Examine the sign of the second derivative of y = sin²x
Examine the sign of the second derivative of $y = \sin^{2}x$.
Explanation of Method
A double-angle identity makes the answer immediate.
Step 1: The derivatives
\[ y' = 2\sin x\cos x = \sin2x, \qquad y'' = 2\cos2x. \]
Step 2: Zeros
\[ y'' = 0 \;\Longleftrightarrow\; \cos2x = 0 \;\Longleftrightarrow\; 2x = \frac{\pi}{2}+n\pi \;\Longleftrightarrow\; x = \frac{\pi}{4}+\frac{n\pi}{2}. \]
Step 3: Sign
The cosine alternates sign on consecutive intervals of length $\dfrac{\pi}{2}$ in $x$: \[y''\gt 0 \ \text{ on } \left(-\frac{\pi}{4}+n\pi,\ \frac{\pi}{4}+n\pi\right), \qquad y''\lt 0 \ \text{ on } \left(\frac{\pi}{4}+n\pi,\ \frac{3\pi}{4}+n\pi\right).\]
Every zero is simple, so each is an inflection.
Step 4: The inflection points
At those points $\sin x = \pm\dfrac{\sqrt2}{2}$, hence
\[ y = \sin^{2}x = \frac12 \qquad\text{at every } x = \frac{\pi}{4}+\frac{n\pi}{2}. \]
Step 5: Verification
\[\sin^{2}x = \frac{1-\cos2x}{2} \;\Longrightarrow\; y'' = 2\cos2x \ \text{ directly} \;\;\checkmark\]
\[y''(0) = 2\gt 0 \ (\text{minimum at } x = 0); \qquad y''\!\left(\tfrac{\pi}{2}\right) = -2\lt 0 \ (\text{maximum}) \;\;\checkmark\]
\[\text{All inflections lie on the line } y = \tfrac12,\ \text{the mean value of } \sin^{2} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y'' = 2\cos2x; \qquad y''>0 \text{ on } \left(-\tfrac{\pi}{4}+n\pi,\tfrac{\pi}{4}+n\pi\right), \quad y''<0 \text{ on } \left(\tfrac{\pi}{4}+n\pi,\tfrac{3\pi}{4}+n\pi\right)\,}\]
Final answer:
\[\boxed{\,\text{inflections } \left(\frac{\pi}{4}+\frac{n\pi}{2},\ \frac12\right), \qquad n\in\mathbb{Z}\,}\]
Remark
Remark. The identity in the first verification line says the curve is a cosine of doubled frequency, shifted and halved; everything about it — extrema at multiples of $\tfrac{\pi}{2}$, inflections exactly halfway between, all at the height $\tfrac12$ — follows from that. The inflections are the points where the graph crosses its own average value.