Is the tensor product of the cone of a morphism $f$ with the cone of the morphism $1-f$ isomorphic to zero in any tensor-triangulated category? I think, if my computation ain't wrong, that if I look at a morphism $f:R\to R$ as a morphism of complexes, in the triangulated (derived, homotopy) category of complexes of $R$-modules, one has that the complex $C(f)\otimes C(\mathrm{id}_R-f)$ is the complex $\cdots\to 0\to R\overset{\mathrm{id}}\to R\to 0\to \cdots$ (in degrees $-1$ and $0$), which is isomorphic (quasi-isomorphic, homotopically equivalent) to zero.
My question is: is this true in a more general context? More precisely, let $(\mathbb{K},\otimes,1)$ be a tensor-triangulated category, $f:1\to 1$ a morphism in $\mathbb{K}$, and consider the completions to distinguished triangles of the morphisms $f$ and $\mathrm{id}-f$
$$1\overset{f}\longrightarrow 1\longrightarrow C\longrightarrow T1$$
and $$1\overset{\mathrm{id}-f}\longrightarrow 1\longrightarrow C'\longrightarrow T1$$
Does it follows that $C\otimes C'\cong 0$?
 A: Yes, it's true generally.
$\require{AMScd}$
Tensoring the triangle
$$1\xrightarrow{1-f}1\rightarrow C'\rightarrow T1$$
with $C$, to get
$$C\xrightarrow{1-f}C\rightarrow C\otimes C'\rightarrow T1,$$
shows that proving that $C\otimes C'=0$ is equivalent to proving that
$C\xrightarrow{1-f}C$ is an isomorphism.
Tensoring the triangle
$$1\xrightarrow{f}1\xrightarrow{\alpha} C\xrightarrow{\beta} T1$$
with $1\xrightarrow{1-f}1$ gives a morphism of triangles
\begin{CD}
  1@>f>>1@>\alpha>>C@>\beta>>T1\\
  @VV1-fV@VV1-fV@VV1-fV@VV1-fV\\
  1@>f>>1@>\alpha>>C@>\beta>>T1.\\
\end{CD}
But also the diagram
\begin{CD}
  1@>f>>1@>\alpha>>C@>\beta>>T1\\
  @VVfV@VVfV@VV0V@VVfV\\
  1@>f>>1@>\alpha>>C@>\beta>>T1.\\
\end{CD}
is a morphism of triangles, since $f\beta$ and $\alpha f$ are compositions of consecutive arrows in triangles, and are therefore zero.
Adding the vertical arrows,
\begin{CD}
  1@>f>>1@>\alpha>>C@>\beta>>T1\\
  @VV1V@VV1V@VV1-fV@VV1V\\
  1@>f>>1@>\alpha>>C@>\beta>>T1.\\
\end{CD}
is a morphism of triangles, and since the first two vertical arrows are
isomorphisms, by the axioms of a triangulated category the third
vertical arrow $C\xrightarrow{1-f}C$ is also an isomorphism, as
required.
