I like Serena said:
The symbol $a$ is introduced here without definition.
I'm guessing it is defined as a solution of $g'(a)=0$.
However, that means that:
$$u(x) = a$$
Writing it as $u=a$ is a short hand notation.
Let's define the original trig function as:
$$G(x) = g(u(x)) \tag 1$$
Then $G(x)$ takes it minimum value when $G'(x)=0$.
That means that:
$$G'(x) = g'(u(x))\cdot u'(x) = 0$$
This happens if $g'(u(x))=0$ or $u'(x) = 0$.
With $a$ as the solution of $g'(a)=0$, we get that $G$ takes a minimum value at $x=u^{-1}(a)$.
The corresponding minimum is:
$$G\Big(u^{-1}(a)\Big) = g\Big(u\big(u^{-1}(a)\big)\Big) = g(a)$$
Okay, so a few things here.
This was a
very, very, very, very, very good explanation. No doubts on that. Kudos to I Like Serena!
So we set $g'(u(x)) = 0$ and supposed it happened at $a$
So,
$g'(u(x)) = 0$ at $u(x) = a$ And like you said
$G(x) = g(u(x))$ which is the original trig equation.
$u(x) = a$ would mean
$x = u^{-1}(a)$
$G(u^{-1}(a)) = g(u(u^{-1}(a)))$
By definition,
$u(u^{-1}(a)) = a$ therefore we get,
$G(u^{-1}(a)) = g(u(u^{-1}(a))) = g(a)$
Wow, that is amazing how this works. Because essentially
We (you) have proved that the minimum is indeed at $u(x) = a$
In problems though, it gets
very confusing.
When people say
$g(u)$ rather than $g(u(x))$
$g(u)$ means a function of $u$ as $u$ being an
independent variable.
$g(u(x))$ means a function of a function $u(x)$ as a
composite function.
Because by definition, a single letter would mean an independent variable.
-- This was the
source of my confusion, which seems to be better, now. Wow, @I Like Serena, that was a million dollar explanation, I wish I could give a million thanks through MHB, excellent. Thank you very much!
(The rest of the problem is quite simple now)!