Why Analyze One Period?
A trigonometric function often repeats the same shape again and again. Instead of analyzing its graph over the entire real line at once, we can study one complete period, record its important features, and repeat that pattern. Critical points tell us where a smooth curve may turn, the first derivative tells us whether it rises or falls, and the second derivative tells us how it bends.
The goal is not to produce a graph by plotting many points. A sketch should be supported by exact values and a consistent account of the function’s behavior between them. For a periodic function, this information from one interval can describe the graph everywhere.
If \(T\) is a period and the function is defined on all real numbers, values and graph features repeat after shifts by \(T\). The interval \([a,a+T]\) is therefore a useful window for a complete cycle. Its two endpoints represent the same point in the repeating pattern, so they should not be mistaken for two different locations in the extended graph.
A Reliable Analysis for One Period
For a differentiable periodic function, find the critical points within one chosen period. A critical point is an interior domain point where the derivative is zero or does not exist. The trigonometric functions and combinations used below are differentiable everywhere, so their critical points are found by solving \(f'(x)=0\). Then check the sign of \(f'\) on the intervals between those points: a positive sign means the function is increasing, and a negative sign means it is decreasing.
Next, calculate \(f''\). Where \(f''(x)>0\), the graph is concave up; where \(f''(x)<0\), it is concave down. A candidate inflection point must be a point on the graph where concavity actually changes. Thus, a zero of \(f''\) is a candidate, not by itself a guarantee of an inflection point.
Find a positive \(T\) for which \(f(x+T)=f(x)\), and select an interval of length \(T\).
Evaluate the function at the interval endpoints and at useful points such as intercepts or familiar angles.
Find critical points, test the derivative signs between them, and classify turning points by the sign changes.
Find where the second derivative is zero or undefined, then check whether its sign changes.
Connect the recorded features consistently, then translate the pattern by integer multiples of the period.
Why the Analysis Repeats
Periodicity of the function also implies periodicity of its derivatives whenever those derivatives exist. This fact justifies repeating the derivative information along with the function values.
To prove the first statement, start with the defining equality \(f(x+T)=f(x)\), which holds for all \(x\). Differentiate both sides with respect to \(x\). The left side has derivative \(f'(x+T)\), by the chain rule, and the right side has derivative \(f'(x)\). Therefore \(f'(x+T)=f'(x)\), as required. If \(f\) is twice differentiable, differentiate that equality once more to obtain \(f''(x+T)=f''(x)\). Hence the signs of the derivatives, the critical-point pattern, and the concavity pattern repeat as well.
Worked Example: Sketching One Period of the Sine Function
Worked Example: Sketching One Period of the Sine Function
Analyze \(f(x)=\sin x\) on \([0,2\pi]\), then describe how to extend the graph.
The sine function has period \(2\pi\), since \(\sin(x+2\pi)=\sin x\). Its endpoint values and familiar central values are
Differentiate:
On \((0,\pi/2)\), \(\cos x>0\), so \(f\) increases. On \((\pi/2,3\pi/2)\), \(\cos x<0\), so \(f\) decreases. On \((3\pi/2,2\pi)\), \(\cos x>0\), so \(f\) increases again. The interior critical points are \(x=\pi/2\) and \(x=3\pi/2\), where the derivative changes from positive to negative and from negative to positive, respectively. Thus \(f(\pi/2)=1\) is a local maximum, and \(f(3\pi/2)=-1\) is a local minimum.
The second derivative is
On \((0,\pi)\), \(\sin x>0\), so \(f''(x)<0\) and the graph is concave down. On \((\pi,2\pi)\), \(\sin x<0\), so \(f''(x)>0\) and the graph is concave up. Concavity changes at \(x=\pi\). On the extended graph it also changes at every integer multiple of \(\pi\), including the period endpoints \(0\) and \(2\pi\).
The resulting cycle passes through \((0,0)\), rises to \((\pi/2,1)\), falls through \((\pi,0)\), reaches \((3\pi/2,-1)\), and returns to \((2\pi,0)\). As a check, the derivative \(\cos x\) is positive, negative, and positive on the three intervals just listed, matching the graph’s rise, fall, and rise. The entire cycle repeats after each shift by \(2\pi\).
Worked Example: A Shorter Period and Its Turning Points
Worked Example: A Shorter Period and Its Turning Points
Sketch \(g(x)=\cos(2x)\) on \([0,\pi]\), and extend the pattern.
The angle \(2x\) increases by \(2\pi\) when \(x\) increases by \(\pi\), so \(\cos(2x)\) has period \(\pi\). At key points,
Applying the chain rule gives
On \((0,\pi/2)\), \(\sin(2x)>0\), so \(g'(x)<0\) and \(g\) decreases. On \((\pi/2,\pi)\), \(\sin(2x)<0\), so \(g'(x)>0\) and \(g\) increases. The interior critical point \(x=\pi/2\) is a local minimum, with value \(-1\). The endpoints both have value \(1\), and match when the period is repeated.
Differentiate again:
On \((0,\pi/4)\), \(\cos(2x)>0\), so the graph is concave down. On \((\pi/4,3\pi/4)\), \(\cos(2x)<0\), so it is concave up. On \((3\pi/4,\pi)\), \(\cos(2x)>0\), so it is concave down. The concavity changes at \(x=\pi/4\) and \(x=3\pi/4\), the two inflection points in this period.
Thus the curve begins at \((0,1)\), descends through \((\pi/4,0)\) to \((\pi/2,-1)\), then rises through \((3\pi/4,0)\) to \((\pi,1)\). As a check, differentiating \(-2\sin(2x)\) gives \(-4\cos(2x)\), agreeing with the second derivative used for the concavity analysis. The same cycle repeats every \(\pi\).
Worked Example: Combining Sine and Cosine
Worked Example: Combining Sine and Cosine
Analyze \(h(x)=\sin x+\cos x\) on \([0,2\pi]\). This example shows how derivative information can organize a curve even when it is not just a single sine or cosine function.
Since each term repeats after \(2\pi\), their sum has period \(2\pi\). The derivative is
To find the critical points, solve \(\cos x-\sin x=0\), or \(\cos x=\sin x\). On \([0,2\pi]\), this gives \(x=\pi/4\) and \(x=5\pi/4\). The derivative is positive at \(x=0\), where \(h'(0)=1\), and negative at \(x=\pi/2\), where \(h'(\pi/2)=-1\). It is positive at \(x=3\pi/2\), where \(h'(3\pi/2)=1\). Therefore \(h\) increases on \((0,\pi/4)\), decreases on \((\pi/4,5\pi/4)\), and increases on \((5\pi/4,2\pi)\).
The values at the critical points are
So these are a local maximum and local minimum, respectively. The endpoint values agree: \(h(0)=1\) and \(h(2\pi)=1\). The second derivative is
It is negative at \(x=0\), positive at \(x=\pi\), and negative at \(x=2\pi\). Solving \(h''(x)=0\) gives \(\sin x+\cos x=0\), which in this period occurs at \(x=3\pi/4\) and \(x=7\pi/4\). The sign changes at both points: the curve is concave down on \((0,3\pi/4)\), concave up on \((3\pi/4,7\pi/4)\), and concave down on \((7\pi/4,2\pi)\). These are inflection points. In fact, \(h\) also equals zero at those two points, since the sine and cosine values cancel there.
The sketch therefore starts at height \(1\), rises to \(\sqrt{2}\), falls to \(-\sqrt{2}\), and rises back to height \(1\), with concavity changing at the two zeros. As a check, the second derivative follows by differentiating \(h'(x)=\cos x-\sin x\), giving \(-\sin x-\cos x\), as used above. The curve repeats after a shift of \(2\pi\).
A Common Pitfall: Forgetting the Chain Rule
When differentiating a trigonometric function whose input is not simply \(x\), the derivative of the inside expression matters. For example, it is incorrect to write
This omits the derivative of \(2x\). The chain rule gives
The omitted factor changes the derivative’s values, even though in this example it does not change its zeros or signs. A sketch should be based on the correct derivatives, not on a formula that happens to preserve some of the qualitative information. Differentiating again also requires the chain rule: the correct second derivative is \(-4\cos(2x)\), not \(-\cos(2x)\).
Putting the Sketch Together
For each period, mark the endpoints, critical points, and any useful intercepts. Label the function values at those points. Then use the first derivative to decide which way the curve moves between marked points and the second derivative to decide how it bends. These pieces of information should agree: a curve that is increasing cannot be drawn as falling, and a curve that changes concavity should not be drawn with one unchanging bend across that point.
Finally, extend the finished period by translating it horizontally by \(T\), \(2T\), and other integer multiples of the period. The result repeats the function’s values and shape. Use a period that is convenient, and, when possible, the fundamental period to avoid analyzing the same cycle more than once.
Check Your Understanding
Use exact values and derivative signs when answering.
- What does it mean for \(T>0\) to be a period of a function?
- For \(g(x)=\cos(2x)\), what is a period, and what is its fundamental period?
- Why is solving \(f''(x)=0\) not sufficient by itself to establish an inflection point?
- What are the first and second derivatives of \(\sin(3x)\)?
- If a differentiable function has period \(T\), what equality relates its derivative values at \(x\) and \(x+T\)?
