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Trigonometry Range of sec^4(x) + csc^4(x)

abender

New member
Jan 8, 2013
4
Hi, fellas. In response to someone's request on another form to find the range of [tex] \sec^4(x)+\csc^4(x)[/tex], I offered the following solution and explanation:

Please help me to find the range of [TEX]sec^{4}(x)+cosec^{4}(x)[/TEX].

Thanks in advance.
[TEX]\sec^4(x) = (\sec^2(x))^2 = (1+\tan^2(x))^2 = \tan^4(x) + 2\tan^2 + 1[/TEX]

[TEX]\csc^4(x) = (\csc^2(x))^2 = (1+\cot^2(x))^2 = \cot^4(x) + 2\cot^2 + 1 [/TEX]


[TEX]\sec^4(x) + \csc^4(x) = \tan^4(x) + 2\tan^2 + 1 + \cot^4(x) + 2\cot^2 + 1 [/TEX][tex] = \tan^4(x) + \cot^4(x) + 2(\tan^2(x)+\cot^2(x)) + 2[/tex]

The range of [TEX]\tan(x)[/TEX] is [TEX](-\infty, \infty)[/TEX]. Likewise, the range of [TEX]\cot(x)[/TEX] is [TEX](-\infty, \infty)[/TEX].

Since [TEX]\tan^4(x)[/TEX] and [TEX]\cot^4(x)[/TEX] are positive even powers, both have range [TEX][0,\infty)[/TEX].

BELOW is where others may disagree with me:

I contend that the range of [TEX]\tan^4(x)+\cot^4(x)[/TEX] is [TEX](0,\infty)[/TEX] as opposed to [TEX][0,\infty)[/TEX], which the sum of the parts may intuitively suggest.
The two ranges differ insofar [TEX](0,\infty)[/TEX] does not contain [TEX]0[/TEX], whereas [TEX][0,\infty)[/TEX] does contain [TEX]0[/TEX].
I believe that the range of [TEX]\tan^4(x)+\cot^4(x)[/TEX] should NOT include 0, i.e., it should be [TEX](0,\infty)[/TEX].

Proof is achieved if we show [TEX]\tan^4(x)+\cot^4(x)>0[/TEX] on the entire domain (reals that are not multiples of [TEX]\pi[/TEX]).

Both terms in [TEX]\tan^4(x)+\cot^4(x)[/TEX] are non-negative in the reals, so clearly the sum itself is non-negative.

[TEX]\tan^4(x)+\cot^4(x)[/TEX] cannot equal 0 because then either [TEX]\tan^4(x)=-\cot^4(x)[/TEX] (which per the line above is not possible) OR [TEX]\tan^4(x)=\cot^4(x)=0[/TEX], which also can't happen because [TEX]\cot^4(x)=\tfrac{1}{\tan^4(x)}[/TEX] and 0 cannot be a denominator.

As such, [TEX]\tan^4(x)+\cot^4(x)>0[/TEX].

And with this new information, [TEX]\sec^4(x) + \csc^4(x) = \underbrace{\left[\tan^4(x) + \cot^4(x)\right]}_{\text{always positive!}} + 2\underbrace{(\tan^2(x)+\cot^2(x))}_{\text{can show positive similarly}} + 2 > 2[/TEX].

It at long last follows that the range of [TEX]\sec^4(x) + \csc^4(x)[/TEX] is [TEX]\left(2,\infty\right)[/TEX].

If someone has an argument for a left bracket instead of my left open parenthesis, I'd love to hear it!

-Andy
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I checked first for the minima at W|A, and it suggests, that given:

$f(x)=\sec^4(x)+\csc^4(x)$

then the global minimum is $f(x)=8$, at:

$x=2(k\pi\pm\tan^{-1}(1\pm\sqrt{2}))$ where $k\in\mathbb{Z}$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Hi abender! :)

You are right that the range will not include zero.
The actual range is a subset of $[0,\infty)$, but also a subset of $(0,\infty)$.

Btw, the range is also not $(2,\infty)$.


What you have, is:

$\sec^4 x + \csc^4 x = \dfrac{1}{\cos^4 x} +\dfrac{1}{\sin^4 x}$

This expression is always positive, and it tends to infinity at every multiple of $\dfrac \pi 2$.
Since cosine and sine are symmetric, it will take its lowest value at $\dfrac \pi 4$, where it takes the value 8.
So the range is $[8, \infty)$.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
$x=2(k\pi\pm\tan^{-1}(1\pm\sqrt{2}))$ where $k\in\mathbb{Z}$
Wow!! That is a pretty complex way of W|A to say $x=\dfrac \pi 4 + k \dfrac \pi 2$. (Giggle)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Wow!! That is a pretty complex way of W|M to say $x=\dfrac \pi 4 + k \dfrac \pi 2$. (Giggle)
It sure is!...and I didn't bother to try to simplify either...(Tmi)

I simply noted that the range was different than the OP suggested and was going to let them go "back to the drawing board." (Smile)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
It sure is!...and I didn't bother to try to simplify either...

I simply noted that the range was different than the OP suggested and was going to let them go "back to the drawing board."
Yes, I'm surprised that W|A did not give a simpler solution.
Either way, it's not easy to simplify.
I didn't know that $\arctan(\sqrt 2 - 1) = \dfrac \pi 8$.
And I wouldn't know how to figure it out in this direction (the other way around I'd use the double angle formulas).
So I learned something new today.
 

abender

New member
Jan 8, 2013
4
Yes, I'm surprised that W|A did not give a simpler solution.
Either way, it's not easy to simplify.
I didn't know that $\arctan(\sqrt 2 - 1) = \dfrac \pi 8$.
And I wouldn't know how to figure it out in this direction (the other way around I'd use the double angle formulas).
So I learned something new today.
You aren't kiddin. Wow.
 

abender

New member
Jan 8, 2013
4
I feel stupid. I responded with a very very very very very rough lower bound. Who does this? "... plus 2 times something positive..." Of course I then end up with a superset of the range. Go Ravens!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Suppose we draw a right triangle, and with respect to one of the two acute angles (we'll call this angle $\theta$), we state:

$\tan(\theta)=\sqrt{2}-1$

and so:

$\displaystyle \sin(\theta)=\frac{\sqrt{2}-1}{\sqrt{2\sqrt{2}(\sqrt{2}-1)}}$

$\displaystyle \cos(\theta)=\frac{1}{\sqrt{2\sqrt{2}(\sqrt{2}-1)}}$

Now we may then state:

$\displaystyle \sin(2\theta)=2\sin(\theta)\cos(\theta)=\frac{2( \sqrt{2}-1)}{2\sqrt{2}(\sqrt{2}-1)}=\frac{1}{\sqrt{2}}$

Hence:

$\displaystyle 2\theta=\frac{\pi}{4}$

$\displaystyle \theta=\frac{\pi}{8}$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
For fun I tried to find the common values for the trigonometric functions.
Wiki did not contain these specific values.
Now it does.
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,685
Find the range of [tex] \sec^4(x)+\csc^4(x)[/tex]......

$\displaystyle \sec^4 x+\csc^4 x=\frac{1}{\cos^4 x}+\frac{1}{\sin^4 x}$

By AM-GM inequality, we have

$\displaystyle \frac{\frac{1}{\cos^4 x}+\frac{1}{\sin^4 x}}{2} \ge \sqrt{\frac{1}{\cos^4 x}.\frac{1}{\sin^4 x}}$

$\displaystyle \frac{\frac{1}{\cos^4 x}+\frac{1}{\sin^4 x}}{2} \ge \frac{1}{\cos^2 x.\sin^2 x}$

$\displaystyle \frac{1}{\cos^4 x}+\frac{1}{\sin^4 x} \ge \frac{2}{\cos^2 x.\sin^2 x}$


$\displaystyle \frac{1}{\cos^4 x}+\frac{1}{\sin^4 x} \ge \frac{2}{(\cos x.\sin x)^2}$

$\displaystyle \frac{1}{\cos^4 x}+\frac{1}{\sin^4 x} \ge \frac{2.2.2}{(2\cos x.\sin x)^2}$

$\displaystyle \frac{1}{\cos^4 x}+\frac{1}{\sin^4 x} \ge \frac{8}{\sin^2 2x}$

$\displaystyle \frac{1}{\cos^4 x}+\frac{1}{\sin^4 x} \ge \frac{8}{\frac{1-\cos 2x}{2}}$

$\displaystyle \frac{1}{\cos^4 x}+\frac{1}{\sin^4 x} \ge \frac{16}{1-\cos 2x}$

Since $\displaystyle 0 \le 1-\cos 2x \le 2 $, we can conclude that $\displaystyle\frac{1}{\cos^4 x}+\frac{1}{\sin^4 x} \ge 8 $, that is, the range of $\displaystyle \sec^4 x+\csc^4 x $ is [8, ∞).