Expert: Josh - 12/15/2008

Question
determine the amplitude, period, and phase shift of the function, y = 1/2 sin(x + pi)        then please try to explain to me how to sketch one complete period of the graph of the function y = 1/2 sin (x + pi)

Answer
Jerry

Let's look at the general equation for a sinusoidal waveform. Consider y=A*sin(kx+q).

AMPLITUDE:= A
PERIOD:=    2*pi/k
PHASE OFFSET:= -q/k

Here, A represents the AMPLITUDE. As you know, a sine function sin(x) oscillates between +1 and -1, so something like A*sin() simply scales the height of the waveform by a factor of A.

Note: A number of definitions may be used. The most common one measures amplitude as the maximum displacement from the equilibrium position (viz.,the center) in the context of simple harmonic motion. Following this definition, the amplitude A would be 1/2 in your question. An alternate definition measures the peak-to-trough distance, i.e., between the highest and lowest point in the waveform A-(-A)=2A. Make sure you know which definition is being used.

PERIOD: The period for sin(x) is 2*pi. The function repeats itself every 2*pi radian. Introducing a "k" in sin(kx), if k>1, we increases its frequency of oscillation and compresses the time axis by a factor of k (relative to the case of sin(x)). If k<1, we reduces its frequency of oscillation and dilates the x-axis by a factor of k.
e.g., sin(2x) has a period of 2*pi/2 = pi. A single period fits into the interval 0<=x<pi, whereas the period is 4*pi for sin(x/2).

NET TRANSLATION and PHASE OFFSET: sin(x+pi/4) shifts the function sin(x) to the left by pi/4 radian. When the coefficient in front of x is 1, this shift also represents the phase offset. However, if we double the frequency of the waveform (consider sin(2x)) and shift it by the same amount (pi/4 radian again), we effectively introduce a lead of 2*(pi/4)=pi/2 radian (twice of what we had before). This is because sin(2x) now has a period of "pi" (instead of 2*pi). Hence, a translation of "pi/4" represents 1/4 of a cycle. Before, sin(x) has a period of 2*pi, so a net shift of "pi/4" only advances the waveform by one eighth "(pi/4)/(2pi)" of a cycle.

The net translation is independent of the frequency, but the phase offset is expressed relative to the frequency of the sinusoidal function.

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Drawing the function:

In the case of f(x)=1/2 sin(x + pi), we draw a sine function bounded between -0.5 and +0.5. It has a period of 2*pi but is shifted (advanced) by pi radian in the negative direction of the x-axis.

Since the function is 2*pi periodic, it's up to you which 2*pi interval to show. If we focus on the interval from x=0 to x=2*pi, we can evaluate it at some key points.

x=0, f(0)=0.5 sin(pi)=0
x=pi/4, f(pi/4)=0.5 sin(5*pi/4)= -0.5*(1/sqrt(2))
x=pi/3, f(pi/3)=0.5 sin(4*pi/3)= -0.5*sqrt(3)/2
x=pi/2, f(pi/2)=0.5 sin(3*pi/2)= -0.5*1
...
keep going until x=2*pi