Graphs of Trigonometric Functions
Read and sketch periodic graphs using amplitude, period, phase shift, and midline.
Learning goals
- Identify key features of a sinusoidal graph.
- Sketch a cycle from five key points.
- Write an equation from graph features.
Periodic functions
Radians measure turns using arc length
The radian measure of a central angle is the length of its intercepted circular arc divided by the radius, with a sign for the direction of rotation when using directed angles. One complete revolution is two pi radians.
This makes angle measures independent of the circle’s size and explains why sine and cosine repeat over an interval of two pi on a radian axis.
Look for a complete pattern that starts repeating, rather than merely two points with the same height. A sine wave reaches the same height at several places within one cycle. The period describes repetition of the whole graph. We will use that distinction when deciding how much of a graph to draw.
A periodic function repeats: \(f(x+T)=f(x)\). The period is the smallest positive repeat length for a nonconstant sine or cosine wave. We use radians on the horizontal axis.
| Function | Period | Range | Five values over one cycle |
|---|---|---|---|
| \(\sin x\) | \(2\pi\) | \([-1,1]\) | \(0,1,0,-1,0\) |
| \(\cos x\) | \(2\pi\) | \([-1,1]\) | \(1,0,-1,0,1\) |
Amplitude, period, and translations
Four different features of a wave
The midline is the horizontal level halfway between a wave’s maximum and minimum. Amplitude is the nonnegative distance from that midline to a peak; period is the horizontal length of one full repeat. Phase shift locates the cycle horizontally.
These features describe different aspects of the graph and should be read separately.
A periodic graph carries several kinds of information at once: its height, its repeating interval, and its position. We will change one feature at a time. Watch which landmarks move and which relationships stay the same.
\[y=a\sin(bx-c)+d\quad\text{or}\quad y=a\cos(bx-c)+d\]\[\text{amplitude}=|a|,\quad T=\frac{2\pi}{|b|},\quad h=\frac cb,\quad\text{midline: }y=d\]Here \(a\ne0\) and \(b\ne0\). Factor the argument as \(b(x-h)\). Positive \(h\) shifts right; negative shifts left. A negative \(a\) reflects the wave across its midline.
Amplitude is a nonnegative distance. The phase shift is c/b, not simply c.
Explore the parameters
Use the graph to test a prediction rather than changing every control at once. Choose one parameter, say what feature you expect to move, and then make the change. Reset or compare with the original curve so that a change in height is not confused with a change in horizontal spacing.
Read an equation
Now connect each part of the equation to a visible feature. Find the midline and the vertical size, then work out the period and horizontal shift. Be especially careful when the input is not already factored: the constant inside the brackets is not automatically the shift. Check where a familiar point in the basic cycle has moved.
A translated sine wave
Identify the amplitude, period, phase shift, midline, and range of this wave.
\(y=2\sin(3x-\pi)\)Show worked solution
The wave shifts right by \(\pi/3\). Its range is \([-2,2]\).
Reflection and a midline
Find the amplitude, period, midline, and range. Determine whether the point at x = 0 is a maximum or minimum.
\(y=-3\cos(2x)+1\)Show worked solution
At zero the cosine is one, so the curve begins at its minimum, negative two.
Five key points
You do not need dozens of plotted values to sketch one cycle accurately. Choose points that reveal the midline crossings and turning points, then place them at the correct spacing. The rest of the curve should follow the smooth pattern between those landmarks.
For \(b>0\), divide one cycle from \(h\) to \(h+T\) into four equal intervals. Calculate values, plot smoothly, then repeat the cycle.
Sketch a cosine graph
Sketch this graph over the stated interval, starting with five key points in one cycle.
\(y=3\cos x,\qquad-\pi\le x\le4\pi\)Show worked solution
| x | \(0\) | \(\pi/2\) | \(\pi\) | \(3\pi/2\) | \(2\pi\) |
|---|---|---|---|---|---|
| \(y\) | \(3\) | \(0\) | \(-3\) | \(0\) | \(3\) |
One cycle starts at a maximum, crosses the axis, reaches a minimum, crosses again, and returns to a maximum. Repeat left and right over the requested interval.
Read time intervals from a wave
Once a wave model is known, it can answer more than a height-at-one-time question. To find when a height crosses a threshold, first subtract the midline and divide by the amplitude. Solve the resulting trigonometric condition within one cycle, then translate those phase angles back into time. Equal heights can occur twice in a cycle; a repeated height alone does not establish the period.
When is a platform high enough?
A platform follows \(h(t)=6+4\cos(\pi t/6)\) metres, where t is in seconds. During \(0\le t\le12\), when is it at least \(8\) m high, and for how long in total?
Show worked solution
The inequality becomes \(\cos(\pi t/6)\ge1/2\). In one phase cycle from \(0\) to \(2\pi\), cosine is at least one half on \([0,\pi/3]\cup[5\pi/3,2\pi]\). Multiply the phase endpoints by \(6/\pi\): \(t\in[0,2]\cup[10,12]\). These intervals have total length \(4\) seconds. Counting only the first crossing at \(2\) seconds misses the high part at the end of the cycle.
The period is the repeat length of the entire pattern, including its direction of motion. Consecutive midline crossings are half a period apart and have opposite crossing directions.
Two sign changes
Explain why \(-3\sin(-2x)+1\) and \(3\sin(2x)+1\) give the same graph.
Show worked solution
Sine is odd: \(\sin(-u)=-\sin u\). Thus the negative argument changes the sine sign, and the outside negative sign changes it back. In contrast cosine is even, so \(-3\cos(-2x)+1=-3\cos(2x)+1\) keeps the vertical reflection. Reading a negative coefficient without considering the whole expression can misidentify a graph.
Build an equation
So far an equation has told us what a wave looks like. Now imagine that the graph is the observation and the formula is unknown. We will extract its midline, amplitude, period, and a convenient starting point, then check that the proposed equation reproduces those features.
From maximum and minimum
Show worked solution
A shifted starting point
Show worked solution
A positive sine curve starts at its midline rising. Other phase-equivalent equations can describe the same graph.
Practice and interpretation
Find the features
Find the amplitude, period, phase shift, midline, and range.
\(y=4\sin(2x-\pi)+3\)Show worked solution
Recover a period
Show worked solution
Measure horizontally between matching points in consecutive cycles, not between a peak and a trough.
A cycle in context
Show worked solution
Quarter-period step
Find the period and the horizontal spacing between successive quarter-cycle key points.
\(y=2\sin(4x)\)Show worked solution
Use this spacing between the five key points.
Additional practice: mixed exercises
Read graph features
For each function give amplitude, period, a phase shift, vertical shift, and midline. Angles are in radians.
- \(y=-4\sin x\)
- \(y=2\cos(x/2)-3\)
- \(y=\sin(4x-\pi)\)
- \(y=8\cos(-5(x-\pi))-7\)
- Sketch \(y=3\cos x\) on \([-\pi,4\pi]\).
Show worked solution
- Amplitude \(4\); period \(2\pi\); phase shift \(0\); vertical shift \(0\); midline \(y=0\). The negative coefficient reflects the sine graph.
- Amplitude \(2\); period \(2\pi/(1/2)=4\pi\); phase shift \(0\); vertical shift \(-3\); midline \(y=-3\).
- Write \(4x-\pi=4(x-\pi/4)\). Amplitude \(1\); period \(\pi/2\); shift right \(\pi/4\); vertical shift \(0\); midline \(y=0\).
- Cosine is even. Amplitude \(8\); period \(2\pi/5\); a phase shift is right \(\pi\); vertical shift \(-7\); midline \(y=-7\). Shifts differing by whole periods describe the same graph.
- One cycle passes through \((0,3),(\pi/2,0),(\pi,-3),(3\pi/2,0),(2\pi,3)\). Repeat with period \(2\pi\) to cover the interval.
More shifts and periodic patterns
Give amplitude, period, phase shift, vertical shift, and midline for each sine or cosine function. Angles are in radians.
- \(y=5\sin(-4x+8\pi)+6\)
- \(y=2\sin(-3x+9\pi)+4\)
- \(y=-7\cos(4x+8\pi)-9\)
- \(y=3\cos(x+5)-6\)
- \(y=-7\sin(3(x+2))+8\)
- A repeating triangular wave crosses upward through \((0,0)\), reaches \((2,2)\), then \((4,-2)\), and crosses upward again at \((5,0)\). Its next peak is \((7,2)\). Find its period.
- A square wave alternates between \(y=3\) and \(y=-3\). It is high on \([2,4)\) and \([6,8)\), and low on the intervening intervals. Find its period and amplitude.
Show worked solution
- Factor the argument as \(-4(x-2\pi)\). Amplitude \(5\); period \(2\pi/4=\pi/2\); a shift right by \(2\pi\); vertical shift \(6\); midline \(y=6\). The negative input coefficient reverses the horizontal orientation.
- The argument is \(-3(x-3\pi)\). Amplitude \(2\); period \(2\pi/3\); a shift right by \(3\pi\); vertical shift \(4\); midline \(y=4\).
- The argument is \(4(x+2\pi)\). Amplitude \(7\); period \(\pi/2\); a shift left by \(2\pi\); vertical shift \(-9\); midline \(y=-9\).
- Amplitude \(3\); period \(2\pi\); shift left by \(5\); vertical shift \(-6\); midline \(y=-6\).
- Amplitude \(7\); period \(2\pi/3\); shift left by \(2\); vertical shift \(8\); midline \(y=8\). The negative outside factor reflects outputs about the midline.
- Consecutive matching peaks occur at \(x=2\) and \(x=7\), so the period is \(7-2=5\). Matching upward zero crossings confirm the same difference.
- The high interval repeats after \(4\) units, so \(T=4\). Amplitude is half the peak-to-trough difference: \((3+3)/2=3\). Its midline is \(y=0\).
Takeaways and connections
- The midline and amplitude determine vertical extent.
- The period and phase shift determine horizontal placement.
- A sketch needs a smooth repeated cycle, not straight line segments.
Reasoning, connections and deeper practice
A model must match the timing of a motion as well as its maximum and minimum. Similar-looking equations can represent the same wave, and a few observations may leave more than one possible period.
Modeling · Match a wheel to its starting motion
A wheel seat ranges from \(1\) to \(9\) m above ground and completes one revolution in \(20\) seconds. At \(t=0\) it crosses the midline upward. Give a sinusoidal height model, its first maximum time, and its height at \(t=15\).
Hint
Find the average and half-difference of the extreme heights. A positive sine starts by rising from its midline.
Show worked solution
The midline is \(5\), amplitude \(4\), and angular frequency \(2\pi/20=\pi/10\). Thus \(h(t)=5+4\sin(\pi t/10)\). The first maximum occurs after a quarter-period, at \(t=5\) seconds, with height \(9\) m. At \(t=15\), sine is \(-1\), giving height \(1\) m. A cosine without a phase adjustment would start at an extreme and fail the given initial condition.
Advanced / Honors · Sampling does not identify a period
A sensor records a wave only at integer times \(t=n\). Prove that \(f(t)=\sin(2\pi t)\) and \(g(t)=\sin(4\pi t)\) give identical recorded samples but are not the same function. State their fundamental periods and give a time that distinguishes them.
Hint
Evaluate both at an integer and then at a noninteger such as one quarter.
Show worked solution
For every integer \(n\), both sines are zero. Nevertheless, \(f(1/4)=1\) and \(g(1/4)=0\). Their fundamental periods are \(1\) and \(1/2\), respectively. The equality of sampled values does not establish equality of continuous graphs or determine their periods. More closely spaced measurements or additional modeling assumptions are needed.
