Proof that a function composed by more than one trigonometric identity is even, odd, or neither.

Question

For example: how would you answer if you were asked if a function $f\left(x\right)=$sin$\left(2x\right)-$cos$\left(2x\right)$ is even, odd or neither? I'm mostly interest in a thorough algebraic proof, if possible.

Answer 1

A function $f$ is even iff $f(-x)=f(x)$ for all $x$ in the domain of $f$ and odd if $f(-x)=-f(x)$ for all $x$ in the domain of $f$. Letting $f(x)=\sin(2x)-\cos(2x)$, we see that $$ f(-x)=\sin(-2x)-\cos(-2x)=-\sin(2x)-\cos(2x), $$ since $\sin$ is odd and $\cos$ is even. This proves that $f$ is not even, since otherwise it would mean that for each $x$ $$ -\sin(2x)-\cos(2x)=f(-x)=f(x)=\sin(2x)-\cos(2x)\implies \sin(2x)=0, $$ which is not true. Similarly, $f$ is not odd since otherwise it would mean that for every $x$ $$ -\sin(2x)+\cos(2x)=-f(x)=f(-x)=-\sin(2x)-\cos(2x)\implies \cos(2x)=0, $$ which is also not true.

Therefore, $f$ is neither even nor odd.