Cone triangles satisfy TR2 with the declared rotation sign

Write the standard cone triangle as $C\stackrel{f}{\to }D\stackrel{j}{\to }\mathrm{Cone}(f)\stackrel{q}{\to }C[1]\mathrm{.}$ A direct cone calculation for $j$ splits $\mathrm{Cone}(j)$ as $C[1]$ together with the contractible cone of $1_D$. Projection onto $C[1]$ is therefore a homotopy equivalence.

Under that projection the standard triangle of $j$ becomes $D\stackrel{j}{\to }\mathrm{Cone}(f)\stackrel{q}{\to }C[1]\stackrel{-f[1]}{\to }D[1];$ the minus sign is the one forced by the cone differential. Thus the left rotation of every standard cone triangle is distinguished.

The converse uses the right rotation, not merely the inverse rotation of the particular triangle in step 2.1. A second direct cone calculation splits $\mathrm{Cone}(-q[-1])\simeq D\oplus \mathrm{Cone}(1_C)$ compatibly with the canonical maps. Since the identity cone is contractible, the standard cone triangle of $-q[-1]$ is therefore isomorphic in $K(\mathcal{A})$ to $\mathrm{Cone}(f)[-1]\stackrel{-q[-1]}{\to }C\stackrel{f}{\to }D\stackrel{j}{\to }\mathrm{Cone}(f),$ the declared right rotation of the original standard triangle.

Every distinguished cone triangle is isomorphic to a standard one, and left and right rotation preserve isomorphisms of triangles. Steps 2.1 and 3.1 therefore prove both directions of the TR2 equivalence for an arbitrary triangle.