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Trigonometric Function Toolkit · Tutorial 127 of 1495

Amplitude, Period, and Phase Shift

Read amplitude, period, and horizontal phase shift from a sinusoid, then use its key points to sketch one cycle correctly.

Beginner 9 min read

What You'll Learn

  • Identify the amplitude and horizontal phase shift in the stated sine and cosine forms.
  • Calculate the fundamental period using the absolute value of the frequency coefficient.
  • Build five key points for one cycle, including when the frequency is negative.
  • Sketch and check a sinusoidal graph from its equation.
  • Recognize when a constant equation has no fundamental period.

Reading the Parameters of a Sinusoid

A sine or cosine graph repeats, but its equation can change how tall the graph is, how quickly it repeats, and where a cycle begins along the horizontal axis. In the forms considered here, those features are encoded directly by three parameters.

$$ y=A\sin\bigl(B(x-C)\bigr) \qquad\text{or}\qquad y=A\cos\bigl(B(x-C)\bigr). $$

Here \(A\), \(B\), and \(C\) are real numbers. The coefficient \(A\) scales the output, \(B\) scales the input angle, and \(C\) specifies a horizontal displacement. Reading these parameters carefully gives a reliable way to plot one complete cycle and continue the repeating curve.

Definition: For a nonconstant sinusoid in either stated form, the amplitude is \(|A|\), the fundamental period is \(\frac{2\pi}{|B|}\), and the horizontal phase shift is \(C\). The period formula requires \(B\ne 0\); for a nonconstant sinusoid we also require \(A\ne 0\).

Amplitude, Period, and Phase Shift

The sine and cosine functions have values between \(-1\) and \(1\). Multiplying either function by \(A\) makes its output range from \(-|A|\) to \(|A|\). Thus the amplitude is a nonnegative distance from the midline \(y=0\) to an extreme value. It is not necessarily the signed coefficient \(A\): for example, if \(A=-3\), the amplitude is \(3\).

The period describes the horizontal distance for one full repetition. The input to sine or cosine is \(B(x-C)\), so changing \(x\) by \(T\) changes that input by \(BT\). One full cycle occurs when the input changes by \(2\pi\) in magnitude. Therefore \(T=2\pi/|B|\). The absolute value matters: a period is positive even if \(B\) is negative.

The phase shift is read from the grouped expression \(x-C\). If \(C>0\), the shift is \(C\) units to the right; if \(C<0\), it is \(|C|\) units to the left. For instance, \(x+\frac{\pi}{4}\) can be written \(x-(-\frac{\pi}{4})\), so its shift is \(-\frac{\pi}{4}\), or left by \(\frac{\pi}{4}\). The location \(x=C\) is where the argument \(B(x-C)\) equals zero.

These parameter readings describe the equation as written. Since the graph repeats every \(T\), shifting \(C\) by an integer multiple of \(T\) gives an equivalent horizontal placement of the same repeating graph.

Why the Period Formula Works

For either sine or cosine, adding \(2\pi\) or subtracting \(2\pi\) to the angle leaves the function value unchanged. If \(T=2\pi/|B|\), then \(BT=2\pi\) when \(B>0\) and \(BT=-2\pi\) when \(B<0\). Consequently, increasing \(x\) by \(T\) changes the angle by one full cycle, in either direction.

$$ B\bigl((x+T)-C\bigr) =B(x-C)+BT =B(x-C)+2\pi\,\operatorname{sgn}(B), $$

where \(\operatorname{sgn}(B)=1\) for \(B>0\) and \(\operatorname{sgn}(B)=-1\) for \(B<0\). Periodicity of sine and cosine then gives the same output at \(x+T\) as at \(x\). The earlier material on periodicity establishes that their fundamental positive period in the angle is \(2\pi\). Thus the smallest positive horizontal period is \(2\pi/|B|\).

If \(A=0\), the equation is the constant function \(y=0\); if \(B=0\), it is constant as well. A constant function repeats for every positive distance, so it does not have a unique fundamental period. In those cases, do not use the period formula to claim a unique cycle length.

Five Key Points for One Cycle

For a nonconstant graph, let \(T=2\pi/|B|\). The five \(x\)-coordinates from \(C\) through one full cycle are \(C,C+T/4,C+T/2,C+3T/4,C+T\). The spacing is \(T/4\), and the points are in increasing \(x\)-order. Their output values depend on whether the function is sine or cosine and, for sine, on the sign of \(B\).

$$ \begin{array}{c|ccccc} & C & C+\frac{T}{4} & C+\frac{T}{2} & C+\frac{3T}{4} & C+T\\ \hline A\sin(B(x-C)) & 0 & A\,\operatorname{sgn}(B) & 0 & -A\,\operatorname{sgn}(B) & 0\\ A\cos(B(x-C)) & A & 0 & -A & 0 & A \end{array} $$

To see the sine entries, substitute \(x=C+jT/4\), where \(j=0,1,2,3,4\). The angle becomes \(B jT/4=j\pi\,\operatorname{sgn}(B)/2\). Evaluating sine at these angles gives the displayed values, including the reversal of the middle two nonzero values when \(B<0\). For cosine, changing the sign of an angle does not change its cosine, so the listed sequence is the same for either sign of \(B\).

Connect the plotted points with a smooth sinusoidal curve and continue the pattern every \(T\) units. The table gives the points, not straight-line segments: sine and cosine graphs are smooth curves.

Worked Examples

Worked Example: A Sine Graph with Positive \(B\)

Sketch one cycle of \(y=3\sin\bigl(2(x-\frac{\pi}{6})\bigr)\), and identify its amplitude, period, and phase shift.

Comparing with \(A\sin(B(x-C))\) gives \(A=3\), \(B=2\), and \(C=\frac{\pi}{6}\). Therefore the amplitude is \(|3|=3\), and the phase shift is \(\frac{\pi}{6}\) to the right. The period is

$$T=\frac{2\pi}{|2|}=\pi.$$

One cycle begins at \(C=\frac{\pi}{6}\). Each key-point step is \(T/4=\frac{\pi}{4}\), so the five \(x\)-coordinates are

$$\frac{\pi}{6},\quad \frac{5\pi}{12},\quad \frac{2\pi}{3},\quad \frac{11\pi}{12},\quad \frac{7\pi}{6}.$$

Since \(B>0\), the sine values in order are \(0,3,0,-3,0\). For example, at \(x=\frac{5\pi}{12}\), the angle is \(2(\frac{5\pi}{12}-\frac{\pi}{6})=2(\frac{\pi}{4})=\frac{\pi}{2}\), giving \(y=3\sin(\frac{\pi}{2})=3\). At \(x=\frac{11\pi}{12}\), the angle is \(2(\frac{3\pi}{4})=\frac{3\pi}{2}\), giving \(y=3\sin(\frac{3\pi}{2})=-3\). The other three angles are \(0,\pi,2\pi\), so their outputs are \(0,0,0\), respectively. The resulting points are \((\frac{\pi}{6},0),(\frac{5\pi}{12},3),(\frac{2\pi}{3},0),(\frac{11\pi}{12},-3),(\frac{7\pi}{6},0)\). The first and last points agree in height and are one period apart, as required.

Worked Example: A Sine Graph with Negative \(B\)

Sketch one cycle of \(y=2\sin\bigl(-3(x-\frac{\pi}{4})\bigr)\). The parameters are \(A=2\), \(B=-3\), and \(C=\frac{\pi}{4}\). The amplitude is \(2\), and the phase shift is \(\frac{\pi}{4}\) to the right. The period is positive:

$$T=\frac{2\pi}{|-3|}=\frac{2\pi}{3},\qquad \frac{T}{4}=\frac{\pi}{6}.$$

The key \(x\)-coordinates are \(\frac{\pi}{4},\frac{5\pi}{12},\frac{7\pi}{12},\frac{3\pi}{4},\frac{11\pi}{12}\). Because \(B<0\), the sine pattern is \(0,-2,0,2,0\). Check the second point: its angle is \(-3(\frac{5\pi}{12}-\frac{\pi}{4})=-3(\frac{\pi}{6})=-\frac{\pi}{2}\), so its output is \(2\sin(-\frac{\pi}{2})=-2\). At the fourth point, the angle is \(-3(\frac{\pi}{2})=-\frac{3\pi}{2}\), and the output is \(2\sin(-\frac{3\pi}{2})=2\). At the first, third, and fifth points, the angles are \(0,-\pi,-2\pi\), and the outputs are all \(0\). Thus the five points have the stated heights. The first and last differ in \(x\) by \(\frac{11\pi}{12}-\frac{\pi}{4}=\frac{2\pi}{3}=T\), confirming one complete cycle.

Worked Example: A Cosine Graph Shifted Left

Sketch one cycle of \(y=4\cos\bigl(\frac{1}{2}(x+\frac{\pi}{3})\bigr)\). Write the input as \(\frac{1}{2}(x-(-\frac{\pi}{3}))\). Hence \(A=4\), \(B=\frac12\), and \(C=-\frac{\pi}{3}\). The amplitude is \(4\), the phase shift is \(\frac{\pi}{3}\) to the left, and

$$T=\frac{2\pi}{|1/2|}=4\pi,\qquad \frac{T}{4}=\pi.$$

The five \(x\)-coordinates are \(-\frac{\pi}{3},\frac{2\pi}{3},\frac{5\pi}{3},\frac{8\pi}{3},\frac{11\pi}{3}\). The cosine heights are \(4,0,-4,0,4\). To check them, the angles at these coordinates are \(0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi\), respectively. Multiplying their cosine values by \(4\) gives \(4,0,-4,0,4\). The first and last \(x\)-coordinates differ by \(4\pi\), the calculated period, and give the same height.

A Common Mistake: Ignoring the Sign of \(B\)

A frequent error is to use the usual increasing-angle sine pattern even when \(B\) is negative. Consider \(y=\sin(-x)\), with \(A=1\), \(B=-1\), \(C=0\), and \(T=2\pi\). The five coordinates are \(0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi\). The correct heights are \(0,-1,0,1,0\), because the angle is \(-x\). Using the positive-\(B\) pattern would incorrectly give \(0,1,0,-1,0\). Check the second and fourth values directly: \(\sin(-\frac{\pi}{2})=-1\), while \(\sin(-\frac{3\pi}{2})=1\). Use the sign-aware table whenever you sketch a sine graph.

Another useful check is to verify the units of your calculation in the horizontal direction: the period must be positive, and the five key points must be separated by exactly \(T/4\). A negative result from \(2\pi/B\) signals that the absolute value in the period formula was omitted.

Why These Parameters Matter

Amplitude, period, and phase shift let you compare sinusoidal graphs without plotting many points first. Amplitude tells you the size of the vertical range, period tells you the horizontal length of a cycle, and phase shift locates the cycle. Together with the correct sine or cosine key-point pattern, they turn an equation into a dependable sketch.

Takeaway: For \(y=A\sin(B(x-C))\) or \(y=A\cos(B(x-C))\), read amplitude as \(|A|\), period as \(\frac{2\pi}{|B|}\) when \(B\ne0\), and horizontal phase shift as \(C\). For sine, account for the sign of \(B\) when assigning the key-point heights.

Check Your Understanding

Use the parameter definitions and the five-point method to answer these questions.

  1. For \(y=5\sin(4(x-\frac{\pi}{3}))\), what are the amplitude, fundamental period, and horizontal phase shift?
  2. For \(y=3\cos(-2(x+\frac{\pi}{6}))\), write the phase shift as a signed value of \(C\), and calculate the period.
  3. For \(y=2\sin(-x)\), list the five heights over one cycle starting at \(x=0\).
  4. For \(y=-4\cos(\frac{1}{2}(x-\pi))\), give the amplitude and the five heights over one cycle starting at \(x=\pi\).
  5. Why is there no unique fundamental period for \(y=7\cos(0(x-2))\)?

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