Show that between any two roots of $e^x\cos(x)=1$ there exists atleast one root of $e^x\sin(x)-1$

Question

Show that between any two roots of $e^x\cos(x)=1$ there exists atleast one root of $e^x\sin(x)-1$

My Attempt:

Let $f(x)=e^x\cos(x)-1$ and $g(x)=e^{x}\sin(x)-1$

Let $a$ and $b$ be two consecutive roots of $f(x)$.

By Rolle's theorem, there exists a $c \in (a,b)$ such that $f'(c)=0$ $\implies e^c(\cos(c)-\sin(c))=0 \implies e^c\cos(c)-1=e^c\sin(c)-1$.

$c$ is not a root of $e^x\cos(x)-1$. So, I don't see how there can be a root of $e^{x}\sin(x)-1$ in $[a,b]$.

Am I going wrong somewhere?

Answer 1

Let $f(x)=\cos(x)-e^{-x}$.

If $a$ and $b$ are two consective roots of $f$ then there exists a $c\in(a,b)$ such that $-\sin(c)+e^{-c}=0$ from which the quoted claim follows.

Answer 2

To elaborate on this a bit more:

$$e^x \cos(x) - 1$$

Multiply everything through by $e^{-x}$:

$$e^{-x}(e^x\cos(x) - 1) = \cos(x) - e^{-x}$$

Notice that $h(x) = cos(x) - e^{-x}$ and $f(x) = e^x\cos(x) - 1$ have the same roots as $h(x) = e^{-x}f(x)$ and $e^{-x}$ is never $0$. So now take the deriviative of $h(x)$:

$$h'(x) = -\sin(x) + e^{-x}$$

Given any two consecutive roots of $h$ (and, well $f$), say $a$ and $b$, by Rolles Theorem there exists $c$ such that $h'(c) = 0$ so:

$$\begin{align} 0 &= -\sin(c) + e^{-c} \\ &= \sin(c) - e^{-c} \\ &= e^c (\sin(c) - e^{-c}) \\ &= e^c \sin(c) - 1 \end{align} $$

Answer 3

Given that $e^x \cos x=1.$

Rearranging we have $\cos x=e^{-x}.$

Take $f(x)=\cos x-e^{-x}.$

Assume that $a$ and $b$ are the roots of $f.$ Then $f(a)=f(b)=0.$

$f$ is continuous on $[a,b]$ and differentiable on $(a,b).$ Hence by Rolle's theorem $\exists ~c \in (a,b)~ \ni^1 f^1(c)=0.$

Now $f^1(x)=-\sin x+e^{-x}.$

i.e.,$-\sin c+e^{-c}=0 $ or $e^c \sin c-1=0.$