Find the stationary points of y = sin³x. Determine which
Find the stationary points of $y = \sin^{3}x$. Determine which of them are extrema, and of what type.
Explanation of Method
Four stationary points per period, but only two are extrema.
Step 1: The derivative
\[ y' = 3\sin^{2}x\cos x. \tag{1} \]
Step 2: Stationary points
A product vanishes when one factor does: \[\sin x = 0 \;\Longrightarrow\; x = k\pi; \qquad \cos x = 0 \;\Longrightarrow\; x = \frac{\pi}{2}+k\pi.\]
Within one period $\bigl[0,2\pi\bigr]$ these are $x = 0,\ \dfrac{\pi}{2},\ \pi,\ \dfrac{3\pi}{2}$.
Step 3: Sign of the derivative
Since $3\sin^{2}x\ge0$, the sign of $y'$ is that of $\cos x$ wherever $\sin x\neq0$: \[\left(0,\tfrac{\pi}{2}\right):\ +\ (\nearrow); \qquad \left(\tfrac{\pi}{2},\pi\right):\ -\ (\searrow); \qquad \left(\pi,\tfrac{3\pi}{2}\right):\ -\ (\searrow); \qquad \left(\tfrac{3\pi}{2},2\pi\right):\ +\ (\nearrow).\]
Note that the two middle intervals carry the same sign: at $x = \pi$ the factor $\sin^{2}x$ touches zero without changing sign, and $\cos x$ is negative on both sides.
Step 4: The points where $\cos x = 0$
At $x = \dfrac{\pi}{2}+2k\pi$ the derivative passes from $+$ to $-$: a local maximum, with $y = \sin^{3}\dfrac{\pi}{2} = 1$.
At $x = \dfrac{3\pi}{2}+2k\pi = -\dfrac{\pi}{2}+2k\pi$ it passes from $-$ to $+$: a local minimum, with $y = -1$.
Step 5: The points where $\sin x = 0$
At $x = k\pi$ we have $y' = 0$, but the sign does not change: near $x = 0$ the derivative is positive on both sides, near $x = \pi$ negative on both sides. These are therefore not extrema but stationary points of inflection, with $y\bigl(k\pi\bigr) = 0$.
The second derivative confirms it: $y'' = 6\sin x\cos^{2}x-3\sin^{3}x = 3\sin x\bigl(2\cos^{2}x-\sin^{2}x\bigr)$, which vanishes at $x = k\pi$ and changes sign there, since the factor $\sin x$ does.
Step 6: Verification
\[y'''(0) = 6\neq0,\ \text{so } x = 0 \text{ is an inflection with a horizontal tangent} \;\;\checkmark\]
\[\sin^{3} \text{ is odd and } 2\pi\text{-periodic, so } \pm1 \text{ are the extreme values, attained where } \sin x = \pm1 \;\;\checkmark\]
\[\text{Near } x = 0:\ \sin^{3}x = x^{3}-\frac{x^{5}}{2}+\cdots,\ \text{the cubic shape of an inflection} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\text{maxima at } x = \frac{\pi}{2}+2k\pi:\ y = 1; \qquad \text{minima at } x = -\frac{\pi}{2}+2k\pi:\ y = -1\,}\]
Remark
Remark. The cube is what creates the inflections: $\sin x$ itself crosses zero with slope $\pm1$ and has no stationary point there, but raising to an odd power flattens the crossing into a horizontal one without destroying the change of concavity. An even power would behave quite differently — $\sin^{2}x$ has minima at $x = k\pi$.