TITLE: If $A,B,C$ are nonsingular, so is $A\sin(t)+B\cos(t)+C$ for some real $t$
QUESTION [5 upvotes]: While trying to answer another question on this site, I found that I needed the following assertion:

If $A,B,C$ are nonsingular complex matrices of the same sizes, then $A\sin(t)+B\cos(t)+C$ is nonsingular for some real number $t$.

Clearly, we may assume that $A$ is the identity matrix and $B+C$ is in Jordan normal form. Then $A\sin(t)+B\cos(t)+C=tI+(B+C)+O(t^2)$. The assertion is clear if $B+C=0$. By using Schur complement, we can then prove the assertion for the case where $B+C$ is a single nilpotent Jordan block, and then for the case where $B+C$ is a nilpotent matrix, and finally for the general case.
Such a proof, however, looks too tedious. Is there any concise proof? Since I intended to use this assertion as part of an elementary proof for another problem statement, the proof for this assertion is preferably elementary as well.

REPLY [3 votes]: It turns out that the problem statement is false. Consider
$$
A=I_2,\ B=\pmatrix{0&-1\\ 1&0},\ C=\pmatrix{0&1\\ 1&0}.
$$
All these three matrices are nonsingular, but
$$
A\sin(t)+B\cos(t)+C = \pmatrix{\sin t& 1-\cos t\\ 1+\cos t&\sin t}
$$
always has zero determinant. I did try to verify the statement in the $2\times2$ case before I posted the question. Unfortunately I made a mistake and thought that the answer for that case is affirmative.