Introduction
Sine and cosine repeat their values, so their critical numbers repeat as well. On a specified interval, the main tasks are to identify which of these periodic critical numbers lie in the interval and then decide whether each one is a local maximum, a local minimum, or neither.
The derivative sign gives the classification: where a function changes from increasing to decreasing, it has a local maximum; where it changes from decreasing to increasing, it has a local minimum. Trigonometric functions provide a clear setting for using this test because their derivatives and periods are familiar, and their critical numbers can be described exactly.
Critical Numbers and Local Extrema
A number \(c\) in the domain of a function \(f\) is a critical number if \(f'(c)=0\) or \(f'(c)\) does not exist. The corresponding point on the graph is \((c,f(c))\). Since sine, cosine, and their sums and constant multiples are differentiable everywhere, their critical numbers are precisely the numbers where their derivatives are zero.
An interior point \(c\) is a local maximum if \(f(c)\geq f(x)\) for all domain points \(x\) sufficiently close to \(c\). It is a local minimum if \(f(c)\leq f(x)\) for all such \(x\). If the inequalities are strict for nearby \(x\ne c\), the extremum is strict.
A critical number is only a candidate: derivative zero does not by itself guarantee an extremum. The First Derivative Test classifies a candidate by comparing the signs of \(f'\) immediately to its left and right.
| Derivative sign to the left | Derivative sign to the right | Classification |
|---|---|---|
| Positive | Negative | Local maximum |
| Negative | Positive | Local minimum |
| Same sign on both sides | Same sign on both sides | No local extremum from this sign test |
The reason for the test is that a positive derivative means the function is increasing, while a negative derivative means it is decreasing. Thus a positive-to-negative change means the function rises toward the point and falls afterward. A negative-to-positive change means it falls toward the point and rises afterward. These monotonicity conclusions apply on the intervals where the derivative has the indicated sign.
Local Extrema of Sine
For \(f(x)=\sin x\), the derivative is \(f'(x)=\cos x\). Therefore, the critical numbers satisfy \(\cos x=0\), which gives
This formula gives all the critical numbers on the real line. On a specified interval, retain only the values that lie inside it. The integer \(k\) is not restricted to positive values; negative integers are needed when the interval includes negative angles.
At \(x=\frac{\pi}{2}+k\pi\), the sine value alternates between \(1\) and \(-1\). When \(k\) is even, the value is \(1\), giving a local maximum; when \(k\) is odd, the value is \(-1\), giving a local minimum. Equivalently, the derivative \(\cos x\) changes from positive to negative at the maxima and from negative to positive at the minima.
Worked Example: Classifying Sine Extrema on an Interval
Find and classify the interior local extrema of \(f(x)=\sin x\) on \([-\pi,2\pi]\).
First differentiate and set the derivative equal to zero:
Now keep only the values in \([-\pi,2\pi]\). For \(k=-1\), \(x=-\frac{\pi}{2}\); for \(k=0\), \(x=\frac{\pi}{2}\); and for \(k=1\), \(x=\frac{3\pi}{2}\). The next values, for \(k=-2\) and \(k=2\), are \(-\frac{3\pi}{2}\) and \(\frac{5\pi}{2}\), both outside the interval.
At \(x=-\frac{\pi}{2}\), the derivative is negative immediately to the left and positive immediately to the right, so this is a local minimum. Its value is \(\sin(-\frac{\pi}{2})=-1\).
At \(x=\frac{\pi}{2}\), the derivative changes from positive to negative, so this is a local maximum. Its value is \(\sin(\frac{\pi}{2})=1\).
At \(x=\frac{3\pi}{2}\), the derivative changes from negative to positive, so this is a local minimum. Its value is \(\sin(\frac{3\pi}{2})=-1\).
Check the classifications using the sine values: the listed points have values \(-1,1,-1\), respectively, as required for the alternating minima and maxima of the sine curve.
Local Extrema of Cosine
For \(f(x)=\cos x\), the derivative is \(f'(x)=-\sin x\). The critical-number equation is \(-\sin x=0\), so
At an even multiple of \(\pi\), \(\cos(k\pi)=1\), so the point is a local maximum. At an odd multiple, \(\cos(k\pi)=-1\), so it is a local minimum. The derivative signs confirm this: \(-\sin x\) changes from positive to negative at even multiples and from negative to positive at odd multiples.
Worked Example: Classifying Cosine Extrema on an Interval
Find and classify the interior local extrema of \(g(x)=\cos x\) on \([-\frac{\pi}{2},\frac{5\pi}{2}]\).
Differentiate and solve:
The critical numbers strictly inside the interval are \(0\), \(\pi\), and \(2\pi\). The values \(-\frac{\pi}{2}\) and \(\frac{5\pi}{2}\) are endpoints and are not included in this interior classification.
For \(x=0\), \(-\sin x\) is positive just to the left and negative just to the right. Thus \(x=0\) is a local maximum, with \(g(0)=\cos 0=1\).
For \(x=\pi\), \(-\sin x\) is negative just to the left and positive just to the right. Thus \(x=\pi\) is a local minimum, with \(g(\pi)=\cos\pi=-1\).
For \(x=2\pi\), \(-\sin x\) changes from positive to negative. Thus \(x=2\pi\) is a local maximum, with \(g(2\pi)=\cos(2\pi)=1\).
The classifications agree with the cosine values at the critical numbers: \(1,-1,1\).
Combining Sine and Cosine
A function of the form \(a\sin x+b\cos x\) can be handled by rewriting it as a shifted sine wave. This makes its critical numbers and extrema visible at once. Suppose \(a\) and \(b\) are not both zero, and define
Then \(R>0\). Choose an angle \(\phi\) such that \(\cos\phi=\frac{a}{R}\) and \(\sin\phi=\frac{b}{R}\). The sine addition identity gives
This is a complete conversion to a single sinusoid: its amplitude is \(R\), and its phase is shifted by \(\phi\). Differentiating the rewritten function gives \(R\cos(x+\phi)\). Since \(R>0\), the critical numbers solve \(\cos(x+\phi)=0\), or
At these points the function alternates between \(R\) and \(-R\). The derivative sign changes from positive to negative at a value \(R\), producing a local maximum, and from negative to positive at a value \(-R\), producing a local minimum. If \(a=b=0\), the function is constant instead, so every point is both a non-strict local maximum and a non-strict local minimum.
Worked Example: Classifying a Sine-Cosine Combination
Find and classify the interior local extrema of \(h(x)=3\sin x+4\cos x\) on \([-\pi,2\pi]\).
Here \(a=3\), \(b=4\), and
Choose \(\phi\) with \(\cos\phi=\frac35\), \(\sin\phi=\frac45\), and \(0<\phi<\frac{\pi}{2}\). Then
Set the derivative equal to zero:
Let \(x_0=\frac{\pi}{2}-\phi\). Since \(0<\phi<\frac{\pi}{2}\), we have \(0<x_0<\frac{\pi}{2}\). The critical numbers in \([-\pi,2\pi]\) are \(x_0-\pi\), \(x_0\), and \(x_0+\pi\): the first is greater than \(-\pi\), the third is less than \(\frac{3\pi}{2}<2\pi\), while the neighboring candidates \(x_0-2\pi\) and \(x_0+2\pi\) fall outside the interval.
At \(x_0-\pi\), the phase \(x+\phi\) equals \(-\frac{\pi}{2}\). The derivative \(5\cos(x+\phi)\) changes from negative to positive there, so this is a local minimum. Its function value is \(5\sin(-\frac{\pi}{2})=-5\).
At \(x_0\), the phase is \(\frac{\pi}{2}\). The derivative changes from positive to negative, so this is a local maximum, with value \(5\sin(\frac{\pi}{2})=5\).
At \(x_0+\pi\), the phase is \(\frac{3\pi}{2}\). The derivative changes from negative to positive, so this is a local minimum, with value \(5\sin(\frac{3\pi}{2})=-5\).
As a check, the original expression at each critical number must match these values. For example, at \(x_0\), the phase identity gives \(3\sin x_0+4\cos x_0=5\sin(x_0+\phi)=5\). At \(x_0-\pi\) and \(x_0+\pi\), it gives \(-5\) in each case.
A Common Mistake: Confusing Function Values with Critical Numbers
A function’s value at a point does not determine whether that point is critical. Critical numbers are found from the derivative, not by looking for familiar function values. For example, \(\sin 0=0\), but this does not make \(0\) a critical number of sine.
In fact, \(\cos x\) is positive in a neighborhood of \(0\), so sine is increasing there; \(x=0\) is neither a local maximum nor a local minimum. The correction is to solve \(f'(x)=0\) first, then use derivative signs to classify each candidate. A second common error is to list only one period’s critical numbers without checking whether more values from the periodic formula lie in the specified interval.
A Reliable Procedure
Write the derivative carefully, including any negative sign from differentiating cosine.
Use the periodic solution formula and retain every solution in the interval’s interior.
Determine whether the derivative changes from positive to negative, negative to positive, or keeps the same sign.
State the critical number, its classification, and the function value; check that the value agrees with the trig formula or phase-shift form.
Periodicity is useful because it supplies all possible candidates, but it does not remove the need to filter by the interval or check derivative signs. On an interval spanning several periods, there may be several alternating maxima and minima. On a shorter interval, some members of the periodic sequence may not be present at all.
These local classifications describe behavior near interior points. They do not by themselves determine the largest or smallest function value across an entire interval; that requires comparing candidates and, when appropriate, endpoints.
Check Your Understanding
For each question, identify the critical numbers in the stated interval before classifying them.
- Find and classify the interior local extrema of \(\sin x\) on \([0,2\pi]\).
- Find and classify the interior local extrema of \(\cos x\) on \([0,3\pi]\).
- Why is \(x=0\) not a critical number of \(f(x)=\sin x\), even though \(\sin 0=0\)?
- Rewrite \(5\sin x+12\cos x\) as a phase-shifted sine function and state its amplitude.
- For a sine-cosine combination with nonzero coefficients, what derivative-sign changes distinguish its local maxima from its local minima?
