Find the inflection points of the curve y = sin x
Find the inflection points of the curve $y = \sin x$.
Explanation of Method
Every zero of the sine is an inflection of its graph.
Step 1: The second derivative
\[ y' = \cos x, \qquad y'' = -\sin x. \]
Step 2: Candidates
\[ y'' = 0 \;\Longleftrightarrow\; \sin x = 0 \;\Longleftrightarrow\; x = n\pi, \qquad n\in\mathbb{Z}. \]
Step 3: Sign of $y''$
On each interval between consecutive multiples of $\pi$ the sine keeps a constant sign, alternating from one interval to the next: \[y'' = -\sin x\lt 0 \ \text{ on } \bigl(2k\pi,\,\bigl(2k+1\bigr)\pi\bigr), \qquad y''\gt 0 \ \text{ on } \bigl(\bigl(2k-1\bigr)\pi,\,2k\pi\bigr).\]
So the concavity reverses at every point $x = n\pi$, and each is a genuine inflection.
Step 4: Verification
\[y''\!\left(\tfrac{\pi}{2}\right) = -1\lt 0; \qquad y''\!\left(\tfrac{3\pi}{2}\right) = +1\gt 0 \;\;\checkmark\]
\[y'\bigl(n\pi\bigr) = \cos n\pi = \pm1\neq0:\ \text{the tangents there are oblique, of slope } \pm1 \;\;\checkmark\]
\[\text{Each inflection sits on the } x\text{-axis, since } \sin n\pi = 0 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\bigl(n\pi,\;0\bigr), \qquad n\in\mathbb{Z}\,}\]
Remark
Remark. For the sine the inflections coincide with the zeros, because $y'' = -y$: the curve is concave exactly where it is above the axis and convex where it is below. That identity — the differential equation $y''+y = 0$ — is what makes the picture so symmetric.