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

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.

• If a statement is not true, a counterexample suffices. For instance, substitute $\frac{\pi}{6}$ and $-\frac{\pi}{6}$ to show the function is neither even nor odd. – N. F. Taussig Nov 8 '18 at 22:13

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.