calculusreal-analysis
Let $f(x) = \cos(x)$, $g(x)=x$, both functions are continuous. $f(0)=1$, $f(\pi/2)=0$, so, by the Intermediate Value Theorem, for any $z\in [0,1]$, there exists $c\in [0,\pi/2]$ such that $f(x)=z$.
This should be simple to prove, but for some reason I have a problem with IVT, don't know why. Would appreciate some help.
I think that the fact that $g(x)=x$ is onto $[0,1]\subset [0, \pi/2]$ should be used here somehow.
The IVT is not the only ingredient here. The way the theorem works is by setting up this square:
where the line in the middle is $y = x$. A function from $[0, 1]$ to $[0, 1]$ that intersects this line will have a fixed point at the point of intersection.
The IVT kicks in when we have a function whose graph enters the top triangle and the bottom triangle at various points, e.g.
The fact is, by the IVT, the function has to cut the line somewhere, i.e. it must have a fixed point.
But, this makes an assumption! The function may only exist in one triangle or the other, but not in both. That is, why can we not have $f(x) > x$ for all $x$ or $f(x) < x$ for all $x$?
The picture above illustrates it. Due to the function having a full domain of $[0, 1]$, there's a squeeze happening. The green function is about as close as we can have to a function satisfying $f(x) > x$. Similarly, the red function attempts to have $f(x) < x$. But, both are pinched towards the diagonal line. The green function must have $f(1) = 1$, and the red function must have $f(0) = 0$.
This illustrates the necessity of defining all the way to $0$ and $1$. Removing either of these points means that the functions are not squeezed to a fixed point (we'd only ensure that $f(x)$ and $x$ become arbitrarily close).
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...
Best Answer
Consider $h(x)=\cos(x)-x$. Then, $h(0)=1$ and $h\left( \displaystyle\frac{\pi}{2}\right)=-\displaystyle\frac{\pi}{2}$. Hence, $h(0)>h\left( \displaystyle\frac{\pi}{2}\right)$ and, clearly, $h(x)$ is continuous. For the IVT, there exist a $z\in\left( 0,\displaystyle\frac{\pi}{2}\right)$ such that $h(z)=0$.
$h(z)=0$ if and only if $\cos(z)-z=0$ if and only if $\cos(z)=z$