Here is a way we can develop the proof...
hence we need not be concerned with
and the maximum value of is <1.
We only need consider both functions >0 which is in quadrant 1, where
In quadrant 1,
Hence we are examining
so we want to know if
since in quadrant 1,
?
Hence we find the maximum value of by differentiating...
which happens at radians.
The maximum value of
This tread is getting old, but Mr. Ryan still seemed confused.
This may be a Lemma to BobP's theorem.
I hope you can agree that: If two functions, f(x) and g(x) are continuous, and if they are not equal for any x, then one of them is greater than the other for all x.
In this case let's see if we can solve:
.
,
so .
Then . (O.K., you could add integer multiple of to this, but that won't help.)
Use the identity: .
This gives us
The maximum value of the cosine is 1, and , so has no solution.
Since (It's definitely negative!)
and ,
it must be that , for all x.
Maybe simplest and without reference to graphs is:
This is the difference of cosines.
One could go further with trigonometry here.
An alternative is to examine the contents of the square brackets on the previous line.
We can examine the maximum and minimum values of these expressions.
Referring to the unit-circle, this occurs when
(1) maximum value within the first square brackets
(2) minimum value within the first square brackets
(3) maximum value within the second square brackets
(4) minimum value within the second square brackets
Therefore, it follows that