Introduction
Sine and cosine repeat their values in regular cycles, so their graphs alternate between bending upward and bending downward. To describe that behavior precisely across any number of periods, we can use the signs of their second derivatives.
The main task is not simply to find where a second derivative is zero. We must also determine its sign on the intervals between those zeros. A zero of the second derivative is an inflection point only when the concavity actually changes there.
Concavity and the Second Derivative
For a twice-differentiable function on an interval, a positive second derivative means the function is concave up there; a negative second derivative means it is concave down. Intuitively, concave up means the slopes increase as \(x\) increases, while concave down means they decrease.
An inflection point is a point on the graph where the function changes concavity. Thus, a point where \(f''=0\) is not automatically an inflection point: the sign of \(f''\) must change across it. Sine and cosine are defined and twice differentiable for every real \(x\), so their candidate inflection points can be checked by comparing signs on the intervals to either side.
Why the Sign of the Second Derivative Determines Concavity
Here is a justification of the criterion. Suppose \(f''>0\) throughout an interval. For any \(u<v\) in that interval, the Mean Value Theorem applied to \(f'\) gives a point \(c\) between \(u\) and \(v\) such that \(f'(v)-f'(u)=f''(c)(v-u)>0\). Therefore \(f'\) is increasing.
To connect increasing slopes to concavity, take any three points \(x<y<z\) in the interval. The Mean Value Theorem applied to \(f\) on \([x,y]\) and \([y,z]\) gives points \(\xi_1\in(x,y)\) and \(\xi_2\in(y,z)\) such that the two secant slopes are \(f'(\xi_1)\) and \(f'(\xi_2)\). Since \(\xi_1<\xi_2\) and \(f'\) is increasing, the first slope is less than the second:
Multiplying by the positive denominators and rearranging gives
This says the graph at \(y\) lies below the chord joining the graph at \(x\) and \(z\), which is the defining chord property of concavity up. When \(f''<0\), the same argument gives decreasing slopes and the reversed chord inequality, so the function is concave down.
The Second Derivatives of Sine and Cosine
Differentiate each function twice. The first derivatives are \((\sin x)'=\cos x\) and \((\cos x)'=-\sin x\). Differentiating once more gives:
For sine, the second derivative is zero at \(x=n\pi\), where \(n\) is any integer. Between consecutive multiples of \(\pi\), sine has a constant sign: it is positive on \((2k\pi,(2k+1)\pi)\) and negative on \(((2k-1)\pi,2k\pi)\), for any integer \(k\). Since the second derivative is \(-\sin x\), its sign is the opposite of the sign of sine.
For cosine, the second derivative is zero at \(x=\frac{\pi}{2}+n\pi\), for any integer \(n\). Cosine is positive on \((-\frac{\pi}{2}+2k\pi,\frac{\pi}{2}+2k\pi)\) and negative on \((\frac{\pi}{2}+2k\pi,\frac{3\pi}{2}+2k\pi)\). Again, the second derivative has the opposite sign.
Worked Example: Sine from \(-\pi\) to \(2\pi\)
Worked Example: Sine from \(-\pi\) to \(2\pi\)
Find the concavity intervals of \(f(x)=\sin x\) on \((-\pi,2\pi)\), and identify the inflection points in that interval.
First differentiate twice:
The candidate locations solve \(-\sin x=0\), or \(\sin x=0\). The values in the interval are \(x=0\) and \(x=\pi\). These divide the interval into \((-\pi,0)\), \((0,\pi)\), and \((\pi,2\pi)\).
On \((-\pi,0)\), sine is negative, so \(f''=-\sin x>0\): the function is concave up. On \((0,\pi)\), sine is positive, so \(f''<0\): it is concave down. On \((\pi,2\pi)\), sine is negative, so \(f''>0\): it is concave up.
The sign changes from positive to negative at \(0\), and from negative to positive at \(\pi\). Since \(f(0)=0\) and \(f(\pi)=0\), the inflection points in the interval are \((0,0)\) and \((\pi,0)\). Substitution checks their locations: \(f''(0)=-\sin 0=0\) and \(f''(\pi)=-\sin\pi=0\), with the sign changes just established.
Worked Example: Cosine from \(-\pi\) to \(2\pi\)
Worked Example: Cosine from \(-\pi\) to \(2\pi\)
Determine the concavity of \(g(x)=\cos x\) on \((-\pi,2\pi)\), and find its inflection points there.
The second derivative is
Solving \(g''(x)=0\) gives \(\cos x=0\), so the candidates in the interval are \(-\frac{\pi}{2}\), \(\frac{\pi}{2}\), and \(\frac{3\pi}{2}\). Test one value in each interval:
These test values show that \(g\) is concave up on \((-\pi,-\frac{\pi}{2})\) and \((\frac{\pi}{2},\frac{3\pi}{2})\), and concave down on \((-\frac{\pi}{2},\frac{\pi}{2})\) and \((\frac{3\pi}{2},2\pi)\).
The signs alternate at each candidate, so all three candidates are inflection points. Because cosine is zero at each one, the points are \((-\frac{\pi}{2},0)\), \((\frac{\pi}{2},0)\), and \((\frac{3\pi}{2},0)\). Substitution confirms that \(g''\) is zero at each candidate, and the test values confirm the required changes of sign.
Worked Example: Sine on an Interval Not Aligned with a Period
Worked Example: Sine on an Interval Not Aligned with a Period
Find the concavity intervals and inflection points of \(h(x)=\sin x\) on \((\frac{\pi}{4},\frac{9\pi}{4})\).
As before, \(h''(x)=-\sin x\). The zeros of this second derivative in the given interval are \(x=\pi\) and \(x=2\pi\). They split the interval into \((\frac{\pi}{4},\pi)\), \((\pi,2\pi)\), and \((2\pi,\frac{9\pi}{4})\).
Sine is positive on \((\frac{\pi}{4},\pi)\), so \(h''<0\) there and the function is concave down. Sine is negative on \((\pi,2\pi)\), so \(h''>0\) there and the function is concave up. On \((2\pi,\frac{9\pi}{4})\), sine is positive, so \(h''<0\) and the function is concave down.
The sign changes at both interior candidates. Since \(h(\pi)=0\) and \(h(2\pi)=0\), the inflection points are \((\pi,0)\) and \((2\pi,0)\). The endpoints of the chosen interval are not included in the requested interval; the concavity classification here concerns the open intervals between the endpoints and candidates.
Reading the Patterns Across All Periods
The integer-indexed formulas in the theorem let us classify concavity without listing periods one at a time. For sine, each interval between consecutive multiples of \(\pi\) has length \(\pi\), and the concavity alternates at each multiple. For cosine, the zeros of its second derivative are also spaced \(\pi\) apart, and its concavity alternates at each of those points.
At every listed candidate, the function itself is zero as well: \(\sin(n\pi)=0\), and \(\cos(\frac{\pi}{2}+n\pi)=0\). The sign changes on either side verify that all these candidates are inflection points, not merely points where the second derivative vanishes.
Intervals of concavity are usually stated as open intervals because the second derivative is zero at their separating endpoints. This does not prevent the endpoint from being an inflection point. It means that concavity is being described separately on either side of that point.
A Common Sign-Chart Mistake
A frequent error is to forget the minus sign in the second derivative of cosine. For example, on \((\frac{\pi}{2},\frac{3\pi}{2})\), cosine is negative. It is tempting to conclude that cosine is concave down there by reading the sign of the function itself. But the second derivative is \(-\cos x\), not \(\cos x\), so it is positive on that interval. The correct conclusion is concave up.
A related check prevents this mistake: use a test point directly in the second derivative. At \(x=\pi\), \(g''(\pi)=-\cos\pi=1>0\), confirming concavity up on the interval containing \(\pi\). Always evaluate the second derivative, rather than inferring concavity from whether the original function is above or below the horizontal axis.
Check Your Understanding
Use the second derivative and its sign on the intervals between candidates.
- On which intervals is \(\sin x\) concave up between \(0\) and \(3\pi\)?
- Find the inflection points of \(\sin x\) in \((-2\pi,2\pi)\).
- Determine the concavity of \(\cos x\) on \((\frac{\pi}{2},\frac{3\pi}{2})\).
- Find all inflection points of \(\cos x\) in \((-\pi,2\pi)\).
- Why is checking only where the second derivative is zero insufficient to establish an inflection point?
