(2pi),
its a vertical shift of +pi
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.
Shift = +2
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.
the period is 2pi. period is 2pi/b and the formula is y=AsinBx.
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
Shift = +2
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.
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.
the period is 2pi. period is 2pi/b and the formula is y=AsinBx.
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
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
2 sin(x) + 1 = 0 2 sin(x) = -1 sin(x) = -1/2 x = 210° and 330°
sine wave, with a period of 2pi/w
y = A sin Bx y = A Cos BxAmplitude = APeriod = 2pi/Bso....Amplitude = 1Period = 2pi/8
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.
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).
If this is a homework question, please consider trying to answer it on your own first, otherwise the value of reinforcement of the lesson will be lost on you. To determine the trigonometry function of sin, with a period of pi, and amplitude of 1, and a vertical shift of +1, start simple and expand. The period of sin(x) is 2 pi, so to halve that period you need sin(2x). The amplitude of sin(2x) is 2, so to halve that amplitude you need 1/2 sin(2x). To shift any function up by 1, simply add 1 to it, so the final answer is 1/2 sin(2x) + 1. Note: This is very simple when you take it step by step.