Prove between any two real roots of $e^x\sin x =1$ there is at least one real roots of $e^x\cos x=-1$
Tried to use Rolle's theorem, chose
$$f(x)=e^x\sin x-1$$
Let $a$ and $b$ are two roots for this function, where $a<b$, so
$$f(a)=0$$
and
$$f(b)=0$$
So there exist at least one points $\mu \in (a,b)$ s.t.
$$f'(\mu)=e^{\mu}\sin \mu + e^{\mu}\cos \mu=0$$
So
$$e^{\mu}\cos \mu=-e^{\mu}\sin \mu$$
However, $e^{x}\sin x=1$ was obtained at the end point of $[a,b]$, but $\mu \in (a,b)$.
Any ideas about how to go further? Thanks~
Best Answer
HINT: Consider two consecutive zeroes $a$ and $b$ of $f(x)=e^x\sin x - 1$. Then note that $f'(a)$ and $f'(b)$ have opposite signs. What does this tell you about $e^a\cos a$ and $e^b\cos b$?