Trigonometric Absolute Extrema on Closed Intervals

Why Periodicity Matters

Sine and cosine repeat their values, so an interval can contain several peaks and troughs. To find an absolute maximum or minimum on a closed interval, it is not enough to inspect the endpoints or to examine only one cycle. We must locate every interior critical number in the interval and compare the function values there with the endpoint values.

This tutorial uses radians. The derivative formulas are \((\sin x)'=\cos x\) and \((\cos x)'=-\sin x\). Since sine and cosine are continuous and differentiable for every real \(x\), their critical numbers occur where the derivative is zero.

The Candidate Method on a Closed Interval

For a continuous function on a closed interval \([a,b]\), the Extreme Value Theorem guarantees that the function attains both an absolute maximum and an absolute minimum. If the function is also differentiable in the interval's interior, every interior location of an absolute extremum must be a critical number. Consequently, the endpoints and the interior critical numbers form a complete candidate list.

Here is why the candidate list is complete. By the Extreme Value Theorem, an absolute maximum is attained at some point \(c\in[a,b]\). If \(c=a\) or \(c=b\), it is already an endpoint candidate. If \(a<c<b\), it is an interior local maximum. For small positive \(h\), the maximum property gives \(f(c+h)-f(c)\leq 0\), so the right-hand difference quotient is nonpositive. For small negative \(h\), the quotient \((f(c+h)-f(c))/h\) is nonnegative. If \(f'(c)\) exists, both one-sided limits equal that derivative, so \(f'(c)\leq 0\) and \(f'(c)\geq 0\). Thus \(f'(c)=0\). The same argument applies to an absolute minimum. Therefore, every absolute extremum is represented among the endpoints and interior critical numbers.

For trigonometric functions, the main additional task is solving the derivative equation correctly. The solutions usually repeat periodically. Write the full family of solutions first, then keep exactly those that lie strictly inside the interval. Endpoints are included separately, even if the derivative also vanishes there.

Sine and Cosine: Critical-Point Patterns

For \(f(x)=\sin x\), the derivative is \(f'(x)=\cos x\). Its critical numbers satisfy \(\cos x=0\), so

At these points, sine alternates between \(1\) and \(-1\). For \(g(x)=\cos x\), the derivative is \(g'(x)=-\sin x\), and its critical numbers satisfy

At even multiples of \(\pi\), cosine is \(1\); at odd multiples, it is \(-1\). These patterns identify where to look, but the interval still determines which solutions belong in the candidate list. The function values must then be compared with both endpoint values.

Simple Sine-Cosine Combinations

A common combination is \(f(x)=A\sin x+B\cos x+C\), where \(A\), \(B\), and \(C\) are constants and \(A\) and \(B\) are not both zero. Its derivative is

Thus its interior critical numbers solve \(A\cos x-B\sin x=0\). The equation is periodic, but its solutions need to be filtered through the given interval just as for sine and cosine alone. Avoid dividing by \(\cos x\) until checking whether \(\cos x=0\) could solve the original equation. Once the candidates are identified, compare the original function values, including the constant term \(C\).

For instance, when \(A\neq 0\), an equation \(A\cos x-B\sin x=0\) can be rewritten as \(\tan x=A/B\) only if \(B\neq 0\) and \(\cos x\neq 0\). The original derivative equation remains the reliable starting point, especially when one coefficient is zero.

A Reliable Workflow

A careful solution makes the interval restriction visible. In particular, listing a general periodic family is not the same as identifying which members are inside the interval.

Common Pitfall: Solving the Wrong Equation

Critical numbers come from setting the derivative equal to zero, not from setting the function equal to zero. For example, on \([0,\pi]\), an incorrect approach to \(f(x)=\sin x\) might solve \(\sin x=0\), find only the endpoints, and conclude that the largest value is \(0\). But those are zeros of the function, not its critical numbers.

The correction is to differentiate first: \(f'(x)=\cos x\). The interior critical equation is \(\cos x=0\), which gives \(x=\pi/2\) in \((0,\pi)\). The candidate values are \(\sin 0=0\), \(\sin(\pi/2)=1\), and \(\sin\pi=0\). Therefore the absolute maximum is \(1\) at \(x=\pi/2\), while the absolute minimum is \(0\) at both endpoints.

A related error is to identify a correct periodic family but test only one of its solutions. On a longer interval, several members may occur, and their function values may represent different extrema or repeated occurrences of the same extremum. Always restrict the full solution family to the interval before making the comparison.

Why the Comparison Is Essential

A critical point is only a candidate; the derivative equation alone does not tell you whether its value is the greatest or least on the whole interval. An endpoint can be more extreme than every interior critical value, and more than one candidate can attain the same absolute value. The conclusion comes from evaluating the original function at the complete candidate list, not merely from naming a point where the derivative vanishes.

For trigonometric functions, periodicity makes this final comparison especially important. A closed interval may contain several critical points, or only part of a cycle, and the endpoints need not fall at familiar angles. Keep the interval, the complete candidate list, and the evaluated function values together in the solution.