In Undergraduate Analysis on p. 90, Lang assumes the existence of two functions $f$ (sine) and $g$ (cosine) satisfying the conditions $f(0)=0$, $g(0)=1$, $f'=g$ and $g'=-f$. He then goes on to show that there exists $x>0$ such that $\cos x=0$.
With your help, I would like to make some of the steps in his proof more explicit.
Suppose that no such number exists. Since $\cos$ is continuous, we
conclude that $\cos x$ cannot be negative for any value $x>0$ (by
intermediate value theorem). Hence $\sin$ is strictly increasing for
all $x>0$, and $\cos x$ is strictly decreasing for all $x>0$…
It's clear why $\sin$ is strictly increasing on the interval $(0,\infty)$. Why is $\cos$ strictly decreasing on the interval $(0,\infty)$? This would require $\sin x>0$ for all $x\in(0,\infty)$. But how can I show this?
… Let $a>0$. Then $0<\cos 2a=\cos^2a-\sin^2a<\cos^2a$. By induction,
we see that $\cos(2^n a)<(\cos a)^{2^n}$ for all positive integers
$n$. Hence $\cos(2^na)$ approaches $0$ as $n$ becomes large, because
$0<\cos a<1$. Since $\cos$ is strictly decreasing for $x>0$, it
follows that $\cos x$ approaches $0$ as $x$ becomes large, and hence
$\sin x$ approaches $1$. In particular, there exists a number $b>0$
such that $$\cos b<\frac{1}{4}\text{ and }\sin b>\frac{1}{2}.$$
If $\lim_{n\rightarrow\infty}\cos(2^na)=0$ for all $a>0$, how can I conclude that $\lim_{x\rightarrow\infty}\cos x=0$? Let $\epsilon>0$. Then I would have to show that there exists $s\in\mathbb{R}$ such that
$$(\forall x)(x\in(s,\infty)\implies|\cos x|<\epsilon).$$
It's not immediately clear how to find such an $s$. I would have to use the fact that $\lim_{n\rightarrow\infty}\cos(2^na)=0$ for all $a>0$, but I don't know how.
Best Answer
To the first question : if sine starts from zero and increases, it is positive.
So the derivative of cosine $g'=-f $ is negative.
Hence cosine decreases.
Added
To the second question: as cosine starts from $1$ at $x=0$ and decreases, it either reaches zero at some $x_0 > 0$, or not.
So there exists $a$ such that $1 = \cos 0 > \cos a > 0$ (that would be any $a>0$ in the latter case, or some $0<a<x_0$ in the former).
Then, the positive value less than $1$, when raised to a positive power $2^n$, results in decreasing values as $n$ grows (we get a fraction of the same fraction of the fraction, and so on).
However, it never gets negative, as a product of two positive values $(\cos a)^{2^n}\,\cdot\,(\cos a)^2$ makes a positive result.
As a result we get a strictly decreasing positive sequence of cosine's values at $x_n=2^na$.
But we know the function is strictly decreasing, so its values between $x_n<x_{n+1}$ are between $\cos(x_n)$ and $\cos(x_{n+1})$, hence positive, too, and bounded from above by decreasing $\cos(x_n) = (\cos a)^{2^n}$ as $n$ grows.
This 'squeezes' the cosine towards zero.
I wonder. however, how did author get the identity $$g(2a)=(g(a))^2−(f(a))^2$$ just form the presented assumptions...