Proving sin x(1 + cos x) Reaches Its Maximum at x = π/3
Finding the maximum of a function involving both sin x and cos x requires a smart substitution. Once we expand and use the identity \(\sin^2 x = 1 - \cos^2 x\), the derivative becomes a quadratic in cos x — something we can factor and solve exactly.
The Problem 4
Show that \(f(x) = \sin x\,(1 + \cos x)\) is maximum when \(x = \dfrac{\pi}{3}\).
Step 1
Find the Critical Values
Let \(f(x) = \sin x\,(1 + \cos x)\). We expand and differentiate using the product rule:
$$f(x) = \sin x + \sin x \cos x$$
$$f'(x) = \cos x + \bigl[\cos x \cdot \cos x + \sin x \cdot (-\sin x)\bigr]$$
$$= \cos x + \cos^2 x - \sin^2 x$$
$$= \cos x + \cos^2 x - (1 - \cos^2 x) \qquad \bigl(\text{using } \sin^2 x = 1 - \cos^2 x\bigr)$$
$$f'(x) = 2\cos^2 x + \cos x - 1$$
Setting \(f'(x) = 0\) and factoring:
$$2\cos^2 x + \cos x - 1 = 0$$
$$2\cos^2 x + 2\cos x - \cos x - 1 = 0$$
$$2\cos x(\cos x + 1) - 1(\cos x + 1) = 0$$
$$(2\cos x - 1)(\cos x + 1) = 0$$
$$\implies \cos x = \frac{1}{2} \quad \text{or} \quad \cos x = -1$$
Choosing the principal value:
\(\cos x = -1\) gives \(x = \pi\), which is a boundary point.
For a maxima in the principal interval, we take
\(\cos x = \dfrac{1}{2} \implies x = \dfrac{\pi}{3}\).
Critical value: \(x = \dfrac{\pi}{3}\)
Step 2
Second Derivative Test — Confirm Maximum
We differentiate \(f'(x) = 2\cos^2 x + \cos x - 1\) again:
Second Derivative
$$f''(x) = 4\cos x \cdot (-\sin x) - \sin x$$
$$f''(x) = -\sin x\,(4\cos x + 1)$$
Evaluating at \(x = \dfrac{\pi}{3}\):
Evaluating f''(π/3)
$$f''\!\left(\frac{\pi}{3}\right) = -\sin\!\left(\frac{\pi}{3}\right)\!\left[4\cos\!\left(\frac{\pi}{3}\right) + 1\right]$$
$$= -\frac{\sqrt{3}}{2}\left[4 \cdot \frac{1}{2} + 1\right]$$
$$= -\frac{\sqrt{3}}{2}\,[2 + 1]$$
$$= -\frac{\sqrt{3}}{2} \times 3 = -\frac{3\sqrt{3}}{2}$$
\(f''\!\left(\dfrac{\pi}{3}\right) = -\dfrac{3\sqrt{3}}{2} < 0\)
→ \(x = \dfrac{\pi}{3}\) is a Maximum Point
Step 3
Compute the Maximum Value
Substituting \(x = \dfrac{\pi}{3}\) into the original function:
Maximum Value at x = π/3
$$f\!\left(\frac{\pi}{3}\right) = \sin\!\left(\frac{\pi}{3}\right)\!\left[1 + \cos\!\left(\frac{\pi}{3}\right)\right]$$
$$= \frac{\sqrt{3}}{2}\left[1 + \frac{1}{2}\right]$$
$$= \frac{\sqrt{3}}{2} \times \frac{3}{2}$$
$$f\!\left(\frac{\pi}{3}\right) = \frac{3\sqrt{3}}{4}$$
Summary at a Glance
\(\dfrac{\pi}{3}\)
Critical point x
\(-\dfrac{3\sqrt{3}}{2}\)
f''(π/3) — negative ✓
\(\dfrac{3\sqrt{3}}{4}\)
Maximum value
f'(π/3) = 0 && f''(π/3) < 0 → Confirmed maximum
∴ sin x (1 + cos x) attains its maximum value of \(\dfrac{3\sqrt{3}}{4}\) at \(x = \dfrac{\pi}{3}\)
Tutorial Video: Maxima & Minima — Calculus
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