Canonical truncations fit a distinguished triangle

Define $e:\mathrm{Cone}(i)\to C$ by $e(b,a)=q(b)$. It is a termwise epimorphic complex map. Its kernel identifies with $\mathrm{Cone}(1_A)$ via $(a^{\mathrm{\prime }},a)\mapsto (i(a^{\mathrm{\prime }}),a)$; the homotopy $h(a^{\mathrm{\prime }},a)=(0,a^{\mathrm{\prime }})$ contracts this kernel, including when $A=0$. The long exact sequence for kernel, cone and quotient makes $e$ a quasi-isomorphism.

The cone triangle is distinguished and $Q(e)$ is invertible. Transporting it gives the short-exact-sequence triangle with connecting map $Q(p)Q(e{)}^{-1}$, where $p(b,a)=a$ maps to $A[1]$. A map of short exact sequences induces $(b,a)\mapsto (vb,ua)$ on cones, commuting with $e$ and $p$; this proves naturality with the stated signs.

Apply this construction to $0\to {\tau }^{\le n}X\to X\to R\to 0$. The quotient has $R^n=X^n\mathrm{/}\ker d^n\cong \mathrm{im}{d}^n$, $R^i=X^i$ for $i>n$ and zero below. The natural $R\to {\tau }^{\ge n+1}X$ is zero at $n$, the quotient at $n+1$, and identity above. Its kernel is the two-term identity complex on $\mathrm{im}{d}^n$, so it is a quasi-isomorphism. This yields the first triangle.

Apply the first triangle to ${\tau }^{\le n+1}X$ at cut $n$; its lower tail has just $H^{n+1}(X)$ in degree $n+1$. Apply it to ${\tau }^{\ge n}X$ at cut $n$; its upper head is just $H^n(X)$ in degree $n$. The cohomology formulas and natural comparison maps identify the remaining truncations, giving the second and third triangles.