What to Track in Trigonometric Curve Analysis
A trigonometric function’s familiar repeating shape does not remove the need for careful curve analysis. The interval matters: it determines which critical points, inflection points, and vertical asymptotes are relevant. A complete analysis combines the function’s domain with the signs of its first and second derivatives.
Unless stated otherwise, angles are measured in radians. Sine and cosine are defined for every real number. Tangent is defined only where cosine is nonzero, so its domain excludes \(x=\frac{\pi}{2}+k\pi\), where \(k\) is an integer.
A positive first derivative indicates that the function is increasing, and a negative first derivative indicates that it is decreasing. A positive second derivative indicates concavity up; a negative second derivative indicates concavity down. For inflection points, look for a change in concavity—not merely a zero or undefined second derivative.
Why Derivative Signs Reveal the Curve’s Shape
The connection between derivative signs and the graph follows from the Mean Value Theorem. If \(f\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \(f'(x)>0\) throughout \((a,b)\), then for some \(c\in(a,b)\),
Since \(b-a>0\), this implies \(f(b)>f(a)\). Thus a positive derivative throughout an interval guarantees that the function increases there; a negative derivative guarantees that it decreases. The second derivative applies the same reasoning to \(f'\): if \(f''>0\), then \(f'\) increases, which corresponds to the graph bending upward. If \(f''<0\), then \(f'\) decreases, corresponding to bending downward.
For trigonometric functions, periodicity helps organize the sign analysis. Sine and cosine repeat every \(2\pi\), while tangent repeats every \(\pi\). Derivatives inherit these repeating signs. On a new interval, however, identify the relevant zeros and domain breaks rather than assuming that a familiar sketch gives the answer.
Worked Example: Analyze Sine on an Interval
Worked Example: Analyze \(f(x)=2\sin x\) on \([-\pi,2\pi]\)
The domain is all real numbers, so the function is defined throughout the interval. Differentiate:
Critical numbers inside the interval occur where \(f'(x)=0\). Thus \(2\cos x=0\), or \(\cos x=0\). The solutions in \([-\pi,2\pi]\) are \(x=-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}\). Checking the sign of \(2\cos x\) on the resulting intervals gives:
| Interval | Sign of \(f'(x)\) | Behavior |
|---|---|---|
| \((-\pi,-\frac{\pi}{2})\) | Negative | Decreasing |
| \((-\frac{\pi}{2},\frac{\pi}{2})\) | Positive | Increasing |
| \((\frac{\pi}{2},\frac{3\pi}{2})\) | Negative | Decreasing |
| \((\frac{3\pi}{2},2\pi)\) | Positive | Increasing |
The derivative changes from negative to positive at \(-\frac{\pi}{2}\), so that point is a local minimum. It changes from positive to negative at \(\frac{\pi}{2}\), a local maximum, and from negative to positive at \(\frac{3\pi}{2}\), another local minimum. Their function values are:
For concavity, solve \(f''(x)=0\): \(-2\sin x=0\), so \(\sin x=0\). The interior zeros are \(x=0\) and \(x=\pi\). On \((-\pi,0)\), sine is negative, so \(-2\sin x>0\): the graph is concave up. On \((0,\pi)\), sine is positive, so \(-2\sin x<0\): it is concave down. On \((\pi,2\pi)\), sine is negative again, so the graph is concave up. Concavity changes at \(0\) and \(\pi\), giving inflection points \((0,0)\) and \((\pi,0)\).
At the interval endpoints, \(f(-\pi)=2\sin(-\pi)=0\) and \(f(2\pi)=2\sin(2\pi)=0\). Comparing these values with the interior extrema shows that the absolute maximum on the interval is \(2\), at \(x=\frac{\pi}{2}\), and the absolute minimum is \(-2\), at \(x=-\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\).
As a check, differentiating \(2\sin x\) gives \(2\cos x\), and differentiating again gives \(-2\sin x\), as used above. Substituting each critical number into \(f\) gives the stated extremum values; substituting \(0\) and \(\pi\) gives \(f(0)=f(\pi)=0\), the stated inflection-point heights.
Worked Example: A Change in Frequency
Worked Example: Analyze \(g(x)=\cos(2x)\) on \([0,\pi]\)
The function is defined throughout the interval. The chain rule gives:
Within the interval, \(g'(x)=0\) when \(\sin(2x)=0\). Since \(2x\) ranges from \(0\) to \(2\pi\), the solutions are \(x=0,\frac{\pi}{2},\pi\). The endpoints are included in the interval, and the only interior critical number is \(x=\frac{\pi}{2}\). On \((0,\frac{\pi}{2})\), \(2x\) lies in \((0,\pi)\), so \(\sin(2x)>0\) and \(g'(x)<0\). On \((\frac{\pi}{2},\pi)\), \(2x\) lies in \((\pi,2\pi)\), so \(\sin(2x)<0\) and \(g'(x)>0\). Therefore \(g\) decreases and then increases.
The second derivative is zero when \(\cos(2x)=0\). The solutions in the interval are \(x=\frac{\pi}{4}\) and \(x=\frac{3\pi}{4}\). The sign of \(g''\) is negative on \((0,\frac{\pi}{4})\), positive on \((\frac{\pi}{4},\frac{3\pi}{4})\), and negative on \((\frac{3\pi}{4},\pi)\). Thus the curve is concave down, then concave up, then concave down. The concavity changes at both candidates, and
The inflection points are \((\frac{\pi}{4},0)\) and \((\frac{3\pi}{4},0)\). At the endpoints, \(g(0)=\cos(0)=1\) and \(g(\pi)=\cos(2\pi)=1\). At the interior critical number, \(g(\frac{\pi}{2})=\cos(\pi)=-1\). Therefore the absolute maximum on this interval is \(1\) at both endpoints, and the absolute minimum is \(-1\) at \(x=\frac{\pi}{2}\).
To check the chain-rule factors, differentiating \(\cos(2x)\) gives \(-\sin(2x)\cdot 2=-2\sin(2x)\); differentiating that gives \(-2\cos(2x)\cdot 2=-4\cos(2x)\). The critical value and both inflection-point values follow by direct substitution as shown.
Tangent: Domain Breaks and Vertical Asymptotes
Tangent requires special attention because it is undefined where \(\cos x=0\). Its vertical asymptotes occur at \(x=\frac{\pi}{2}+k\pi\), for integers \(k\). These points divide its domain into separate intervals. Analyze each branch on its own; an excluded asymptote is not a point on the graph and cannot be a critical point or an extremum.
Worked Example: Analyze One Branch of Tangent
Worked Example: Analyze \(h(x)=\tan x\) on \((-\frac{\pi}{2},\frac{\pi}{2})\)
The endpoints are excluded because \(\tan x\) is undefined there. On the open interval, \(\cos x\ne 0\), so:
The derivative is positive everywhere on the branch, so tangent is increasing throughout. It has no critical numbers there: its derivative never equals zero, and there are no points in the interval where its derivative is undefined.
Differentiating again gives
Since \(\sec^2 x>0\), the sign of \(h''(x)\) is the sign of \(\tan x\). Tangent is negative on \((-\frac{\pi}{2},0)\) and positive on \((0,\frac{\pi}{2})\). Thus the graph is concave down on the first interval and concave up on the second. The concavity changes at \(x=0\), where \(h(0)=0\), so \((0,0)\) is an inflection point.
To find the behavior near the ends, use \(\tan x=\frac{\sin x}{\cos x}\). As \(x\) approaches \(-\frac{\pi}{2}\) from the right, sine approaches \(-1\) and cosine approaches \(0\) through positive values. Therefore \(\tan x\) tends to \(-\infty\). As \(x\) approaches \(\frac{\pi}{2}\) from the left, sine approaches \(1\) and cosine approaches \(0\) through positive values, so \(\tan x\) tends to \(+\infty\):
The branch therefore has vertical asymptotes \(x=-\frac{\pi}{2}\) and \(x=\frac{\pi}{2}\). As a check, differentiating \(\tan x\) gives \(\sec^2 x\), and differentiating again gives \(2\sec^2 x\tan x\). The second derivative is zero at \(x=0\), and its sign changes from negative to positive there, confirming the inflection point.
A Common Mistake: Treating an Excluded Point as Critical
A critical number must be a number in the function’s domain where the derivative is zero or does not exist. For example, a tempting but incorrect argument for tangent is:
The conclusion is wrong because \(\tan(\frac{\pi}{2})\) is itself undefined. That point is not in the function’s domain, so it cannot be a critical number or an extremum. It is a vertical asymptote. Analyze the branch to the left and the branch to the right separately.
A related caution applies to inflection points: solving \(f''(x)=0\) identifies candidates, not automatically inflection points. Check whether concavity actually changes across each candidate, and confirm that the function is defined there.
Check Your Understanding
For each question, use the specified interval and justify conclusions with derivative signs, concavity, or one-sided behavior as appropriate.
- For \(f(x)=\sin x\) on \([0,2\pi]\), where is the function increasing, decreasing, and concave up?
- For \(g(x)=\cos x\) on \([-\pi,\pi]\), identify the interior critical numbers and classify their local extrema.
- Why is \(x=\frac{\pi}{2}\) not a critical number of \(\tan x\)?
- For \(h(x)=\tan x\) on \((0,\frac{\pi}{2})\), determine its concavity and its limit as \(x\) approaches \(\frac{\pi}{2}\) from the left.
- What must be checked before a zero of a trigonometric function’s second derivative can be called an inflection point?
