MHB What is the connection between roots of f and g using Rolle's Theorem?

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Using Rolle's Theorem, it can be shown that between any two roots of the function f(x) = 1 - e^(x)*sin(x), there exists at least one root of the function g(x) = 1 + e^(x)*cos(x). When two successive roots of f are considered, there exists a point c in the interval (a, b) where the derivative f'(c) equals zero, indicating a turning point. At this turning point, the equations derived show that f(c) equals g(c), confirming that they intersect. Consequently, g must have at least one root between the two successive roots of f. This establishes a direct connection between the roots of f and g as dictated by Rolle's Theorem.
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[h=1][/h]I'm doing a question and I am getting stuck and need help

Question:
Consider the continuous functions f(x) = 1 - e^(x)*sin(x) and g(x) = 1 + e^(x)*cos(x). Using Rolle's Theorem, prove that between any two roots of f there exists at least one root of g.

Hint
Remember that, a root of f is a point x in the domain of f such that f(x) = 0.

Can someone provide a natural language proof of this?
 
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Let a and b be two successive roots of f. Observe that by Rolle's theorem, we must have some c in (a,b) such that f'(c)=0. This means we must have:

$\displaystyle -e^c\cos(c)-e^c\sin(c)=0$

$\displaystyle -e^c\sin(c)=e^c\cos(c)$

$\displaystyle 1-e^c\sin(c)=1+e^c\cos(c)$

$\displaystyle f(c)=g(c)$

So, we find that f and g meet at the turning points of f. This means g must have at least one root between two successive turning points of f.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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