Let $f : \mathbb R \to \mathbb R$ be a continuous $2\pi$-periodic function, i.e. for every $t \in \mathbb R$, we have $f(t) = f(t+2\pi)$. Prove that there exists $t_0\in \mathbb R$ such that $$f(t_0) = f\big(t_0 +\frac{\pi}{2}\big)$$
My attempt: I take $$g(t)= f(t)-f\big(t+\frac{\pi}{2}\big)$$ Then how to find out zero of continuous function $g$ using intermediate value theoem?
Or is there any other way to solve this problem?
$g$ is continuous. If it is never $0$ the it is always positive or always negative. Suppose $g(t) <0$ for all $t$. Then $f(t+2\pi) >f(t+\frac 3 2 \pi)>f(t+\pi)>f(t) $ which is a contradiction. Similarly, $g(t) >0$ for all $t$ leads to a contradiction.